3.5.26 \(\int \frac {x^{7/2} (A+B x)}{(a+c x^2)^3} \, dx\)

Optimal. Leaf size=320 \[ \frac {\left (21 \sqrt {a} B-5 A \sqrt {c}\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{64 \sqrt {2} a^{3/4} c^{11/4}}-\frac {\left (21 \sqrt {a} B-5 A \sqrt {c}\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{64 \sqrt {2} a^{3/4} c^{11/4}}-\frac {\left (21 \sqrt {a} B+5 A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{3/4} c^{11/4}}+\frac {\left (21 \sqrt {a} B+5 A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{3/4} c^{11/4}}-\frac {\sqrt {x} (5 A+7 B x)}{16 c^2 \left (a+c x^2\right )}-\frac {x^{5/2} (A+B x)}{4 c \left (a+c x^2\right )^2} \]

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Rubi [A]  time = 0.31, antiderivative size = 320, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {819, 827, 1168, 1162, 617, 204, 1165, 628} \begin {gather*} \frac {\left (21 \sqrt {a} B-5 A \sqrt {c}\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{64 \sqrt {2} a^{3/4} c^{11/4}}-\frac {\left (21 \sqrt {a} B-5 A \sqrt {c}\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{64 \sqrt {2} a^{3/4} c^{11/4}}-\frac {\left (21 \sqrt {a} B+5 A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{3/4} c^{11/4}}+\frac {\left (21 \sqrt {a} B+5 A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{3/4} c^{11/4}}-\frac {\sqrt {x} (5 A+7 B x)}{16 c^2 \left (a+c x^2\right )}-\frac {x^{5/2} (A+B x)}{4 c \left (a+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(7/2)*(A + B*x))/(a + c*x^2)^3,x]

[Out]

-(x^(5/2)*(A + B*x))/(4*c*(a + c*x^2)^2) - (Sqrt[x]*(5*A + 7*B*x))/(16*c^2*(a + c*x^2)) - ((21*Sqrt[a]*B + 5*A
*Sqrt[c])*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(3/4)*c^(11/4)) + ((21*Sqrt[a]*B + 5*A*
Sqrt[c])*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(3/4)*c^(11/4)) + ((21*Sqrt[a]*B - 5*A*S
qrt[c])*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*a^(3/4)*c^(11/4)) - ((21*Sqrt[
a]*B - 5*A*Sqrt[c])*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*a^(3/4)*c^(11/4))

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 827

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2, Subst[Int[(e*f
- d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

Rubi steps

\begin {align*} \int \frac {x^{7/2} (A+B x)}{\left (a+c x^2\right )^3} \, dx &=-\frac {x^{5/2} (A+B x)}{4 c \left (a+c x^2\right )^2}+\frac {\int \frac {x^{3/2} \left (\frac {5 a A}{2}+\frac {7 a B x}{2}\right )}{\left (a+c x^2\right )^2} \, dx}{4 a c}\\ &=-\frac {x^{5/2} (A+B x)}{4 c \left (a+c x^2\right )^2}-\frac {\sqrt {x} (5 A+7 B x)}{16 c^2 \left (a+c x^2\right )}+\frac {\int \frac {\frac {5 a^2 A}{4}+\frac {21}{4} a^2 B x}{\sqrt {x} \left (a+c x^2\right )} \, dx}{8 a^2 c^2}\\ &=-\frac {x^{5/2} (A+B x)}{4 c \left (a+c x^2\right )^2}-\frac {\sqrt {x} (5 A+7 B x)}{16 c^2 \left (a+c x^2\right )}+\frac {\operatorname {Subst}\left (\int \frac {\frac {5 a^2 A}{4}+\frac {21}{4} a^2 B x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{4 a^2 c^2}\\ &=-\frac {x^{5/2} (A+B x)}{4 c \left (a+c x^2\right )^2}-\frac {\sqrt {x} (5 A+7 B x)}{16 c^2 \left (a+c x^2\right )}-\frac {\left (21 B-\frac {5 A \sqrt {c}}{\sqrt {a}}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a} \sqrt {c}-c x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{32 c^3}+\frac {\left (21 B+\frac {5 A \sqrt {c}}{\sqrt {a}}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a} \sqrt {c}+c x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{32 c^3}\\ &=-\frac {x^{5/2} (A+B x)}{4 c \left (a+c x^2\right )^2}-\frac {\sqrt {x} (5 A+7 B x)}{16 c^2 \left (a+c x^2\right )}+\frac {\left (21 B+\frac {5 A \sqrt {c}}{\sqrt {a}}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt {x}\right )}{64 c^3}+\frac {\left (21 B+\frac {5 A \sqrt {c}}{\sqrt {a}}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt {x}\right )}{64 c^3}+\frac {\left (21 \sqrt {a} B-5 A \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{c}}+2 x}{-\frac {\sqrt {a}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt {x}\right )}{64 \sqrt {2} a^{3/4} c^{11/4}}+\frac {\left (21 \sqrt {a} B-5 A \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{c}}-2 x}{-\frac {\sqrt {a}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt {x}\right )}{64 \sqrt {2} a^{3/4} c^{11/4}}\\ &=-\frac {x^{5/2} (A+B x)}{4 c \left (a+c x^2\right )^2}-\frac {\sqrt {x} (5 A+7 B x)}{16 c^2 \left (a+c x^2\right )}+\frac {\left (21 \sqrt {a} B-5 A \sqrt {c}\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} a^{3/4} c^{11/4}}-\frac {\left (21 \sqrt {a} B-5 A \sqrt {c}\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} a^{3/4} c^{11/4}}+\frac {\left (21 \sqrt {a} B+5 A \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{3/4} c^{11/4}}-\frac {\left (21 \sqrt {a} B+5 A \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{3/4} c^{11/4}}\\ &=-\frac {x^{5/2} (A+B x)}{4 c \left (a+c x^2\right )^2}-\frac {\sqrt {x} (5 A+7 B x)}{16 c^2 \left (a+c x^2\right )}-\frac {\left (21 \sqrt {a} B+5 A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{3/4} c^{11/4}}+\frac {\left (21 \sqrt {a} B+5 A \sqrt {c}\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{3/4} c^{11/4}}+\frac {\left (21 \sqrt {a} B-5 A \sqrt {c}\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} a^{3/4} c^{11/4}}-\frac {\left (21 \sqrt {a} B-5 A \sqrt {c}\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} a^{3/4} c^{11/4}}\\ \end {align*}

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Mathematica [A]  time = 0.48, size = 385, normalized size = 1.20 \begin {gather*} \frac {-\frac {5 \sqrt {2} a^{5/4} A \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{c^{9/4}}+\frac {5 \sqrt {2} a^{5/4} A \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{c^{9/4}}-\frac {10 \sqrt {2} a^{5/4} A \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{c^{9/4}}+\frac {10 \sqrt {2} a^{5/4} A \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{c^{9/4}}-\frac {40 a A \sqrt {x}}{c^2}-\frac {8 A x^{9/2}}{a+c x^2}+\frac {32 a A x^{9/2}}{\left (a+c x^2\right )^2}+\frac {84 (-a)^{7/4} B \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-a}}\right )}{c^{11/4}}+\frac {84 (-a)^{3/4} a B \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-a}}\right )}{c^{11/4}}-\frac {56 a B x^{3/2}}{c^2}-\frac {24 B x^{11/2}}{a+c x^2}+\frac {32 a B x^{11/2}}{\left (a+c x^2\right )^2}+\frac {8 A x^{5/2}}{c}+\frac {24 B x^{7/2}}{c}}{128 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(7/2)*(A + B*x))/(a + c*x^2)^3,x]

[Out]

((-40*a*A*Sqrt[x])/c^2 - (56*a*B*x^(3/2))/c^2 + (8*A*x^(5/2))/c + (24*B*x^(7/2))/c + (32*a*A*x^(9/2))/(a + c*x
^2)^2 + (32*a*B*x^(11/2))/(a + c*x^2)^2 - (8*A*x^(9/2))/(a + c*x^2) - (24*B*x^(11/2))/(a + c*x^2) - (10*Sqrt[2
]*a^(5/4)*A*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/c^(9/4) + (10*Sqrt[2]*a^(5/4)*A*ArcTan[1 + (Sqrt[2]
*c^(1/4)*Sqrt[x])/a^(1/4)])/c^(9/4) + (84*(-a)^(7/4)*B*ArcTan[(c^(1/4)*Sqrt[x])/(-a)^(1/4)])/c^(11/4) + (84*(-
a)^(3/4)*a*B*ArcTanh[(c^(1/4)*Sqrt[x])/(-a)^(1/4)])/c^(11/4) - (5*Sqrt[2]*a^(5/4)*A*Log[Sqrt[a] - Sqrt[2]*a^(1
/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/c^(9/4) + (5*Sqrt[2]*a^(5/4)*A*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x]
+ Sqrt[c]*x])/c^(9/4))/(128*a^2)

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IntegrateAlgebraic [A]  time = 1.19, size = 191, normalized size = 0.60 \begin {gather*} -\frac {\left (21 \sqrt {a} B+5 A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}}\right )}{32 \sqrt {2} a^{3/4} c^{11/4}}-\frac {\left (21 \sqrt {a} B-5 A \sqrt {c}\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}}{\sqrt {a}+\sqrt {c} x}\right )}{32 \sqrt {2} a^{3/4} c^{11/4}}-\frac {\sqrt {x} \left (5 a A+7 a B x+9 A c x^2+11 B c x^3\right )}{16 c^2 \left (a+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^(7/2)*(A + B*x))/(a + c*x^2)^3,x]

[Out]

-1/16*(Sqrt[x]*(5*a*A + 7*a*B*x + 9*A*c*x^2 + 11*B*c*x^3))/(c^2*(a + c*x^2)^2) - ((21*Sqrt[a]*B + 5*A*Sqrt[c])
*ArcTan[(Sqrt[a] - Sqrt[c]*x)/(Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x])])/(32*Sqrt[2]*a^(3/4)*c^(11/4)) - ((21*Sqrt[a]
*B - 5*A*Sqrt[c])*ArcTanh[(Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[c]*x)])/(32*Sqrt[2]*a^(3/4)*c^(11/
4))

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fricas [B]  time = 0.46, size = 986, normalized size = 3.08 \begin {gather*} \frac {{\left (c^{4} x^{4} + 2 \, a c^{3} x^{2} + a^{2} c^{2}\right )} \sqrt {-\frac {a c^{5} \sqrt {-\frac {194481 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{3} c^{11}}} + 210 \, A B}{a c^{5}}} \log \left (-{\left (194481 \, B^{4} a^{2} - 625 \, A^{4} c^{2}\right )} \sqrt {x} + {\left (21 \, B a^{3} c^{8} \sqrt {-\frac {194481 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{3} c^{11}}} - 2205 \, A B^{2} a^{2} c^{3} + 125 \, A^{3} a c^{4}\right )} \sqrt {-\frac {a c^{5} \sqrt {-\frac {194481 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{3} c^{11}}} + 210 \, A B}{a c^{5}}}\right ) - {\left (c^{4} x^{4} + 2 \, a c^{3} x^{2} + a^{2} c^{2}\right )} \sqrt {-\frac {a c^{5} \sqrt {-\frac {194481 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{3} c^{11}}} + 210 \, A B}{a c^{5}}} \log \left (-{\left (194481 \, B^{4} a^{2} - 625 \, A^{4} c^{2}\right )} \sqrt {x} - {\left (21 \, B a^{3} c^{8} \sqrt {-\frac {194481 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{3} c^{11}}} - 2205 \, A B^{2} a^{2} c^{3} + 125 \, A^{3} a c^{4}\right )} \sqrt {-\frac {a c^{5} \sqrt {-\frac {194481 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{3} c^{11}}} + 210 \, A B}{a c^{5}}}\right ) - {\left (c^{4} x^{4} + 2 \, a c^{3} x^{2} + a^{2} c^{2}\right )} \sqrt {\frac {a c^{5} \sqrt {-\frac {194481 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{3} c^{11}}} - 210 \, A B}{a c^{5}}} \log \left (-{\left (194481 \, B^{4} a^{2} - 625 \, A^{4} c^{2}\right )} \sqrt {x} + {\left (21 \, B a^{3} c^{8} \sqrt {-\frac {194481 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{3} c^{11}}} + 2205 \, A B^{2} a^{2} c^{3} - 125 \, A^{3} a c^{4}\right )} \sqrt {\frac {a c^{5} \sqrt {-\frac {194481 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{3} c^{11}}} - 210 \, A B}{a c^{5}}}\right ) + {\left (c^{4} x^{4} + 2 \, a c^{3} x^{2} + a^{2} c^{2}\right )} \sqrt {\frac {a c^{5} \sqrt {-\frac {194481 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{3} c^{11}}} - 210 \, A B}{a c^{5}}} \log \left (-{\left (194481 \, B^{4} a^{2} - 625 \, A^{4} c^{2}\right )} \sqrt {x} - {\left (21 \, B a^{3} c^{8} \sqrt {-\frac {194481 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{3} c^{11}}} + 2205 \, A B^{2} a^{2} c^{3} - 125 \, A^{3} a c^{4}\right )} \sqrt {\frac {a c^{5} \sqrt {-\frac {194481 \, B^{4} a^{2} - 22050 \, A^{2} B^{2} a c + 625 \, A^{4} c^{2}}{a^{3} c^{11}}} - 210 \, A B}{a c^{5}}}\right ) - 4 \, {\left (11 \, B c x^{3} + 9 \, A c x^{2} + 7 \, B a x + 5 \, A a\right )} \sqrt {x}}{64 \, {\left (c^{4} x^{4} + 2 \, a c^{3} x^{2} + a^{2} c^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)/(c*x^2+a)^3,x, algorithm="fricas")

[Out]

1/64*((c^4*x^4 + 2*a*c^3*x^2 + a^2*c^2)*sqrt(-(a*c^5*sqrt(-(194481*B^4*a^2 - 22050*A^2*B^2*a*c + 625*A^4*c^2)/
(a^3*c^11)) + 210*A*B)/(a*c^5))*log(-(194481*B^4*a^2 - 625*A^4*c^2)*sqrt(x) + (21*B*a^3*c^8*sqrt(-(194481*B^4*
a^2 - 22050*A^2*B^2*a*c + 625*A^4*c^2)/(a^3*c^11)) - 2205*A*B^2*a^2*c^3 + 125*A^3*a*c^4)*sqrt(-(a*c^5*sqrt(-(1
94481*B^4*a^2 - 22050*A^2*B^2*a*c + 625*A^4*c^2)/(a^3*c^11)) + 210*A*B)/(a*c^5))) - (c^4*x^4 + 2*a*c^3*x^2 + a
^2*c^2)*sqrt(-(a*c^5*sqrt(-(194481*B^4*a^2 - 22050*A^2*B^2*a*c + 625*A^4*c^2)/(a^3*c^11)) + 210*A*B)/(a*c^5))*
log(-(194481*B^4*a^2 - 625*A^4*c^2)*sqrt(x) - (21*B*a^3*c^8*sqrt(-(194481*B^4*a^2 - 22050*A^2*B^2*a*c + 625*A^
4*c^2)/(a^3*c^11)) - 2205*A*B^2*a^2*c^3 + 125*A^3*a*c^4)*sqrt(-(a*c^5*sqrt(-(194481*B^4*a^2 - 22050*A^2*B^2*a*
c + 625*A^4*c^2)/(a^3*c^11)) + 210*A*B)/(a*c^5))) - (c^4*x^4 + 2*a*c^3*x^2 + a^2*c^2)*sqrt((a*c^5*sqrt(-(19448
1*B^4*a^2 - 22050*A^2*B^2*a*c + 625*A^4*c^2)/(a^3*c^11)) - 210*A*B)/(a*c^5))*log(-(194481*B^4*a^2 - 625*A^4*c^
2)*sqrt(x) + (21*B*a^3*c^8*sqrt(-(194481*B^4*a^2 - 22050*A^2*B^2*a*c + 625*A^4*c^2)/(a^3*c^11)) + 2205*A*B^2*a
^2*c^3 - 125*A^3*a*c^4)*sqrt((a*c^5*sqrt(-(194481*B^4*a^2 - 22050*A^2*B^2*a*c + 625*A^4*c^2)/(a^3*c^11)) - 210
*A*B)/(a*c^5))) + (c^4*x^4 + 2*a*c^3*x^2 + a^2*c^2)*sqrt((a*c^5*sqrt(-(194481*B^4*a^2 - 22050*A^2*B^2*a*c + 62
5*A^4*c^2)/(a^3*c^11)) - 210*A*B)/(a*c^5))*log(-(194481*B^4*a^2 - 625*A^4*c^2)*sqrt(x) - (21*B*a^3*c^8*sqrt(-(
194481*B^4*a^2 - 22050*A^2*B^2*a*c + 625*A^4*c^2)/(a^3*c^11)) + 2205*A*B^2*a^2*c^3 - 125*A^3*a*c^4)*sqrt((a*c^
5*sqrt(-(194481*B^4*a^2 - 22050*A^2*B^2*a*c + 625*A^4*c^2)/(a^3*c^11)) - 210*A*B)/(a*c^5))) - 4*(11*B*c*x^3 +
9*A*c*x^2 + 7*B*a*x + 5*A*a)*sqrt(x))/(c^4*x^4 + 2*a*c^3*x^2 + a^2*c^2)

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giac [A]  time = 0.22, size = 293, normalized size = 0.92 \begin {gather*} -\frac {11 \, B c x^{\frac {7}{2}} + 9 \, A c x^{\frac {5}{2}} + 7 \, B a x^{\frac {3}{2}} + 5 \, A a \sqrt {x}}{16 \, {\left (c x^{2} + a\right )}^{2} c^{2}} + \frac {\sqrt {2} {\left (5 \, \left (a c^{3}\right )^{\frac {1}{4}} A c^{2} + 21 \, \left (a c^{3}\right )^{\frac {3}{4}} B\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{64 \, a c^{5}} + \frac {\sqrt {2} {\left (5 \, \left (a c^{3}\right )^{\frac {1}{4}} A c^{2} + 21 \, \left (a c^{3}\right )^{\frac {3}{4}} B\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{64 \, a c^{5}} + \frac {\sqrt {2} {\left (5 \, \left (a c^{3}\right )^{\frac {1}{4}} A c^{2} - 21 \, \left (a c^{3}\right )^{\frac {3}{4}} B\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{c}}\right )}{128 \, a c^{5}} - \frac {\sqrt {2} {\left (5 \, \left (a c^{3}\right )^{\frac {1}{4}} A c^{2} - 21 \, \left (a c^{3}\right )^{\frac {3}{4}} B\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{c}}\right )}{128 \, a c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)/(c*x^2+a)^3,x, algorithm="giac")

[Out]

-1/16*(11*B*c*x^(7/2) + 9*A*c*x^(5/2) + 7*B*a*x^(3/2) + 5*A*a*sqrt(x))/((c*x^2 + a)^2*c^2) + 1/64*sqrt(2)*(5*(
a*c^3)^(1/4)*A*c^2 + 21*(a*c^3)^(3/4)*B)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/c)^(1/4) + 2*sqrt(x))/(a/c)^(1/4))/(a*
c^5) + 1/64*sqrt(2)*(5*(a*c^3)^(1/4)*A*c^2 + 21*(a*c^3)^(3/4)*B)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/c)^(1/4) - 2*
sqrt(x))/(a/c)^(1/4))/(a*c^5) + 1/128*sqrt(2)*(5*(a*c^3)^(1/4)*A*c^2 - 21*(a*c^3)^(3/4)*B)*log(sqrt(2)*sqrt(x)
*(a/c)^(1/4) + x + sqrt(a/c))/(a*c^5) - 1/128*sqrt(2)*(5*(a*c^3)^(1/4)*A*c^2 - 21*(a*c^3)^(3/4)*B)*log(-sqrt(2
)*sqrt(x)*(a/c)^(1/4) + x + sqrt(a/c))/(a*c^5)

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maple [A]  time = 0.06, size = 327, normalized size = 1.02 \begin {gather*} \frac {5 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{64 a \,c^{2}}+\frac {5 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{64 a \,c^{2}}+\frac {5 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \ln \left (\frac {x +\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{c}}}{x -\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{c}}}\right )}{128 a \,c^{2}}+\frac {21 \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{64 \left (\frac {a}{c}\right )^{\frac {1}{4}} c^{3}}+\frac {21 \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{64 \left (\frac {a}{c}\right )^{\frac {1}{4}} c^{3}}+\frac {21 \sqrt {2}\, B \ln \left (\frac {x -\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{c}}}{x +\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{c}}}\right )}{128 \left (\frac {a}{c}\right )^{\frac {1}{4}} c^{3}}+\frac {-\frac {11 B \,x^{\frac {7}{2}}}{16 c}-\frac {9 A \,x^{\frac {5}{2}}}{16 c}-\frac {7 B a \,x^{\frac {3}{2}}}{16 c^{2}}-\frac {5 A a \sqrt {x}}{16 c^{2}}}{\left (c \,x^{2}+a \right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(B*x+A)/(c*x^2+a)^3,x)

[Out]

2*(-11/32*B/c*x^(7/2)-9/32*A/c*x^(5/2)-7/32*B*a/c^2*x^(3/2)-5/32*a*A/c^2*x^(1/2))/(c*x^2+a)^2+5/128/c^2*A*(a/c
)^(1/4)/a*2^(1/2)*ln((x+(a/c)^(1/4)*2^(1/2)*x^(1/2)+(a/c)^(1/2))/(x-(a/c)^(1/4)*2^(1/2)*x^(1/2)+(a/c)^(1/2)))+
5/64/c^2*A*(a/c)^(1/4)/a*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)+1)+5/64/c^2*A*(a/c)^(1/4)/a*2^(1/2)*arctan
(2^(1/2)/(a/c)^(1/4)*x^(1/2)-1)+21/128/c^3*B/(a/c)^(1/4)*2^(1/2)*ln((x-(a/c)^(1/4)*2^(1/2)*x^(1/2)+(a/c)^(1/2)
)/(x+(a/c)^(1/4)*2^(1/2)*x^(1/2)+(a/c)^(1/2)))+21/64/c^3*B/(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1
/2)+1)+21/64/c^3*B/(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)-1)

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maxima [A]  time = 1.33, size = 292, normalized size = 0.91 \begin {gather*} -\frac {11 \, B c x^{\frac {7}{2}} + 9 \, A c x^{\frac {5}{2}} + 7 \, B a x^{\frac {3}{2}} + 5 \, A a \sqrt {x}}{16 \, {\left (c^{4} x^{4} + 2 \, a c^{3} x^{2} + a^{2} c^{2}\right )}} + \frac {\frac {2 \, \sqrt {2} {\left (21 \, B \sqrt {a} + 5 \, A \sqrt {c}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} {\left (21 \, B \sqrt {a} + 5 \, A \sqrt {c}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} {\left (21 \, B \sqrt {a} - 5 \, A \sqrt {c}\right )} \log \left (\sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} {\left (21 \, B \sqrt {a} - 5 \, A \sqrt {c}\right )} \log \left (-\sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {3}{4}}}}{128 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)/(c*x^2+a)^3,x, algorithm="maxima")

[Out]

-1/16*(11*B*c*x^(7/2) + 9*A*c*x^(5/2) + 7*B*a*x^(3/2) + 5*A*a*sqrt(x))/(c^4*x^4 + 2*a*c^3*x^2 + a^2*c^2) + 1/1
28*(2*sqrt(2)*(21*B*sqrt(a) + 5*A*sqrt(c))*arctan(1/2*sqrt(2)*(sqrt(2)*a^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x))/sq
rt(sqrt(a)*sqrt(c)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(c))*sqrt(c)) + 2*sqrt(2)*(21*B*sqrt(a) + 5*A*sqrt(c))*arctan(-
1/2*sqrt(2)*(sqrt(2)*a^(1/4)*c^(1/4) - 2*sqrt(c)*sqrt(x))/sqrt(sqrt(a)*sqrt(c)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(c)
)*sqrt(c)) - sqrt(2)*(21*B*sqrt(a) - 5*A*sqrt(c))*log(sqrt(2)*a^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(a))/(
a^(3/4)*c^(3/4)) + sqrt(2)*(21*B*sqrt(a) - 5*A*sqrt(c))*log(-sqrt(2)*a^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqr
t(a))/(a^(3/4)*c^(3/4)))/c^2

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mupad [B]  time = 1.29, size = 686, normalized size = 2.14 \begin {gather*} -2\,\mathrm {atanh}\left (\frac {25\,A^2\,\sqrt {x}\,\sqrt {\frac {441\,B^2\,\sqrt {-a^3\,c^{11}}}{4096\,a^2\,c^{11}}-\frac {25\,A^2\,\sqrt {-a^3\,c^{11}}}{4096\,a^3\,c^{10}}-\frac {105\,A\,B}{2048\,a\,c^5}}}{32\,\left (\frac {525\,A^2\,B}{2048\,c^3}-\frac {9261\,B^3\,a}{2048\,c^4}+\frac {125\,A^3\,\sqrt {-a^3\,c^{11}}}{2048\,a^2\,c^8}-\frac {2205\,A\,B^2\,\sqrt {-a^3\,c^{11}}}{2048\,a\,c^9}\right )}-\frac {441\,B^2\,a\,\sqrt {x}\,\sqrt {\frac {441\,B^2\,\sqrt {-a^3\,c^{11}}}{4096\,a^2\,c^{11}}-\frac {25\,A^2\,\sqrt {-a^3\,c^{11}}}{4096\,a^3\,c^{10}}-\frac {105\,A\,B}{2048\,a\,c^5}}}{32\,\left (\frac {525\,A^2\,B}{2048\,c^2}-\frac {9261\,B^3\,a}{2048\,c^3}+\frac {125\,A^3\,\sqrt {-a^3\,c^{11}}}{2048\,a^2\,c^7}-\frac {2205\,A\,B^2\,\sqrt {-a^3\,c^{11}}}{2048\,a\,c^8}\right )}\right )\,\sqrt {-\frac {25\,A^2\,c\,\sqrt {-a^3\,c^{11}}-441\,B^2\,a\,\sqrt {-a^3\,c^{11}}+210\,A\,B\,a^2\,c^6}{4096\,a^3\,c^{11}}}-2\,\mathrm {atanh}\left (\frac {25\,A^2\,\sqrt {x}\,\sqrt {\frac {25\,A^2\,\sqrt {-a^3\,c^{11}}}{4096\,a^3\,c^{10}}-\frac {105\,A\,B}{2048\,a\,c^5}-\frac {441\,B^2\,\sqrt {-a^3\,c^{11}}}{4096\,a^2\,c^{11}}}}{32\,\left (\frac {525\,A^2\,B}{2048\,c^3}-\frac {9261\,B^3\,a}{2048\,c^4}-\frac {125\,A^3\,\sqrt {-a^3\,c^{11}}}{2048\,a^2\,c^8}+\frac {2205\,A\,B^2\,\sqrt {-a^3\,c^{11}}}{2048\,a\,c^9}\right )}-\frac {441\,B^2\,a\,\sqrt {x}\,\sqrt {\frac {25\,A^2\,\sqrt {-a^3\,c^{11}}}{4096\,a^3\,c^{10}}-\frac {105\,A\,B}{2048\,a\,c^5}-\frac {441\,B^2\,\sqrt {-a^3\,c^{11}}}{4096\,a^2\,c^{11}}}}{32\,\left (\frac {525\,A^2\,B}{2048\,c^2}-\frac {9261\,B^3\,a}{2048\,c^3}-\frac {125\,A^3\,\sqrt {-a^3\,c^{11}}}{2048\,a^2\,c^7}+\frac {2205\,A\,B^2\,\sqrt {-a^3\,c^{11}}}{2048\,a\,c^8}\right )}\right )\,\sqrt {-\frac {441\,B^2\,a\,\sqrt {-a^3\,c^{11}}-25\,A^2\,c\,\sqrt {-a^3\,c^{11}}+210\,A\,B\,a^2\,c^6}{4096\,a^3\,c^{11}}}-\frac {\frac {9\,A\,x^{5/2}}{16\,c}+\frac {11\,B\,x^{7/2}}{16\,c}+\frac {5\,A\,a\,\sqrt {x}}{16\,c^2}+\frac {7\,B\,a\,x^{3/2}}{16\,c^2}}{a^2+2\,a\,c\,x^2+c^2\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(7/2)*(A + B*x))/(a + c*x^2)^3,x)

[Out]

- 2*atanh((25*A^2*x^(1/2)*((441*B^2*(-a^3*c^11)^(1/2))/(4096*a^2*c^11) - (25*A^2*(-a^3*c^11)^(1/2))/(4096*a^3*
c^10) - (105*A*B)/(2048*a*c^5))^(1/2))/(32*((525*A^2*B)/(2048*c^3) - (9261*B^3*a)/(2048*c^4) + (125*A^3*(-a^3*
c^11)^(1/2))/(2048*a^2*c^8) - (2205*A*B^2*(-a^3*c^11)^(1/2))/(2048*a*c^9))) - (441*B^2*a*x^(1/2)*((441*B^2*(-a
^3*c^11)^(1/2))/(4096*a^2*c^11) - (25*A^2*(-a^3*c^11)^(1/2))/(4096*a^3*c^10) - (105*A*B)/(2048*a*c^5))^(1/2))/
(32*((525*A^2*B)/(2048*c^2) - (9261*B^3*a)/(2048*c^3) + (125*A^3*(-a^3*c^11)^(1/2))/(2048*a^2*c^7) - (2205*A*B
^2*(-a^3*c^11)^(1/2))/(2048*a*c^8))))*(-(25*A^2*c*(-a^3*c^11)^(1/2) - 441*B^2*a*(-a^3*c^11)^(1/2) + 210*A*B*a^
2*c^6)/(4096*a^3*c^11))^(1/2) - 2*atanh((25*A^2*x^(1/2)*((25*A^2*(-a^3*c^11)^(1/2))/(4096*a^3*c^10) - (105*A*B
)/(2048*a*c^5) - (441*B^2*(-a^3*c^11)^(1/2))/(4096*a^2*c^11))^(1/2))/(32*((525*A^2*B)/(2048*c^3) - (9261*B^3*a
)/(2048*c^4) - (125*A^3*(-a^3*c^11)^(1/2))/(2048*a^2*c^8) + (2205*A*B^2*(-a^3*c^11)^(1/2))/(2048*a*c^9))) - (4
41*B^2*a*x^(1/2)*((25*A^2*(-a^3*c^11)^(1/2))/(4096*a^3*c^10) - (105*A*B)/(2048*a*c^5) - (441*B^2*(-a^3*c^11)^(
1/2))/(4096*a^2*c^11))^(1/2))/(32*((525*A^2*B)/(2048*c^2) - (9261*B^3*a)/(2048*c^3) - (125*A^3*(-a^3*c^11)^(1/
2))/(2048*a^2*c^7) + (2205*A*B^2*(-a^3*c^11)^(1/2))/(2048*a*c^8))))*(-(441*B^2*a*(-a^3*c^11)^(1/2) - 25*A^2*c*
(-a^3*c^11)^(1/2) + 210*A*B*a^2*c^6)/(4096*a^3*c^11))^(1/2) - ((9*A*x^(5/2))/(16*c) + (11*B*x^(7/2))/(16*c) +
(5*A*a*x^(1/2))/(16*c^2) + (7*B*a*x^(3/2))/(16*c^2))/(a^2 + c^2*x^4 + 2*a*c*x^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(7/2)*(B*x+A)/(c*x**2+a)**3,x)

[Out]

Timed out

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