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

Optimal. Leaf size=292 \[ \frac {c^{3/4} \left (\sqrt {a} B+A \sqrt {c}\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{2 \sqrt {2} a^{9/4}}-\frac {c^{3/4} \left (\sqrt {a} B+A \sqrt {c}\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{2 \sqrt {2} a^{9/4}}+\frac {c^{3/4} \left (\sqrt {a} B-A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{9/4}}-\frac {c^{3/4} \left (\sqrt {a} B-A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} a^{9/4}}+\frac {2 A c}{a^2 \sqrt {x}}-\frac {2 A}{5 a x^{5/2}}-\frac {2 B}{3 a x^{3/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.32, antiderivative size = 292, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {829, 827, 1168, 1162, 617, 204, 1165, 628} \begin {gather*} \frac {c^{3/4} \left (\sqrt {a} B+A \sqrt {c}\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{2 \sqrt {2} a^{9/4}}-\frac {c^{3/4} \left (\sqrt {a} B+A \sqrt {c}\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {a}+\sqrt {c} x\right )}{2 \sqrt {2} a^{9/4}}+\frac {c^{3/4} \left (\sqrt {a} B-A \sqrt {c}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{9/4}}-\frac {c^{3/4} \left (\sqrt {a} B-A \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} a^{9/4}}+\frac {2 A c}{a^2 \sqrt {x}}-\frac {2 A}{5 a x^{5/2}}-\frac {2 B}{3 a x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(5*a*x^(5/2)) - (2*B)/(3*a*x^(3/2)) + (2*A*c)/(a^2*Sqrt[x]) + ((Sqrt[a]*B - A*Sqrt[c])*c^(3/4)*ArcTan[1
- (Sqrt[2]*c^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(9/4)) - ((Sqrt[a]*B - A*Sqrt[c])*c^(3/4)*ArcTan[1 + (Sqrt[2
]*c^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(9/4)) + ((Sqrt[a]*B + A*Sqrt[c])*c^(3/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4
)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*a^(9/4)) - ((Sqrt[a]*B + A*Sqrt[c])*c^(3/4)*Log[Sqrt[a] + Sqrt[2]*a
^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*a^(9/4))

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 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 829

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

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 {A+B x}{x^{7/2} \left (a+c x^2\right )} \, dx &=-\frac {2 A}{5 a x^{5/2}}+\frac {\int \frac {a B-A c x}{x^{5/2} \left (a+c x^2\right )} \, dx}{a}\\ &=-\frac {2 A}{5 a x^{5/2}}-\frac {2 B}{3 a x^{3/2}}+\frac {\int \frac {-a A c-a B c x}{x^{3/2} \left (a+c x^2\right )} \, dx}{a^2}\\ &=-\frac {2 A}{5 a x^{5/2}}-\frac {2 B}{3 a x^{3/2}}+\frac {2 A c}{a^2 \sqrt {x}}+\frac {\int \frac {-a^2 B c+a A c^2 x}{\sqrt {x} \left (a+c x^2\right )} \, dx}{a^3}\\ &=-\frac {2 A}{5 a x^{5/2}}-\frac {2 B}{3 a x^{3/2}}+\frac {2 A c}{a^2 \sqrt {x}}+\frac {2 \operatorname {Subst}\left (\int \frac {-a^2 B c+a A c^2 x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{a^3}\\ &=-\frac {2 A}{5 a x^{5/2}}-\frac {2 B}{3 a x^{3/2}}+\frac {2 A c}{a^2 \sqrt {x}}-\frac {\left (\left (\sqrt {a} B-A \sqrt {c}\right ) \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a} \sqrt {c}+c x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{a^2}-\frac {\left (\left (\sqrt {a} B+A \sqrt {c}\right ) \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a} \sqrt {c}-c x^2}{a+c x^4} \, dx,x,\sqrt {x}\right )}{a^2}\\ &=-\frac {2 A}{5 a x^{5/2}}-\frac {2 B}{3 a x^{3/2}}+\frac {2 A c}{a^2 \sqrt {x}}-\frac {\left (\left (\sqrt {a} B-A \sqrt {c}\right ) \sqrt {c}\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 )}{2 a^2}-\frac {\left (\left (\sqrt {a} B-A \sqrt {c}\right ) \sqrt {c}\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 )}{2 a^2}+\frac {\left (\left (\sqrt {a} B+A \sqrt {c}\right ) c^{3/4}\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 )}{2 \sqrt {2} a^{9/4}}+\frac {\left (\left (\sqrt {a} B+A \sqrt {c}\right ) c^{3/4}\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 )}{2 \sqrt {2} a^{9/4}}\\ &=-\frac {2 A}{5 a x^{5/2}}-\frac {2 B}{3 a x^{3/2}}+\frac {2 A c}{a^2 \sqrt {x}}+\frac {\left (\sqrt {a} B+A \sqrt {c}\right ) c^{3/4} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} a^{9/4}}-\frac {\left (\sqrt {a} B+A \sqrt {c}\right ) c^{3/4} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} a^{9/4}}-\frac {\left (\left (\sqrt {a} B-A \sqrt {c}\right ) c^{3/4}\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 )}{\sqrt {2} a^{9/4}}+\frac {\left (\left (\sqrt {a} B-A \sqrt {c}\right ) c^{3/4}\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 )}{\sqrt {2} a^{9/4}}\\ &=-\frac {2 A}{5 a x^{5/2}}-\frac {2 B}{3 a x^{3/2}}+\frac {2 A c}{a^2 \sqrt {x}}+\frac {\left (\sqrt {a} B-A \sqrt {c}\right ) c^{3/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{9/4}}-\frac {\left (\sqrt {a} B-A \sqrt {c}\right ) c^{3/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{9/4}}+\frac {\left (\sqrt {a} B+A \sqrt {c}\right ) c^{3/4} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} a^{9/4}}-\frac {\left (\sqrt {a} B+A \sqrt {c}\right ) c^{3/4} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} a^{9/4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.01, size = 54, normalized size = 0.18 \begin {gather*} -\frac {2 \left (3 A \, _2F_1\left (-\frac {5}{4},1;-\frac {1}{4};-\frac {c x^2}{a}\right )+5 B x \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};-\frac {c x^2}{a}\right )\right )}{15 a x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(3*A*Hypergeometric2F1[-5/4, 1, -1/4, -((c*x^2)/a)] + 5*B*x*Hypergeometric2F1[-3/4, 1, 1/4, -((c*x^2)/a)])
)/(15*a*x^(5/2))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.37, size = 167, normalized size = 0.57 \begin {gather*} \frac {\left (\sqrt {a} B c^{3/4}-A c^{5/4}\right ) \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}}\right )}{\sqrt {2} a^{9/4}}-\frac {\left (\sqrt {a} B c^{3/4}+A c^{5/4}\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} \sqrt {x}}{\sqrt {a}+\sqrt {c} x}\right )}{\sqrt {2} a^{9/4}}-\frac {2 \left (3 a A+5 a B x-15 A c x^2\right )}{15 a^2 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(3*a*A + 5*a*B*x - 15*A*c*x^2))/(15*a^2*x^(5/2)) + ((Sqrt[a]*B*c^(3/4) - A*c^(5/4))*ArcTan[(Sqrt[a] - Sqrt
[c]*x)/(Sqrt[2]*a^(1/4)*c^(1/4)*Sqrt[x])])/(Sqrt[2]*a^(9/4)) - ((Sqrt[a]*B*c^(3/4) + A*c^(5/4))*ArcTanh[(Sqrt[
2]*a^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[c]*x)])/(Sqrt[2]*a^(9/4))

________________________________________________________________________________________

fricas [B]  time = 0.44, size = 869, normalized size = 2.98 \begin {gather*} \frac {15 \, a^{2} x^{3} \sqrt {\frac {a^{4} \sqrt {-\frac {B^{4} a^{2} c^{3} - 2 \, A^{2} B^{2} a c^{4} + A^{4} c^{5}}{a^{9}}} + 2 \, A B c^{2}}{a^{4}}} \log \left (-{\left (B^{4} a^{2} c^{2} - A^{4} c^{4}\right )} \sqrt {x} + {\left (A a^{7} \sqrt {-\frac {B^{4} a^{2} c^{3} - 2 \, A^{2} B^{2} a c^{4} + A^{4} c^{5}}{a^{9}}} + B^{3} a^{4} c - A^{2} B a^{3} c^{2}\right )} \sqrt {\frac {a^{4} \sqrt {-\frac {B^{4} a^{2} c^{3} - 2 \, A^{2} B^{2} a c^{4} + A^{4} c^{5}}{a^{9}}} + 2 \, A B c^{2}}{a^{4}}}\right ) - 15 \, a^{2} x^{3} \sqrt {\frac {a^{4} \sqrt {-\frac {B^{4} a^{2} c^{3} - 2 \, A^{2} B^{2} a c^{4} + A^{4} c^{5}}{a^{9}}} + 2 \, A B c^{2}}{a^{4}}} \log \left (-{\left (B^{4} a^{2} c^{2} - A^{4} c^{4}\right )} \sqrt {x} - {\left (A a^{7} \sqrt {-\frac {B^{4} a^{2} c^{3} - 2 \, A^{2} B^{2} a c^{4} + A^{4} c^{5}}{a^{9}}} + B^{3} a^{4} c - A^{2} B a^{3} c^{2}\right )} \sqrt {\frac {a^{4} \sqrt {-\frac {B^{4} a^{2} c^{3} - 2 \, A^{2} B^{2} a c^{4} + A^{4} c^{5}}{a^{9}}} + 2 \, A B c^{2}}{a^{4}}}\right ) - 15 \, a^{2} x^{3} \sqrt {-\frac {a^{4} \sqrt {-\frac {B^{4} a^{2} c^{3} - 2 \, A^{2} B^{2} a c^{4} + A^{4} c^{5}}{a^{9}}} - 2 \, A B c^{2}}{a^{4}}} \log \left (-{\left (B^{4} a^{2} c^{2} - A^{4} c^{4}\right )} \sqrt {x} + {\left (A a^{7} \sqrt {-\frac {B^{4} a^{2} c^{3} - 2 \, A^{2} B^{2} a c^{4} + A^{4} c^{5}}{a^{9}}} - B^{3} a^{4} c + A^{2} B a^{3} c^{2}\right )} \sqrt {-\frac {a^{4} \sqrt {-\frac {B^{4} a^{2} c^{3} - 2 \, A^{2} B^{2} a c^{4} + A^{4} c^{5}}{a^{9}}} - 2 \, A B c^{2}}{a^{4}}}\right ) + 15 \, a^{2} x^{3} \sqrt {-\frac {a^{4} \sqrt {-\frac {B^{4} a^{2} c^{3} - 2 \, A^{2} B^{2} a c^{4} + A^{4} c^{5}}{a^{9}}} - 2 \, A B c^{2}}{a^{4}}} \log \left (-{\left (B^{4} a^{2} c^{2} - A^{4} c^{4}\right )} \sqrt {x} - {\left (A a^{7} \sqrt {-\frac {B^{4} a^{2} c^{3} - 2 \, A^{2} B^{2} a c^{4} + A^{4} c^{5}}{a^{9}}} - B^{3} a^{4} c + A^{2} B a^{3} c^{2}\right )} \sqrt {-\frac {a^{4} \sqrt {-\frac {B^{4} a^{2} c^{3} - 2 \, A^{2} B^{2} a c^{4} + A^{4} c^{5}}{a^{9}}} - 2 \, A B c^{2}}{a^{4}}}\right ) + 4 \, {\left (15 \, A c x^{2} - 5 \, B a x - 3 \, A a\right )} \sqrt {x}}{30 \, a^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(15*a^2*x^3*sqrt((a^4*sqrt(-(B^4*a^2*c^3 - 2*A^2*B^2*a*c^4 + A^4*c^5)/a^9) + 2*A*B*c^2)/a^4)*log(-(B^4*a^
2*c^2 - A^4*c^4)*sqrt(x) + (A*a^7*sqrt(-(B^4*a^2*c^3 - 2*A^2*B^2*a*c^4 + A^4*c^5)/a^9) + B^3*a^4*c - A^2*B*a^3
*c^2)*sqrt((a^4*sqrt(-(B^4*a^2*c^3 - 2*A^2*B^2*a*c^4 + A^4*c^5)/a^9) + 2*A*B*c^2)/a^4)) - 15*a^2*x^3*sqrt((a^4
*sqrt(-(B^4*a^2*c^3 - 2*A^2*B^2*a*c^4 + A^4*c^5)/a^9) + 2*A*B*c^2)/a^4)*log(-(B^4*a^2*c^2 - A^4*c^4)*sqrt(x) -
(A*a^7*sqrt(-(B^4*a^2*c^3 - 2*A^2*B^2*a*c^4 + A^4*c^5)/a^9) + B^3*a^4*c - A^2*B*a^3*c^2)*sqrt((a^4*sqrt(-(B^4
*a^2*c^3 - 2*A^2*B^2*a*c^4 + A^4*c^5)/a^9) + 2*A*B*c^2)/a^4)) - 15*a^2*x^3*sqrt(-(a^4*sqrt(-(B^4*a^2*c^3 - 2*A
^2*B^2*a*c^4 + A^4*c^5)/a^9) - 2*A*B*c^2)/a^4)*log(-(B^4*a^2*c^2 - A^4*c^4)*sqrt(x) + (A*a^7*sqrt(-(B^4*a^2*c^
3 - 2*A^2*B^2*a*c^4 + A^4*c^5)/a^9) - B^3*a^4*c + A^2*B*a^3*c^2)*sqrt(-(a^4*sqrt(-(B^4*a^2*c^3 - 2*A^2*B^2*a*c
^4 + A^4*c^5)/a^9) - 2*A*B*c^2)/a^4)) + 15*a^2*x^3*sqrt(-(a^4*sqrt(-(B^4*a^2*c^3 - 2*A^2*B^2*a*c^4 + A^4*c^5)/
a^9) - 2*A*B*c^2)/a^4)*log(-(B^4*a^2*c^2 - A^4*c^4)*sqrt(x) - (A*a^7*sqrt(-(B^4*a^2*c^3 - 2*A^2*B^2*a*c^4 + A^
4*c^5)/a^9) - B^3*a^4*c + A^2*B*a^3*c^2)*sqrt(-(a^4*sqrt(-(B^4*a^2*c^3 - 2*A^2*B^2*a*c^4 + A^4*c^5)/a^9) - 2*A
*B*c^2)/a^4)) + 4*(15*A*c*x^2 - 5*B*a*x - 3*A*a)*sqrt(x))/(a^2*x^3)

________________________________________________________________________________________

giac [A]  time = 0.19, size = 265, normalized size = 0.91 \begin {gather*} -\frac {\sqrt {2} {\left (\left (a c^{3}\right )^{\frac {1}{4}} B a c - \left (a c^{3}\right )^{\frac {3}{4}} A\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 )}{2 \, a^{3} c} - \frac {\sqrt {2} {\left (\left (a c^{3}\right )^{\frac {1}{4}} B a c - \left (a c^{3}\right )^{\frac {3}{4}} A\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 )}{2 \, a^{3} c} - \frac {\sqrt {2} {\left (\left (a c^{3}\right )^{\frac {1}{4}} B a c + \left (a c^{3}\right )^{\frac {3}{4}} A\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{c}}\right )}{4 \, a^{3} c} + \frac {\sqrt {2} {\left (\left (a c^{3}\right )^{\frac {1}{4}} B a c + \left (a c^{3}\right )^{\frac {3}{4}} A\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{c}}\right )}{4 \, a^{3} c} + \frac {2 \, {\left (15 \, A c x^{2} - 5 \, B a x - 3 \, A a\right )}}{15 \, a^{2} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(2)*((a*c^3)^(1/4)*B*a*c - (a*c^3)^(3/4)*A)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/c)^(1/4) + 2*sqrt(x))/(a/c
)^(1/4))/(a^3*c) - 1/2*sqrt(2)*((a*c^3)^(1/4)*B*a*c - (a*c^3)^(3/4)*A)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/c)^(1/4
) - 2*sqrt(x))/(a/c)^(1/4))/(a^3*c) - 1/4*sqrt(2)*((a*c^3)^(1/4)*B*a*c + (a*c^3)^(3/4)*A)*log(sqrt(2)*sqrt(x)*
(a/c)^(1/4) + x + sqrt(a/c))/(a^3*c) + 1/4*sqrt(2)*((a*c^3)^(1/4)*B*a*c + (a*c^3)^(3/4)*A)*log(-sqrt(2)*sqrt(x
)*(a/c)^(1/4) + x + sqrt(a/c))/(a^3*c) + 2/15*(15*A*c*x^2 - 5*B*a*x - 3*A*a)/(a^2*x^(5/2))

________________________________________________________________________________________

maple [A]  time = 0.06, size = 302, normalized size = 1.03 \begin {gather*} \frac {\sqrt {2}\, A c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{2 \left (\frac {a}{c}\right )^{\frac {1}{4}} a^{2}}+\frac {\sqrt {2}\, A c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{2 \left (\frac {a}{c}\right )^{\frac {1}{4}} a^{2}}+\frac {\sqrt {2}\, A c \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 )}{4 \left (\frac {a}{c}\right )^{\frac {1}{4}} a^{2}}-\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{2 a^{2}}-\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{2 a^{2}}-\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B c \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 )}{4 a^{2}}+\frac {2 A c}{a^{2} \sqrt {x}}-\frac {2 B}{3 a \,x^{\frac {3}{2}}}-\frac {2 A}{5 a \,x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c/a^2*B*(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)+1)-1/2*c/a^2*B*(a/c)^(1/4)*2^(1/2)*arctan(
2^(1/2)/(a/c)^(1/4)*x^(1/2)-1)-1/4*c/a^2*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)))+1/4*c/a^2*A/(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)))+1/2*c/a^2*A/(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c
)^(1/4)*x^(1/2)+1)+1/2*c/a^2*A/(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x^(1/2)-1)-2/5*A/a/x^(5/2)-2/3*B
/a/x^(3/2)+2*A*c/a^2/x^(1/2)

________________________________________________________________________________________

maxima [A]  time = 1.28, size = 263, normalized size = 0.90 \begin {gather*} -\frac {c {\left (\frac {2 \, \sqrt {2} {\left (B a \sqrt {c} - A \sqrt {a} 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 (B a \sqrt {c} - A \sqrt {a} 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 (B a \sqrt {c} + A \sqrt {a} 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 (B a \sqrt {c} + A \sqrt {a} 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}}}\right )}}{4 \, a^{2}} + \frac {2 \, {\left (15 \, A c x^{2} - 5 \, B a x - 3 \, A a\right )}}{15 \, a^{2} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*c*(2*sqrt(2)*(B*a*sqrt(c) - A*sqrt(a)*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)) + 2*sqrt(2)*(B*a*sqrt(c) - A*sqrt(a)*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)*(B*a*sqrt(c) + A*sqrt(a)*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)*(B*a*sqrt(c) + A*sqrt(a)*c)*log(-sqrt(2)*a^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqr
t(a))/(a^(3/4)*c^(3/4)))/a^2 + 2/15*(15*A*c*x^2 - 5*B*a*x - 3*A*a)/(a^2*x^(5/2))

________________________________________________________________________________________

mupad [B]  time = 1.25, size = 664, normalized size = 2.27 \begin {gather*} -2\,\mathrm {atanh}\left (\frac {32\,A^2\,a^7\,c^6\,\sqrt {x}\,\sqrt {\frac {A^2\,c\,\sqrt {-a^9\,c^3}}{4\,a^9}-\frac {B^2\,\sqrt {-a^9\,c^3}}{4\,a^8}+\frac {A\,B\,c^2}{2\,a^4}}}{16\,A^3\,a^5\,c^7-16\,A\,B^2\,a^6\,c^6+16\,B^3\,a^2\,c^4\,\sqrt {-a^9\,c^3}-16\,A^2\,B\,a\,c^5\,\sqrt {-a^9\,c^3}}-\frac {32\,B^2\,a^8\,c^5\,\sqrt {x}\,\sqrt {\frac {A^2\,c\,\sqrt {-a^9\,c^3}}{4\,a^9}-\frac {B^2\,\sqrt {-a^9\,c^3}}{4\,a^8}+\frac {A\,B\,c^2}{2\,a^4}}}{16\,A^3\,a^5\,c^7-16\,A\,B^2\,a^6\,c^6+16\,B^3\,a^2\,c^4\,\sqrt {-a^9\,c^3}-16\,A^2\,B\,a\,c^5\,\sqrt {-a^9\,c^3}}\right )\,\sqrt {\frac {A^2\,c\,\sqrt {-a^9\,c^3}-B^2\,a\,\sqrt {-a^9\,c^3}+2\,A\,B\,a^5\,c^2}{4\,a^9}}-2\,\mathrm {atanh}\left (\frac {32\,A^2\,a^7\,c^6\,\sqrt {x}\,\sqrt {\frac {B^2\,\sqrt {-a^9\,c^3}}{4\,a^8}-\frac {A^2\,c\,\sqrt {-a^9\,c^3}}{4\,a^9}+\frac {A\,B\,c^2}{2\,a^4}}}{16\,A^3\,a^5\,c^7-16\,A\,B^2\,a^6\,c^6-16\,B^3\,a^2\,c^4\,\sqrt {-a^9\,c^3}+16\,A^2\,B\,a\,c^5\,\sqrt {-a^9\,c^3}}-\frac {32\,B^2\,a^8\,c^5\,\sqrt {x}\,\sqrt {\frac {B^2\,\sqrt {-a^9\,c^3}}{4\,a^8}-\frac {A^2\,c\,\sqrt {-a^9\,c^3}}{4\,a^9}+\frac {A\,B\,c^2}{2\,a^4}}}{16\,A^3\,a^5\,c^7-16\,A\,B^2\,a^6\,c^6-16\,B^3\,a^2\,c^4\,\sqrt {-a^9\,c^3}+16\,A^2\,B\,a\,c^5\,\sqrt {-a^9\,c^3}}\right )\,\sqrt {\frac {B^2\,a\,\sqrt {-a^9\,c^3}-A^2\,c\,\sqrt {-a^9\,c^3}+2\,A\,B\,a^5\,c^2}{4\,a^9}}-\frac {\frac {2\,A}{5\,a}+\frac {2\,B\,x}{3\,a}-\frac {2\,A\,c\,x^2}{a^2}}{x^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- 2*atanh((32*A^2*a^7*c^6*x^(1/2)*((A^2*c*(-a^9*c^3)^(1/2))/(4*a^9) - (B^2*(-a^9*c^3)^(1/2))/(4*a^8) + (A*B*c^
2)/(2*a^4))^(1/2))/(16*A^3*a^5*c^7 - 16*A*B^2*a^6*c^6 + 16*B^3*a^2*c^4*(-a^9*c^3)^(1/2) - 16*A^2*B*a*c^5*(-a^9
*c^3)^(1/2)) - (32*B^2*a^8*c^5*x^(1/2)*((A^2*c*(-a^9*c^3)^(1/2))/(4*a^9) - (B^2*(-a^9*c^3)^(1/2))/(4*a^8) + (A
*B*c^2)/(2*a^4))^(1/2))/(16*A^3*a^5*c^7 - 16*A*B^2*a^6*c^6 + 16*B^3*a^2*c^4*(-a^9*c^3)^(1/2) - 16*A^2*B*a*c^5*
(-a^9*c^3)^(1/2)))*((A^2*c*(-a^9*c^3)^(1/2) - B^2*a*(-a^9*c^3)^(1/2) + 2*A*B*a^5*c^2)/(4*a^9))^(1/2) - 2*atanh
((32*A^2*a^7*c^6*x^(1/2)*((B^2*(-a^9*c^3)^(1/2))/(4*a^8) - (A^2*c*(-a^9*c^3)^(1/2))/(4*a^9) + (A*B*c^2)/(2*a^4
))^(1/2))/(16*A^3*a^5*c^7 - 16*A*B^2*a^6*c^6 - 16*B^3*a^2*c^4*(-a^9*c^3)^(1/2) + 16*A^2*B*a*c^5*(-a^9*c^3)^(1/
2)) - (32*B^2*a^8*c^5*x^(1/2)*((B^2*(-a^9*c^3)^(1/2))/(4*a^8) - (A^2*c*(-a^9*c^3)^(1/2))/(4*a^9) + (A*B*c^2)/(
2*a^4))^(1/2))/(16*A^3*a^5*c^7 - 16*A*B^2*a^6*c^6 - 16*B^3*a^2*c^4*(-a^9*c^3)^(1/2) + 16*A^2*B*a*c^5*(-a^9*c^3
)^(1/2)))*((B^2*a*(-a^9*c^3)^(1/2) - A^2*c*(-a^9*c^3)^(1/2) + 2*A*B*a^5*c^2)/(4*a^9))^(1/2) - ((2*A)/(5*a) + (
2*B*x)/(3*a) - (2*A*c*x^2)/a^2)/x^(5/2)

________________________________________________________________________________________

sympy [A]  time = 108.65, size = 398, normalized size = 1.36 \begin {gather*} \begin {cases} \tilde {\infty } \left (- \frac {2 A}{9 x^{\frac {9}{2}}} - \frac {2 B}{7 x^{\frac {7}{2}}}\right ) & \text {for}\: a = 0 \wedge c = 0 \\\frac {- \frac {2 A}{5 x^{\frac {5}{2}}} - \frac {2 B}{3 x^{\frac {3}{2}}}}{a} & \text {for}\: c = 0 \\\frac {- \frac {2 A}{9 x^{\frac {9}{2}}} - \frac {2 B}{7 x^{\frac {7}{2}}}}{c} & \text {for}\: a = 0 \\- \frac {2 A}{5 a x^{\frac {5}{2}}} + \frac {2 A c}{a^{2} \sqrt {x}} - \frac {\left (-1\right )^{\frac {3}{4}} A c \log {\left (- \sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 a^{\frac {9}{4}} \sqrt [4]{\frac {1}{c}}} + \frac {\left (-1\right )^{\frac {3}{4}} A c \log {\left (\sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 a^{\frac {9}{4}} \sqrt [4]{\frac {1}{c}}} + \frac {\left (-1\right )^{\frac {3}{4}} A c \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{a} \sqrt [4]{\frac {1}{c}}} \right )}}{a^{\frac {9}{4}} \sqrt [4]{\frac {1}{c}}} - \frac {2 B}{3 a x^{\frac {3}{2}}} + \frac {\sqrt [4]{-1} B c \sqrt [4]{\frac {1}{c}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 a^{\frac {7}{4}}} - \frac {\sqrt [4]{-1} B c \sqrt [4]{\frac {1}{c}} \log {\left (\sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 a^{\frac {7}{4}}} + \frac {\sqrt [4]{-1} B c \sqrt [4]{\frac {1}{c}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{a} \sqrt [4]{\frac {1}{c}}} \right )}}{a^{\frac {7}{4}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(-2*A/(9*x**(9/2)) - 2*B/(7*x**(7/2))), Eq(a, 0) & Eq(c, 0)), ((-2*A/(5*x**(5/2)) - 2*B/(3*x**(
3/2)))/a, Eq(c, 0)), ((-2*A/(9*x**(9/2)) - 2*B/(7*x**(7/2)))/c, Eq(a, 0)), (-2*A/(5*a*x**(5/2)) + 2*A*c/(a**2*
sqrt(x)) - (-1)**(3/4)*A*c*log(-(-1)**(1/4)*a**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*a**(9/4)*(1/c)**(1/4)) + (-1)*
*(3/4)*A*c*log((-1)**(1/4)*a**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*a**(9/4)*(1/c)**(1/4)) + (-1)**(3/4)*A*c*atan((
-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/c)**(1/4)))/(a**(9/4)*(1/c)**(1/4)) - 2*B/(3*a*x**(3/2)) + (-1)**(1/4)*B*c*(1/
c)**(1/4)*log(-(-1)**(1/4)*a**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*a**(7/4)) - (-1)**(1/4)*B*c*(1/c)**(1/4)*log((-
1)**(1/4)*a**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*a**(7/4)) + (-1)**(1/4)*B*c*(1/c)**(1/4)*atan((-1)**(3/4)*sqrt(x
)/(a**(1/4)*(1/c)**(1/4)))/a**(7/4), True))

________________________________________________________________________________________