3.242 \(\int \frac{\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{15/2}} \, dx\)

Optimal. Leaf size=204 \[ \frac{4 c^{3/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (3 A c+7 b B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{21 \sqrt [4]{b} \sqrt{b x^2+c x^4}}+\frac{4 c \sqrt{b x^2+c x^4} (3 A c+7 b B)}{21 b \sqrt{x}}-\frac{2 \left (b x^2+c x^4\right )^{3/2} (3 A c+7 b B)}{21 b x^{9/2}}-\frac{2 A \left (b x^2+c x^4\right )^{5/2}}{7 b x^{17/2}} \]

[Out]

(4*c*(7*b*B + 3*A*c)*Sqrt[b*x^2 + c*x^4])/(21*b*Sqrt[x]) - (2*(7*b*B + 3*A*c)*(b
*x^2 + c*x^4)^(3/2))/(21*b*x^(9/2)) - (2*A*(b*x^2 + c*x^4)^(5/2))/(7*b*x^(17/2))
+ (4*c^(3/4)*(7*b*B + 3*A*c)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b]
+ Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(21*b^(1/4)
*Sqrt[b*x^2 + c*x^4])

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Rubi [A]  time = 0.556767, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214 \[ \frac{4 c^{3/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (3 A c+7 b B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{21 \sqrt [4]{b} \sqrt{b x^2+c x^4}}+\frac{4 c \sqrt{b x^2+c x^4} (3 A c+7 b B)}{21 b \sqrt{x}}-\frac{2 \left (b x^2+c x^4\right )^{3/2} (3 A c+7 b B)}{21 b x^{9/2}}-\frac{2 A \left (b x^2+c x^4\right )^{5/2}}{7 b x^{17/2}} \]

Antiderivative was successfully verified.

[In]  Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^(15/2),x]

[Out]

(4*c*(7*b*B + 3*A*c)*Sqrt[b*x^2 + c*x^4])/(21*b*Sqrt[x]) - (2*(7*b*B + 3*A*c)*(b
*x^2 + c*x^4)^(3/2))/(21*b*x^(9/2)) - (2*A*(b*x^2 + c*x^4)^(5/2))/(7*b*x^(17/2))
+ (4*c^(3/4)*(7*b*B + 3*A*c)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b]
+ Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(21*b^(1/4)
*Sqrt[b*x^2 + c*x^4])

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Rubi in Sympy [A]  time = 45.6746, size = 197, normalized size = 0.97 \[ - \frac{2 A \left (b x^{2} + c x^{4}\right )^{\frac{5}{2}}}{7 b x^{\frac{17}{2}}} + \frac{4 c \left (3 A c + 7 B b\right ) \sqrt{b x^{2} + c x^{4}}}{21 b \sqrt{x}} - \frac{2 \left (3 A c + 7 B b\right ) \left (b x^{2} + c x^{4}\right )^{\frac{3}{2}}}{21 b x^{\frac{9}{2}}} + \frac{4 c^{\frac{3}{4}} \sqrt{\frac{b + c x^{2}}{\left (\sqrt{b} + \sqrt{c} x\right )^{2}}} \left (\sqrt{b} + \sqrt{c} x\right ) \left (3 A c + 7 B b\right ) \sqrt{b x^{2} + c x^{4}} F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}} \right )}\middle | \frac{1}{2}\right )}{21 \sqrt [4]{b} x \left (b + c x^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((B*x**2+A)*(c*x**4+b*x**2)**(3/2)/x**(15/2),x)

[Out]

-2*A*(b*x**2 + c*x**4)**(5/2)/(7*b*x**(17/2)) + 4*c*(3*A*c + 7*B*b)*sqrt(b*x**2
+ c*x**4)/(21*b*sqrt(x)) - 2*(3*A*c + 7*B*b)*(b*x**2 + c*x**4)**(3/2)/(21*b*x**(
9/2)) + 4*c**(3/4)*sqrt((b + c*x**2)/(sqrt(b) + sqrt(c)*x)**2)*(sqrt(b) + sqrt(c
)*x)*(3*A*c + 7*B*b)*sqrt(b*x**2 + c*x**4)*elliptic_f(2*atan(c**(1/4)*sqrt(x)/b*
*(1/4)), 1/2)/(21*b**(1/4)*x*(b + c*x**2))

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Mathematica [C]  time = 0.602382, size = 139, normalized size = 0.68 \[ \frac{2}{21} \sqrt{x^2 \left (b+c x^2\right )} \left (\frac{7 B x^2 \left (c x^2-b\right )-3 A \left (b+3 c x^2\right )}{x^{9/2}}+\frac{4 i c \sqrt{\frac{b}{c x^2}+1} (3 A c+7 b B) F\left (\left .i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{b}}{\sqrt{c}}}}{\sqrt{x}}\right )\right |-1\right )}{\sqrt{\frac{i \sqrt{b}}{\sqrt{c}}} \left (b+c x^2\right )}\right ) \]

Antiderivative was successfully verified.

[In]  Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^(15/2),x]

[Out]

(2*Sqrt[x^2*(b + c*x^2)]*((7*B*x^2*(-b + c*x^2) - 3*A*(b + 3*c*x^2))/x^(9/2) + (
(4*I)*c*(7*b*B + 3*A*c)*Sqrt[1 + b/(c*x^2)]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[b])
/Sqrt[c]]/Sqrt[x]], -1])/(Sqrt[(I*Sqrt[b])/Sqrt[c]]*(b + c*x^2))))/21

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Maple [A]  time = 0.025, size = 254, normalized size = 1.3 \[{\frac{2}{21\, \left ( c{x}^{2}+b \right ) ^{2}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( 6\,A\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ) \sqrt{-bc}{x}^{3}c+14\,B\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ) \sqrt{-bc}{x}^{3}b+7\,B{c}^{2}{x}^{6}-9\,A{c}^{2}{x}^{4}-12\,Abc{x}^{2}-7\,B{b}^{2}{x}^{2}-3\,{b}^{2}A \right ){x}^{-{\frac{13}{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^(15/2),x)

[Out]

2/21*(c*x^4+b*x^2)^(3/2)/x^(13/2)/(c*x^2+b)^2*(6*A*((c*x+(-b*c)^(1/2))/(-b*c)^(1
/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^
(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*(-b*c)^(1/2
)*x^3*c+14*B*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2)
)/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-
b*c)^(1/2))^(1/2),1/2*2^(1/2))*(-b*c)^(1/2)*x^3*b+7*B*c^2*x^6-9*A*c^2*x^4-12*A*b
*c*x^2-7*B*b^2*x^2-3*b^2*A)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (c x^{4} + b x^{2}\right )}^{\frac{3}{2}}{\left (B x^{2} + A\right )}}{x^{\frac{15}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^4 + b*x^2)^(3/2)*(B*x^2 + A)/x^(15/2),x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)*(B*x^2 + A)/x^(15/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (B c x^{4} +{\left (B b + A c\right )} x^{2} + A b\right )} \sqrt{c x^{4} + b x^{2}}}{x^{\frac{11}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^4 + b*x^2)^(3/2)*(B*x^2 + A)/x^(15/2),x, algorithm="fricas")

[Out]

integral((B*c*x^4 + (B*b + A*c)*x^2 + A*b)*sqrt(c*x^4 + b*x^2)/x^(11/2), 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((B*x**2+A)*(c*x**4+b*x**2)**(3/2)/x**(15/2),x)

[Out]

Timed out

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (c x^{4} + b x^{2}\right )}^{\frac{3}{2}}{\left (B x^{2} + A\right )}}{x^{\frac{15}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^4 + b*x^2)^(3/2)*(B*x^2 + A)/x^(15/2),x, algorithm="giac")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)*(B*x^2 + A)/x^(15/2), x)