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

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*(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)*ArcT
an[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))

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Rubi [A]  time = 0.53163, antiderivative size = 322, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.423 \[ -\frac{7 (3 b B-11 A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{15/4} \sqrt [4]{c}}+\frac{7 (3 b B-11 A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{15/4} \sqrt [4]{c}}-\frac{7 (3 b B-11 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{15/4} \sqrt [4]{c}}+\frac{7 (3 b B-11 A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt{2} b^{15/4} \sqrt [4]{c}}+\frac{7 (3 b B-11 A c)}{48 b^3 c x^{3/2}}-\frac{3 b B-11 A c}{16 b^2 c x^{3/2} \left (b+c x^2\right )}-\frac{b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2} \]

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)*ArcT
an[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))

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Rubi in Sympy [A]  time = 83.2786, size = 306, normalized size = 0.95 \[ \frac{A c - B b}{4 b c x^{\frac{3}{2}} \left (b + c x^{2}\right )^{2}} + \frac{11 A c - 3 B b}{16 b^{2} c x^{\frac{3}{2}} \left (b + c x^{2}\right )} - \frac{7 \left (11 A c - 3 B b\right )}{48 b^{3} c x^{\frac{3}{2}}} + \frac{7 \sqrt{2} \left (11 A c - 3 B b\right ) \log{\left (- \sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x} + \sqrt{b} + \sqrt{c} x \right )}}{128 b^{\frac{15}{4}} \sqrt [4]{c}} - \frac{7 \sqrt{2} \left (11 A c - 3 B b\right ) \log{\left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x} + \sqrt{b} + \sqrt{c} x \right )}}{128 b^{\frac{15}{4}} \sqrt [4]{c}} + \frac{7 \sqrt{2} \left (11 A c - 3 B b\right ) \operatorname{atan}{\left (1 - \frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}} \right )}}{64 b^{\frac{15}{4}} \sqrt [4]{c}} - \frac{7 \sqrt{2} \left (11 A c - 3 B b\right ) \operatorname{atan}{\left (1 + \frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}} \right )}}{64 b^{\frac{15}{4}} \sqrt [4]{c}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

(A*c - B*b)/(4*b*c*x**(3/2)*(b + c*x**2)**2) + (11*A*c - 3*B*b)/(16*b**2*c*x**(3
/2)*(b + c*x**2)) - 7*(11*A*c - 3*B*b)/(48*b**3*c*x**(3/2)) + 7*sqrt(2)*(11*A*c
- 3*B*b)*log(-sqrt(2)*b**(1/4)*c**(1/4)*sqrt(x) + sqrt(b) + sqrt(c)*x)/(128*b**(
15/4)*c**(1/4)) - 7*sqrt(2)*(11*A*c - 3*B*b)*log(sqrt(2)*b**(1/4)*c**(1/4)*sqrt(
x) + sqrt(b) + sqrt(c)*x)/(128*b**(15/4)*c**(1/4)) + 7*sqrt(2)*(11*A*c - 3*B*b)*
atan(1 - sqrt(2)*c**(1/4)*sqrt(x)/b**(1/4))/(64*b**(15/4)*c**(1/4)) - 7*sqrt(2)*
(11*A*c - 3*B*b)*atan(1 + sqrt(2)*c**(1/4)*sqrt(x)/b**(1/4))/(64*b**(15/4)*c**(1
/4))

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Mathematica [A]  time = 0.530459, size = 286, normalized size = 0.89 \[ \frac{\frac{96 b^{7/4} \sqrt{x} (b B-A c)}{\left (b+c x^2\right )^2}+\frac{24 b^{3/4} \sqrt{x} (7 b B-15 A c)}{b+c x^2}-\frac{256 A b^{3/4}}{x^{3/2}}+\frac{21 \sqrt{2} (11 A c-3 b B) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{\sqrt [4]{c}}+\frac{21 \sqrt{2} (3 b B-11 A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{\sqrt [4]{c}}+\frac{42 \sqrt{2} (11 A c-3 b B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt [4]{c}}+\frac{42 \sqrt{2} (3 b B-11 A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{\sqrt [4]{c}}}{384 b^{15/4}} \]

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^(7/4)*(b*B - A*c)*Sqrt[x])/(b + c*x^2)^2 + (24
*b^(3/4)*(7*b*B - 15*A*c)*Sqrt[x])/(b + c*x^2) + (42*Sqrt[2]*(-3*b*B + 11*A*c)*A
rcTan[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) + (21*Sqrt[2]*(-3*b*
B + 11*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/c^(1/4)
+ (21*Sqrt[2]*(3*b*B - 11*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + S
qrt[c]*x])/c^(1/4))/(384*b^(15/4))

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Maple [A]  time = 0.029, size = 357, normalized size = 1.1 \[ -{\frac{2\,A}{3\,{b}^{3}}{x}^{-{\frac{3}{2}}}}-{\frac{15\,A{c}^{2}}{16\,{b}^{3} \left ( c{x}^{2}+b \right ) ^{2}}{x}^{{\frac{5}{2}}}}+{\frac{7\,Bc}{16\,{b}^{2} \left ( c{x}^{2}+b \right ) ^{2}}{x}^{{\frac{5}{2}}}}-{\frac{19\,Ac}{16\,{b}^{2} \left ( c{x}^{2}+b \right ) ^{2}}\sqrt{x}}+{\frac{11\,B}{16\,b \left ( c{x}^{2}+b \right ) ^{2}}\sqrt{x}}-{\frac{77\,\sqrt{2}Ac}{64\,{b}^{4}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ) }-{\frac{77\,\sqrt{2}Ac}{128\,{b}^{4}}\sqrt [4]{{\frac{b}{c}}}\ln \left ({1 \left ( x+\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) \left ( x-\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) ^{-1}} \right ) }-{\frac{77\,\sqrt{2}Ac}{64\,{b}^{4}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ) }+{\frac{21\,\sqrt{2}B}{64\,{b}^{3}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ) }+{\frac{21\,\sqrt{2}B}{128\,{b}^{3}}\sqrt [4]{{\frac{b}{c}}}\ln \left ({1 \left ( x+\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) \left ( x-\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) ^{-1}} \right ) }+{\frac{21\,\sqrt{2}B}{64\,{b}^{3}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ) } \]

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]

-2/3*A/b^3/x^(3/2)-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)*x^(1/2)*2^(1/2)+(b/c)^(1/2))/(x-(b/c)^(1/4)*x
^(1/2)*2^(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*arctan(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)*x^(1/2)*2^(1/2)+
(b/c)^(1/2))/(x-(b/c)^(1/4)*x^(1/2)*2^(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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.247391, size = 929, normalized size = 2.89 \[ \frac{28 \,{\left (3 \, B b c - 11 \, A c^{2}\right )} x^{4} - 128 \, A b^{2} + 44 \,{\left (3 \, B b^{2} - 11 \, A b c\right )} x^{2} + 84 \,{\left (b^{3} c^{2} x^{5} + 2 \, b^{4} c x^{3} + b^{5} x\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{1}{4}} \arctan \left (-\frac{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}}}{{\left (3 \, B b - 11 \, A c\right )} \sqrt{x} - \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}}\right ) - 21 \,{\left (b^{3} c^{2} x^{5} + 2 \, b^{4} c x^{3} + b^{5} x\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{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^{5} + 2 \, b^{4} c x^{3} + b^{5} x\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{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 )}{192 \,{\left (b^{3} c^{2} x^{5} + 2 \, b^{4} c x^{3} + b^{5} x\right )} \sqrt{x}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/192*(28*(3*B*b*c - 11*A*c^2)*x^4 - 128*A*b^2 + 44*(3*B*b^2 - 11*A*b*c)*x^2 + 8
4*(b^3*c^2*x^5 + 2*b^4*c*x^3 + b^5*x)*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))^(1/4)*arcta
n(-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)/((3*B*b - 11*A*c)*sqrt(x) - 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))) - 21*(b^3*c^2*
x^5 + 2*b^4*c*x^3 + b^5*x)*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))^(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^5 + 2*b^4*c
*x^3 + b^5*x)*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))^(1/4)*log(-7*b^4*(-(81*B^4*b^4 - 11
88*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)))/((b^3*c^2*x^5 + 2*b^4*c*x^3 + b^5*x)*s
qrt(x))

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

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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GIAC/XCAS [A]  time = 0.224499, size = 410, normalized size = 1.27 \[ \frac{7 \, \sqrt{2}{\left (3 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - 11 \, \left (b c^{3}\right )^{\frac{1}{4}} A c\right )} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{b}{c}\right )^{\frac{1}{4}} + 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{b}{c}\right )^{\frac{1}{4}}}\right )}{64 \, b^{4} c} + \frac{7 \, \sqrt{2}{\left (3 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - 11 \, \left (b c^{3}\right )^{\frac{1}{4}} A c\right )} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{b}{c}\right )^{\frac{1}{4}} - 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{b}{c}\right )^{\frac{1}{4}}}\right )}{64 \, b^{4} c} + \frac{7 \, \sqrt{2}{\left (3 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - 11 \, \left (b c^{3}\right )^{\frac{1}{4}} A c\right )}{\rm ln}\left (\sqrt{2} \sqrt{x} \left (\frac{b}{c}\right )^{\frac{1}{4}} + x + \sqrt{\frac{b}{c}}\right )}{128 \, b^{4} c} - \frac{7 \, \sqrt{2}{\left (3 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - 11 \, \left (b c^{3}\right )^{\frac{1}{4}} A c\right )}{\rm ln}\left (-\sqrt{2} \sqrt{x} \left (\frac{b}{c}\right )^{\frac{1}{4}} + x + \sqrt{\frac{b}{c}}\right )}{128 \, b^{4} c} - \frac{2 \, A}{3 \, b^{3} x^{\frac{3}{2}}} + \frac{7 \, B b c x^{\frac{5}{2}} - 15 \, A c^{2} x^{\frac{5}{2}} + 11 \, B b^{2} \sqrt{x} - 19 \, A b c \sqrt{x}}{16 \,{\left (c x^{2} + b\right )}^{2} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)*x^(7/2)/(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)*(sq
rt(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*sqr
t(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)*ln(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)*ln(-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)