3.215 \(\int \frac {x^{7/2} (A+B x^2)}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=322 \[ -\frac {7 (3 b B-11 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{15/4} \sqrt [4]{c}}+\frac {7 (3 b B-11 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{15/4} \sqrt [4]{c}}-\frac {7 (3 b B-11 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{15/4} \sqrt [4]{c}}+\frac {7 (3 b B-11 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} b^{15/4} \sqrt [4]{c}}+\frac {7 (3 b B-11 A c)}{48 b^3 c x^{3/2}}-\frac {3 b B-11 A c}{16 b^2 c x^{3/2} \left (b+c x^2\right )}-\frac {b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2} \]

[Out]

7/48*(-11*A*c+3*B*b)/b^3/c/x^(3/2)+1/4*(A*c-B*b)/b/c/x^(3/2)/(c*x^2+b)^2+1/16*(11*A*c-3*B*b)/b^2/c/x^(3/2)/(c*
x^2+b)-7/64*(-11*A*c+3*B*b)*arctan(1-c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(15/4)/c^(1/4)*2^(1/2)+7/64*(-11*A*c+3
*B*b)*arctan(1+c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(15/4)/c^(1/4)*2^(1/2)-7/128*(-11*A*c+3*B*b)*ln(b^(1/2)+x*c^
(1/2)-b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2))/b^(15/4)/c^(1/4)*2^(1/2)+7/128*(-11*A*c+3*B*b)*ln(b^(1/2)+x*c^(1/2)+b^(
1/4)*c^(1/4)*2^(1/2)*x^(1/2))/b^(15/4)/c^(1/4)*2^(1/2)

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Rubi [A]  time = 0.25, antiderivative size = 322, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {1584, 457, 290, 325, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac {3 b B-11 A c}{16 b^2 c x^{3/2} \left (b+c x^2\right )}+\frac {7 (3 b B-11 A c)}{48 b^3 c x^{3/2}}-\frac {7 (3 b B-11 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{15/4} \sqrt [4]{c}}+\frac {7 (3 b B-11 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{15/4} \sqrt [4]{c}}-\frac {7 (3 b B-11 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{15/4} \sqrt [4]{c}}+\frac {7 (3 b B-11 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} b^{15/4} \sqrt [4]{c}}-\frac {b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^(7/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

(7*(3*b*B - 11*A*c))/(48*b^3*c*x^(3/2)) - (b*B - A*c)/(4*b*c*x^(3/2)*(b + c*x^2)^2) - (3*b*B - 11*A*c)/(16*b^2
*c*x^(3/2)*(b + c*x^2)) - (7*(3*b*B - 11*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(15
/4)*c^(1/4)) + (7*(3*b*B - 11*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(15/4)*c^(1/4)
) - (7*(3*b*B - 11*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(15/4)*c^(1/
4)) + (7*(3*b*B - 11*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(15/4)*c^(
1/4))

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 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

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 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 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int \frac {x^{7/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac {A+B x^2}{x^{5/2} \left (b+c x^2\right )^3} \, dx\\ &=-\frac {b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}+\frac {\left (-\frac {3 b B}{2}+\frac {11 A c}{2}\right ) \int \frac {1}{x^{5/2} \left (b+c x^2\right )^2} \, dx}{4 b c}\\ &=-\frac {b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}-\frac {3 b B-11 A c}{16 b^2 c x^{3/2} \left (b+c x^2\right )}-\frac {(7 (3 b B-11 A c)) \int \frac {1}{x^{5/2} \left (b+c x^2\right )} \, dx}{32 b^2 c}\\ &=\frac {7 (3 b B-11 A c)}{48 b^3 c x^{3/2}}-\frac {b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}-\frac {3 b B-11 A c}{16 b^2 c x^{3/2} \left (b+c x^2\right )}+\frac {(7 (3 b B-11 A c)) \int \frac {1}{\sqrt {x} \left (b+c x^2\right )} \, dx}{32 b^3}\\ &=\frac {7 (3 b B-11 A c)}{48 b^3 c x^{3/2}}-\frac {b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}-\frac {3 b B-11 A c}{16 b^2 c x^{3/2} \left (b+c x^2\right )}+\frac {(7 (3 b B-11 A c)) \operatorname {Subst}\left (\int \frac {1}{b+c x^4} \, dx,x,\sqrt {x}\right )}{16 b^3}\\ &=\frac {7 (3 b B-11 A c)}{48 b^3 c x^{3/2}}-\frac {b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}-\frac {3 b B-11 A c}{16 b^2 c x^{3/2} \left (b+c x^2\right )}+\frac {(7 (3 b B-11 A c)) \operatorname {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{32 b^{7/2}}+\frac {(7 (3 b B-11 A c)) \operatorname {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{32 b^{7/2}}\\ &=\frac {7 (3 b B-11 A c)}{48 b^3 c x^{3/2}}-\frac {b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}-\frac {3 b B-11 A c}{16 b^2 c x^{3/2} \left (b+c x^2\right )}+\frac {(7 (3 b B-11 A c)) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {b}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt {x}\right )}{64 b^{7/2} \sqrt {c}}+\frac {(7 (3 b B-11 A c)) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {b}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt {x}\right )}{64 b^{7/2} \sqrt {c}}-\frac {(7 (3 b B-11 A c)) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{b}}{\sqrt [4]{c}}+2 x}{-\frac {\sqrt {b}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt {x}\right )}{64 \sqrt {2} b^{15/4} \sqrt [4]{c}}-\frac {(7 (3 b B-11 A c)) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{b}}{\sqrt [4]{c}}-2 x}{-\frac {\sqrt {b}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt {x}\right )}{64 \sqrt {2} b^{15/4} \sqrt [4]{c}}\\ &=\frac {7 (3 b B-11 A c)}{48 b^3 c x^{3/2}}-\frac {b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}-\frac {3 b B-11 A c}{16 b^2 c x^{3/2} \left (b+c x^2\right )}-\frac {7 (3 b B-11 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{15/4} \sqrt [4]{c}}+\frac {7 (3 b B-11 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{15/4} \sqrt [4]{c}}+\frac {(7 (3 b B-11 A c)) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{15/4} \sqrt [4]{c}}-\frac {(7 (3 b B-11 A c)) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{15/4} \sqrt [4]{c}}\\ &=\frac {7 (3 b B-11 A c)}{48 b^3 c x^{3/2}}-\frac {b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}-\frac {3 b B-11 A c}{16 b^2 c x^{3/2} \left (b+c x^2\right )}-\frac {7 (3 b B-11 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{15/4} \sqrt [4]{c}}+\frac {7 (3 b B-11 A c) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{15/4} \sqrt [4]{c}}-\frac {7 (3 b B-11 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{15/4} \sqrt [4]{c}}+\frac {7 (3 b B-11 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{15/4} \sqrt [4]{c}}\\ \end {align*}

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Mathematica [A]  time = 0.50, size = 400, normalized size = 1.24 \[ \frac {-\frac {96 A b^{7/4} c \sqrt {x}}{\left (b+c x^2\right )^2}-\frac {360 A b^{3/4} c \sqrt {x}}{b+c x^2}-\frac {256 A b^{3/4}}{x^{3/2}}+\frac {42 \sqrt {2} (11 A c-3 b B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt [4]{c}}+\frac {42 \sqrt {2} (3 b B-11 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt [4]{c}}+231 \sqrt {2} A c^{3/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )-231 \sqrt {2} A c^{3/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )+\frac {96 b^{11/4} B \sqrt {x}}{\left (b+c x^2\right )^2}+\frac {168 b^{7/4} B \sqrt {x}}{b+c x^2}-\frac {63 \sqrt {2} b B \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{\sqrt [4]{c}}+\frac {63 \sqrt {2} b B \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{\sqrt [4]{c}}}{384 b^{15/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(7/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

((-256*A*b^(3/4))/x^(3/2) + (96*b^(11/4)*B*Sqrt[x])/(b + c*x^2)^2 - (96*A*b^(7/4)*c*Sqrt[x])/(b + c*x^2)^2 + (
168*b^(7/4)*B*Sqrt[x])/(b + c*x^2) - (360*A*b^(3/4)*c*Sqrt[x])/(b + c*x^2) + (42*Sqrt[2]*(-3*b*B + 11*A*c)*Arc
Tan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/c^(1/4) + (42*Sqrt[2]*(3*b*B - 11*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)
*Sqrt[x])/b^(1/4)])/c^(1/4) - (63*Sqrt[2]*b*B*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/c^(1
/4) + 231*Sqrt[2]*A*c^(3/4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] + (63*Sqrt[2]*b*B*Log[S
qrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/c^(1/4) - 231*Sqrt[2]*A*c^(3/4)*Log[Sqrt[b] + Sqrt[2]*b
^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(384*b^(15/4))

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fricas [B]  time = 0.85, size = 809, normalized size = 2.51 \[ -\frac {84 \, {\left (b^{3} c^{2} x^{6} + 2 \, b^{4} c x^{4} + b^{5} x^{2}\right )} \left (-\frac {81 \, B^{4} b^{4} - 1188 \, A B^{3} b^{3} c + 6534 \, A^{2} B^{2} b^{2} c^{2} - 15972 \, A^{3} B b c^{3} + 14641 \, A^{4} c^{4}}{b^{15} c}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {b^{8} \sqrt {-\frac {81 \, B^{4} b^{4} - 1188 \, A B^{3} b^{3} c + 6534 \, A^{2} B^{2} b^{2} c^{2} - 15972 \, A^{3} B b c^{3} + 14641 \, A^{4} c^{4}}{b^{15} c}} + {\left (9 \, B^{2} b^{2} - 66 \, A B b c + 121 \, A^{2} c^{2}\right )} x} b^{11} c \left (-\frac {81 \, B^{4} b^{4} - 1188 \, A B^{3} b^{3} c + 6534 \, A^{2} B^{2} b^{2} c^{2} - 15972 \, A^{3} B b c^{3} + 14641 \, A^{4} c^{4}}{b^{15} c}\right )^{\frac {3}{4}} + {\left (3 \, B b^{12} c - 11 \, A b^{11} c^{2}\right )} \sqrt {x} \left (-\frac {81 \, B^{4} b^{4} - 1188 \, A B^{3} b^{3} c + 6534 \, A^{2} B^{2} b^{2} c^{2} - 15972 \, A^{3} B b c^{3} + 14641 \, A^{4} c^{4}}{b^{15} c}\right )^{\frac {3}{4}}}{81 \, B^{4} b^{4} - 1188 \, A B^{3} b^{3} c + 6534 \, A^{2} B^{2} b^{2} c^{2} - 15972 \, A^{3} B b c^{3} + 14641 \, A^{4} c^{4}}\right ) + 21 \, {\left (b^{3} c^{2} x^{6} + 2 \, b^{4} c x^{4} + b^{5} x^{2}\right )} \left (-\frac {81 \, B^{4} b^{4} - 1188 \, A B^{3} b^{3} c + 6534 \, A^{2} B^{2} b^{2} c^{2} - 15972 \, A^{3} B b c^{3} + 14641 \, A^{4} c^{4}}{b^{15} c}\right )^{\frac {1}{4}} \log \left (7 \, b^{4} \left (-\frac {81 \, B^{4} b^{4} - 1188 \, A B^{3} b^{3} c + 6534 \, A^{2} B^{2} b^{2} c^{2} - 15972 \, A^{3} B b c^{3} + 14641 \, A^{4} c^{4}}{b^{15} c}\right )^{\frac {1}{4}} - 7 \, {\left (3 \, B b - 11 \, A c\right )} \sqrt {x}\right ) - 21 \, {\left (b^{3} c^{2} x^{6} + 2 \, b^{4} c x^{4} + b^{5} x^{2}\right )} \left (-\frac {81 \, B^{4} b^{4} - 1188 \, A B^{3} b^{3} c + 6534 \, A^{2} B^{2} b^{2} c^{2} - 15972 \, A^{3} B b c^{3} + 14641 \, A^{4} c^{4}}{b^{15} c}\right )^{\frac {1}{4}} \log \left (-7 \, b^{4} \left (-\frac {81 \, B^{4} b^{4} - 1188 \, A B^{3} b^{3} c + 6534 \, A^{2} B^{2} b^{2} c^{2} - 15972 \, A^{3} B b c^{3} + 14641 \, A^{4} c^{4}}{b^{15} c}\right )^{\frac {1}{4}} - 7 \, {\left (3 \, B b - 11 \, A c\right )} \sqrt {x}\right ) - 4 \, {\left (7 \, {\left (3 \, B b c - 11 \, A c^{2}\right )} x^{4} - 32 \, A b^{2} + 11 \, {\left (3 \, B b^{2} - 11 \, A b c\right )} x^{2}\right )} \sqrt {x}}{192 \, {\left (b^{3} c^{2} x^{6} + 2 \, b^{4} c x^{4} + b^{5} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

-1/192*(84*(b^3*c^2*x^6 + 2*b^4*c*x^4 + b^5*x^2)*(-(81*B^4*b^4 - 1188*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 159
72*A^3*B*b*c^3 + 14641*A^4*c^4)/(b^15*c))^(1/4)*arctan((sqrt(b^8*sqrt(-(81*B^4*b^4 - 1188*A*B^3*b^3*c + 6534*A
^2*B^2*b^2*c^2 - 15972*A^3*B*b*c^3 + 14641*A^4*c^4)/(b^15*c)) + (9*B^2*b^2 - 66*A*B*b*c + 121*A^2*c^2)*x)*b^11
*c*(-(81*B^4*b^4 - 1188*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 15972*A^3*B*b*c^3 + 14641*A^4*c^4)/(b^15*c))^(3/4
) + (3*B*b^12*c - 11*A*b^11*c^2)*sqrt(x)*(-(81*B^4*b^4 - 1188*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 15972*A^3*B
*b*c^3 + 14641*A^4*c^4)/(b^15*c))^(3/4))/(81*B^4*b^4 - 1188*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 15972*A^3*B*b
*c^3 + 14641*A^4*c^4)) + 21*(b^3*c^2*x^6 + 2*b^4*c*x^4 + b^5*x^2)*(-(81*B^4*b^4 - 1188*A*B^3*b^3*c + 6534*A^2*
B^2*b^2*c^2 - 15972*A^3*B*b*c^3 + 14641*A^4*c^4)/(b^15*c))^(1/4)*log(7*b^4*(-(81*B^4*b^4 - 1188*A*B^3*b^3*c +
6534*A^2*B^2*b^2*c^2 - 15972*A^3*B*b*c^3 + 14641*A^4*c^4)/(b^15*c))^(1/4) - 7*(3*B*b - 11*A*c)*sqrt(x)) - 21*(
b^3*c^2*x^6 + 2*b^4*c*x^4 + b^5*x^2)*(-(81*B^4*b^4 - 1188*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 15972*A^3*B*b*c
^3 + 14641*A^4*c^4)/(b^15*c))^(1/4)*log(-7*b^4*(-(81*B^4*b^4 - 1188*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 15972
*A^3*B*b*c^3 + 14641*A^4*c^4)/(b^15*c))^(1/4) - 7*(3*B*b - 11*A*c)*sqrt(x)) - 4*(7*(3*B*b*c - 11*A*c^2)*x^4 -
32*A*b^2 + 11*(3*B*b^2 - 11*A*b*c)*x^2)*sqrt(x))/(b^3*c^2*x^6 + 2*b^4*c*x^4 + b^5*x^2)

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giac [A]  time = 0.20, size = 304, normalized size = 0.94 \[ \frac {7 \, \sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 11 \, \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{4} c} + \frac {7 \, \sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 11 \, \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{4} c} + \frac {7 \, \sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 11 \, \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{4} c} - \frac {7 \, \sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 11 \, \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{4} c} - \frac {2 \, A}{3 \, b^{3} x^{\frac {3}{2}}} + \frac {7 \, B b c x^{\frac {5}{2}} - 15 \, A c^{2} x^{\frac {5}{2}} + 11 \, B b^{2} \sqrt {x} - 19 \, A b c \sqrt {x}}{16 \, {\left (c x^{2} + b\right )}^{2} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

7/64*sqrt(2)*(3*(b*c^3)^(1/4)*B*b - 11*(b*c^3)^(1/4)*A*c)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))
/(b/c)^(1/4))/(b^4*c) + 7/64*sqrt(2)*(3*(b*c^3)^(1/4)*B*b - 11*(b*c^3)^(1/4)*A*c)*arctan(-1/2*sqrt(2)*(sqrt(2)
*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/(b^4*c) + 7/128*sqrt(2)*(3*(b*c^3)^(1/4)*B*b - 11*(b*c^3)^(1/4)*A*c)*lo
g(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^4*c) - 7/128*sqrt(2)*(3*(b*c^3)^(1/4)*B*b - 11*(b*c^3)^(1/4)
*A*c)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^4*c) - 2/3*A/(b^3*x^(3/2)) + 1/16*(7*B*b*c*x^(5/2)
- 15*A*c^2*x^(5/2) + 11*B*b^2*sqrt(x) - 19*A*b*c*sqrt(x))/((c*x^2 + b)^2*b^3)

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maple [A]  time = 0.07, size = 357, normalized size = 1.11 \[ -\frac {15 A \,c^{2} x^{\frac {5}{2}}}{16 \left (c \,x^{2}+b \right )^{2} b^{3}}+\frac {7 B c \,x^{\frac {5}{2}}}{16 \left (c \,x^{2}+b \right )^{2} b^{2}}-\frac {19 A c \sqrt {x}}{16 \left (c \,x^{2}+b \right )^{2} b^{2}}+\frac {11 B \sqrt {x}}{16 \left (c \,x^{2}+b \right )^{2} b}-\frac {77 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{64 b^{4}}-\frac {77 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{64 b^{4}}-\frac {77 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A c \ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{128 b^{4}}+\frac {21 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{64 b^{3}}+\frac {21 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{64 b^{3}}+\frac {21 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{128 b^{3}}-\frac {2 A}{3 b^{3} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x)

[Out]

-15/16/b^3/(c*x^2+b)^2*x^(5/2)*A*c^2+7/16/b^2/(c*x^2+b)^2*x^(5/2)*B*c-19/16/b^2/(c*x^2+b)^2*A*x^(1/2)*c+11/16/
b/(c*x^2+b)^2*B*x^(1/2)-77/64/b^4*(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)*c-77/128/b^4*(b/
c)^(1/4)*2^(1/2)*A*ln((x+(b/c)^(1/4)*2^(1/2)*x^(1/2)+(b/c)^(1/2))/(x-(b/c)^(1/4)*2^(1/2)*x^(1/2)+(b/c)^(1/2)))
*c-77/64/b^4*(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)*c+21/64/b^3*(b/c)^(1/4)*2^(1/2)*B*arc
tan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+21/128/b^3*(b/c)^(1/4)*2^(1/2)*B*ln((x+(b/c)^(1/4)*2^(1/2)*x^(1/2)+(b/c)^(1
/2))/(x-(b/c)^(1/4)*2^(1/2)*x^(1/2)+(b/c)^(1/2)))+21/64/b^3*(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x
^(1/2)+1)-2/3*A/b^3/x^(3/2)

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maxima [A]  time = 3.09, size = 285, normalized size = 0.89 \[ \frac {7 \, {\left (3 \, B b c - 11 \, A c^{2}\right )} x^{4} - 32 \, A b^{2} + 11 \, {\left (3 \, B b^{2} - 11 \, A b c\right )} x^{2}}{48 \, {\left (b^{3} c^{2} x^{\frac {11}{2}} + 2 \, b^{4} c x^{\frac {7}{2}} + b^{5} x^{\frac {3}{2}}\right )}} + \frac {7 \, {\left (\frac {2 \, \sqrt {2} {\left (3 \, B b - 11 \, A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} {\left (3 \, B b - 11 \, A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} {\left (3 \, B b - 11 \, A c\right )} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (3 \, B b - 11 \, A c\right )} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}}\right )}}{128 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

1/48*(7*(3*B*b*c - 11*A*c^2)*x^4 - 32*A*b^2 + 11*(3*B*b^2 - 11*A*b*c)*x^2)/(b^3*c^2*x^(11/2) + 2*b^4*c*x^(7/2)
+ b^5*x^(3/2)) + 7/128*(2*sqrt(2)*(3*B*b - 11*A*c)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sq
rt(x))/sqrt(sqrt(b)*sqrt(c)))/(sqrt(b)*sqrt(sqrt(b)*sqrt(c))) + 2*sqrt(2)*(3*B*b - 11*A*c)*arctan(-1/2*sqrt(2)
*(sqrt(2)*b^(1/4)*c^(1/4) - 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/(sqrt(b)*sqrt(sqrt(b)*sqrt(c))) + sqrt(2
)*(3*B*b - 11*A*c)*log(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(3/4)*c^(1/4)) - sqrt(2)*(3*B
*b - 11*A*c)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(3/4)*c^(1/4)))/b^3

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mupad [B]  time = 0.45, size = 888, normalized size = 2.76 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(7/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x)

[Out]

- ((2*A)/(3*b) + (11*x^2*(11*A*c - 3*B*b))/(48*b^2) + (7*c*x^4*(11*A*c - 3*B*b))/(48*b^3))/(b^2*x^(3/2) + c^2*
x^(11/2) + 2*b*c*x^(7/2)) - (atan((((11*A*c - 3*B*b)*(x^(1/2)*(97140736*A^2*b^9*c^5 + 7225344*B^2*b^11*c^3 - 5
2985856*A*B*b^10*c^4) - (7*(11*A*c - 3*B*b)*(80740352*A*b^13*c^4 - 22020096*B*b^14*c^3))/(64*(-b)^(15/4)*c^(1/
4)))*7i)/(64*(-b)^(15/4)*c^(1/4)) + ((11*A*c - 3*B*b)*(x^(1/2)*(97140736*A^2*b^9*c^5 + 7225344*B^2*b^11*c^3 -
52985856*A*B*b^10*c^4) + (7*(11*A*c - 3*B*b)*(80740352*A*b^13*c^4 - 22020096*B*b^14*c^3))/(64*(-b)^(15/4)*c^(1
/4)))*7i)/(64*(-b)^(15/4)*c^(1/4)))/((7*(11*A*c - 3*B*b)*(x^(1/2)*(97140736*A^2*b^9*c^5 + 7225344*B^2*b^11*c^3
- 52985856*A*B*b^10*c^4) - (7*(11*A*c - 3*B*b)*(80740352*A*b^13*c^4 - 22020096*B*b^14*c^3))/(64*(-b)^(15/4)*c
^(1/4))))/(64*(-b)^(15/4)*c^(1/4)) - (7*(11*A*c - 3*B*b)*(x^(1/2)*(97140736*A^2*b^9*c^5 + 7225344*B^2*b^11*c^3
- 52985856*A*B*b^10*c^4) + (7*(11*A*c - 3*B*b)*(80740352*A*b^13*c^4 - 22020096*B*b^14*c^3))/(64*(-b)^(15/4)*c
^(1/4))))/(64*(-b)^(15/4)*c^(1/4))))*(11*A*c - 3*B*b)*7i)/(32*(-b)^(15/4)*c^(1/4)) - (7*atan(((7*(11*A*c - 3*B
*b)*(x^(1/2)*(97140736*A^2*b^9*c^5 + 7225344*B^2*b^11*c^3 - 52985856*A*B*b^10*c^4) - ((11*A*c - 3*B*b)*(807403
52*A*b^13*c^4 - 22020096*B*b^14*c^3)*7i)/(64*(-b)^(15/4)*c^(1/4))))/(64*(-b)^(15/4)*c^(1/4)) + (7*(11*A*c - 3*
B*b)*(x^(1/2)*(97140736*A^2*b^9*c^5 + 7225344*B^2*b^11*c^3 - 52985856*A*B*b^10*c^4) + ((11*A*c - 3*B*b)*(80740
352*A*b^13*c^4 - 22020096*B*b^14*c^3)*7i)/(64*(-b)^(15/4)*c^(1/4))))/(64*(-b)^(15/4)*c^(1/4)))/(((11*A*c - 3*B
*b)*(x^(1/2)*(97140736*A^2*b^9*c^5 + 7225344*B^2*b^11*c^3 - 52985856*A*B*b^10*c^4) - ((11*A*c - 3*B*b)*(807403
52*A*b^13*c^4 - 22020096*B*b^14*c^3)*7i)/(64*(-b)^(15/4)*c^(1/4)))*7i)/(64*(-b)^(15/4)*c^(1/4)) - ((11*A*c - 3
*B*b)*(x^(1/2)*(97140736*A^2*b^9*c^5 + 7225344*B^2*b^11*c^3 - 52985856*A*B*b^10*c^4) + ((11*A*c - 3*B*b)*(8074
0352*A*b^13*c^4 - 22020096*B*b^14*c^3)*7i)/(64*(-b)^(15/4)*c^(1/4)))*7i)/(64*(-b)^(15/4)*c^(1/4))))*(11*A*c -
3*B*b))/(32*(-b)^(15/4)*c^(1/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(7/2)*(B*x**2+A)/(c*x**4+b*x**2)**3,x)

[Out]

Timed out

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