3.549 \(\int \frac{\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^6} \, dx\)

Optimal. Leaf size=152 \[ \frac{1}{2} b^{3/2} (5 a B+2 A b) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )+\frac{b^2 x \sqrt{a+b x^2} (5 a B+2 A b)}{2 a}-\frac{b \left (a+b x^2\right )^{3/2} (5 a B+2 A b)}{3 a x}-\frac{\left (a+b x^2\right )^{5/2} (5 a B+2 A b)}{15 a x^3}-\frac{A \left (a+b x^2\right )^{7/2}}{5 a x^5} \]

[Out]

(b^2*(2*A*b + 5*a*B)*x*Sqrt[a + b*x^2])/(2*a) - (b*(2*A*b + 5*a*B)*(a + b*x^2)^(
3/2))/(3*a*x) - ((2*A*b + 5*a*B)*(a + b*x^2)^(5/2))/(15*a*x^3) - (A*(a + b*x^2)^
(7/2))/(5*a*x^5) + (b^(3/2)*(2*A*b + 5*a*B)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]]
)/2

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Rubi [A]  time = 0.180825, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227 \[ \frac{1}{2} b^{3/2} (5 a B+2 A b) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )+\frac{b^2 x \sqrt{a+b x^2} (5 a B+2 A b)}{2 a}-\frac{b \left (a+b x^2\right )^{3/2} (5 a B+2 A b)}{3 a x}-\frac{\left (a+b x^2\right )^{5/2} (5 a B+2 A b)}{15 a x^3}-\frac{A \left (a+b x^2\right )^{7/2}}{5 a x^5} \]

Antiderivative was successfully verified.

[In]  Int[((a + b*x^2)^(5/2)*(A + B*x^2))/x^6,x]

[Out]

(b^2*(2*A*b + 5*a*B)*x*Sqrt[a + b*x^2])/(2*a) - (b*(2*A*b + 5*a*B)*(a + b*x^2)^(
3/2))/(3*a*x) - ((2*A*b + 5*a*B)*(a + b*x^2)^(5/2))/(15*a*x^3) - (A*(a + b*x^2)^
(7/2))/(5*a*x^5) + (b^(3/2)*(2*A*b + 5*a*B)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]]
)/2

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Rubi in Sympy [A]  time = 18.2327, size = 136, normalized size = 0.89 \[ - \frac{A \left (a + b x^{2}\right )^{\frac{7}{2}}}{5 a x^{5}} + \frac{b^{\frac{3}{2}} \left (2 A b + 5 B a\right ) \operatorname{atanh}{\left (\frac{\sqrt{b} x}{\sqrt{a + b x^{2}}} \right )}}{2} + \frac{b^{2} x \sqrt{a + b x^{2}} \left (2 A b + 5 B a\right )}{2 a} - \frac{b \left (a + b x^{2}\right )^{\frac{3}{2}} \left (2 A b + 5 B a\right )}{3 a x} - \frac{\left (a + b x^{2}\right )^{\frac{5}{2}} \left (2 A b + 5 B a\right )}{15 a x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((b*x**2+a)**(5/2)*(B*x**2+A)/x**6,x)

[Out]

-A*(a + b*x**2)**(7/2)/(5*a*x**5) + b**(3/2)*(2*A*b + 5*B*a)*atanh(sqrt(b)*x/sqr
t(a + b*x**2))/2 + b**2*x*sqrt(a + b*x**2)*(2*A*b + 5*B*a)/(2*a) - b*(a + b*x**2
)**(3/2)*(2*A*b + 5*B*a)/(3*a*x) - (a + b*x**2)**(5/2)*(2*A*b + 5*B*a)/(15*a*x**
3)

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Mathematica [A]  time = 0.169395, size = 105, normalized size = 0.69 \[ \frac{1}{2} b^{3/2} (5 a B+2 A b) \log \left (\sqrt{b} \sqrt{a+b x^2}+b x\right )-\frac{\sqrt{a+b x^2} \left (6 a^2 A+2 b x^4 (35 a B+23 A b)+2 a x^2 (5 a B+11 A b)-15 b^2 B x^6\right )}{30 x^5} \]

Antiderivative was successfully verified.

[In]  Integrate[((a + b*x^2)^(5/2)*(A + B*x^2))/x^6,x]

[Out]

-(Sqrt[a + b*x^2]*(6*a^2*A + 2*a*(11*A*b + 5*a*B)*x^2 + 2*b*(23*A*b + 35*a*B)*x^
4 - 15*b^2*B*x^6))/(30*x^5) + (b^(3/2)*(2*A*b + 5*a*B)*Log[b*x + Sqrt[b]*Sqrt[a
+ b*x^2]])/2

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Maple [A]  time = 0.013, size = 251, normalized size = 1.7 \[ -{\frac{A}{5\,a{x}^{5}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{2\,Ab}{15\,{a}^{2}{x}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{8\,{b}^{2}A}{15\,{a}^{3}x} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{8\,A{b}^{3}x}{15\,{a}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{2\,A{b}^{3}x}{3\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{A{b}^{3}x}{a}\sqrt{b{x}^{2}+a}}+A{b}^{{\frac{5}{2}}}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ) -{\frac{B}{3\,a{x}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{4\,Bb}{3\,{a}^{2}x} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{4\,{b}^{2}Bx}{3\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{5\,{b}^{2}Bx}{3\,a} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{b}^{2}Bx}{2}\sqrt{b{x}^{2}+a}}+{\frac{5\,Ba}{2}{b}^{{\frac{3}{2}}}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ) } \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((b*x^2+a)^(5/2)*(B*x^2+A)/x^6,x)

[Out]

-1/5*A*(b*x^2+a)^(7/2)/a/x^5-2/15*A*b/a^2/x^3*(b*x^2+a)^(7/2)-8/15*A*b^2/a^3/x*(
b*x^2+a)^(7/2)+8/15*A*b^3/a^3*x*(b*x^2+a)^(5/2)+2/3*A*b^3/a^2*x*(b*x^2+a)^(3/2)+
A*b^3/a*x*(b*x^2+a)^(1/2)+A*b^(5/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))-1/3*B/a/x^3*(b
*x^2+a)^(7/2)-4/3*B*b/a^2/x*(b*x^2+a)^(7/2)+4/3*B*b^2/a^2*x*(b*x^2+a)^(5/2)+5/3*
B*b^2/a*x*(b*x^2+a)^(3/2)+5/2*B*b^2*x*(b*x^2+a)^(1/2)+5/2*B*b^(3/2)*a*ln(x*b^(1/
2)+(b*x^2+a)^(1/2))

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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)*(b*x^2 + a)^(5/2)/x^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.249151, size = 1, normalized size = 0.01 \[ \left [\frac{15 \,{\left (5 \, B a b + 2 \, A b^{2}\right )} \sqrt{b} x^{5} \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) + 2 \,{\left (15 \, B b^{2} x^{6} - 2 \,{\left (35 \, B a b + 23 \, A b^{2}\right )} x^{4} - 6 \, A a^{2} - 2 \,{\left (5 \, B a^{2} + 11 \, A a b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{60 \, x^{5}}, \frac{15 \,{\left (5 \, B a b + 2 \, A b^{2}\right )} \sqrt{-b} x^{5} \arctan \left (\frac{b x}{\sqrt{b x^{2} + a} \sqrt{-b}}\right ) +{\left (15 \, B b^{2} x^{6} - 2 \,{\left (35 \, B a b + 23 \, A b^{2}\right )} x^{4} - 6 \, A a^{2} - 2 \,{\left (5 \, B a^{2} + 11 \, A a b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{30 \, x^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)*(b*x^2 + a)^(5/2)/x^6,x, algorithm="fricas")

[Out]

[1/60*(15*(5*B*a*b + 2*A*b^2)*sqrt(b)*x^5*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(
b)*x - a) + 2*(15*B*b^2*x^6 - 2*(35*B*a*b + 23*A*b^2)*x^4 - 6*A*a^2 - 2*(5*B*a^2
+ 11*A*a*b)*x^2)*sqrt(b*x^2 + a))/x^5, 1/30*(15*(5*B*a*b + 2*A*b^2)*sqrt(-b)*x^
5*arctan(b*x/(sqrt(b*x^2 + a)*sqrt(-b))) + (15*B*b^2*x^6 - 2*(35*B*a*b + 23*A*b^
2)*x^4 - 6*A*a^2 - 2*(5*B*a^2 + 11*A*a*b)*x^2)*sqrt(b*x^2 + a))/x^5]

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Sympy [A]  time = 23.3963, size = 292, normalized size = 1.92 \[ - \frac{A \sqrt{a} b^{2}}{x \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{A a^{2} \sqrt{b} \sqrt{\frac{a}{b x^{2}} + 1}}{5 x^{4}} - \frac{11 A a b^{\frac{3}{2}} \sqrt{\frac{a}{b x^{2}} + 1}}{15 x^{2}} - \frac{8 A b^{\frac{5}{2}} \sqrt{\frac{a}{b x^{2}} + 1}}{15} + A b^{\frac{5}{2}} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )} - \frac{A b^{3} x}{\sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{2 B a^{\frac{3}{2}} b}{x \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{B \sqrt{a} b^{2} x \sqrt{1 + \frac{b x^{2}}{a}}}{2} - \frac{2 B \sqrt{a} b^{2} x}{\sqrt{1 + \frac{b x^{2}}{a}}} - \frac{B a^{2} \sqrt{b} \sqrt{\frac{a}{b x^{2}} + 1}}{3 x^{2}} - \frac{B a b^{\frac{3}{2}} \sqrt{\frac{a}{b x^{2}} + 1}}{3} + \frac{5 B a b^{\frac{3}{2}} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x**2+a)**(5/2)*(B*x**2+A)/x**6,x)

[Out]

-A*sqrt(a)*b**2/(x*sqrt(1 + b*x**2/a)) - A*a**2*sqrt(b)*sqrt(a/(b*x**2) + 1)/(5*
x**4) - 11*A*a*b**(3/2)*sqrt(a/(b*x**2) + 1)/(15*x**2) - 8*A*b**(5/2)*sqrt(a/(b*
x**2) + 1)/15 + A*b**(5/2)*asinh(sqrt(b)*x/sqrt(a)) - A*b**3*x/(sqrt(a)*sqrt(1 +
b*x**2/a)) - 2*B*a**(3/2)*b/(x*sqrt(1 + b*x**2/a)) + B*sqrt(a)*b**2*x*sqrt(1 +
b*x**2/a)/2 - 2*B*sqrt(a)*b**2*x/sqrt(1 + b*x**2/a) - B*a**2*sqrt(b)*sqrt(a/(b*x
**2) + 1)/(3*x**2) - B*a*b**(3/2)*sqrt(a/(b*x**2) + 1)/3 + 5*B*a*b**(3/2)*asinh(
sqrt(b)*x/sqrt(a))/2

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GIAC/XCAS [A]  time = 0.259681, size = 433, normalized size = 2.85 \[ \frac{1}{2} \, \sqrt{b x^{2} + a} B b^{2} x - \frac{1}{4} \,{\left (5 \, B a b^{\frac{3}{2}} + 2 \, A b^{\frac{5}{2}}\right )}{\rm ln}\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2}\right ) + \frac{2 \,{\left (45 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{8} B a^{2} b^{\frac{3}{2}} + 45 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{8} A a b^{\frac{5}{2}} - 150 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{6} B a^{3} b^{\frac{3}{2}} - 90 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{6} A a^{2} b^{\frac{5}{2}} + 200 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{4} B a^{4} b^{\frac{3}{2}} + 140 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{4} A a^{3} b^{\frac{5}{2}} - 130 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} B a^{5} b^{\frac{3}{2}} - 70 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} A a^{4} b^{\frac{5}{2}} + 35 \, B a^{6} b^{\frac{3}{2}} + 23 \, A a^{5} b^{\frac{5}{2}}\right )}}{15 \,{\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} - a\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^2 + A)*(b*x^2 + a)^(5/2)/x^6,x, algorithm="giac")

[Out]

1/2*sqrt(b*x^2 + a)*B*b^2*x - 1/4*(5*B*a*b^(3/2) + 2*A*b^(5/2))*ln((sqrt(b)*x -
sqrt(b*x^2 + a))^2) + 2/15*(45*(sqrt(b)*x - sqrt(b*x^2 + a))^8*B*a^2*b^(3/2) + 4
5*(sqrt(b)*x - sqrt(b*x^2 + a))^8*A*a*b^(5/2) - 150*(sqrt(b)*x - sqrt(b*x^2 + a)
)^6*B*a^3*b^(3/2) - 90*(sqrt(b)*x - sqrt(b*x^2 + a))^6*A*a^2*b^(5/2) + 200*(sqrt
(b)*x - sqrt(b*x^2 + a))^4*B*a^4*b^(3/2) + 140*(sqrt(b)*x - sqrt(b*x^2 + a))^4*A
*a^3*b^(5/2) - 130*(sqrt(b)*x - sqrt(b*x^2 + a))^2*B*a^5*b^(3/2) - 70*(sqrt(b)*x
- sqrt(b*x^2 + a))^2*A*a^4*b^(5/2) + 35*B*a^6*b^(3/2) + 23*A*a^5*b^(5/2))/((sqr
t(b)*x - sqrt(b*x^2 + a))^2 - a)^5