3.500 \(\int (e x)^m \left (a+b x^3\right )^{5/2} \left (A+B x^3\right ) \, dx\)

Optimal. Leaf size=134 \[ \frac{2 B \left (a+b x^3\right )^{7/2} (e x)^{m+1}}{b e (2 m+23)}-\frac{a^2 \sqrt{a+b x^3} (e x)^{m+1} (2 a B (m+1)-A b (2 m+23)) \, _2F_1\left (-\frac{5}{2},\frac{m+1}{3};\frac{m+4}{3};-\frac{b x^3}{a}\right )}{b e (m+1) (2 m+23) \sqrt{\frac{b x^3}{a}+1}} \]

[Out]

(2*B*(e*x)^(1 + m)*(a + b*x^3)^(7/2))/(b*e*(23 + 2*m)) - (a^2*(2*a*B*(1 + m) - A
*b*(23 + 2*m))*(e*x)^(1 + m)*Sqrt[a + b*x^3]*Hypergeometric2F1[-5/2, (1 + m)/3,
(4 + m)/3, -((b*x^3)/a)])/(b*e*(1 + m)*(23 + 2*m)*Sqrt[1 + (b*x^3)/a])

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Rubi [A]  time = 0.269361, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125 \[ \frac{a^2 \sqrt{a+b x^3} (e x)^{m+1} \left (\frac{A}{m+1}-\frac{2 a B}{2 b m+23 b}\right ) \, _2F_1\left (-\frac{5}{2},\frac{m+1}{3};\frac{m+4}{3};-\frac{b x^3}{a}\right )}{e \sqrt{\frac{b x^3}{a}+1}}+\frac{2 B \left (a+b x^3\right )^{7/2} (e x)^{m+1}}{b e (2 m+23)} \]

Antiderivative was successfully verified.

[In]  Int[(e*x)^m*(a + b*x^3)^(5/2)*(A + B*x^3),x]

[Out]

(2*B*(e*x)^(1 + m)*(a + b*x^3)^(7/2))/(b*e*(23 + 2*m)) + (a^2*(A/(1 + m) - (2*a*
B)/(23*b + 2*b*m))*(e*x)^(1 + m)*Sqrt[a + b*x^3]*Hypergeometric2F1[-5/2, (1 + m)
/3, (4 + m)/3, -((b*x^3)/a)])/(e*Sqrt[1 + (b*x^3)/a])

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Rubi in Sympy [A]  time = 18.8742, size = 112, normalized size = 0.84 \[ \frac{2 B \left (e x\right )^{m + 1} \left (a + b x^{3}\right )^{\frac{7}{2}}}{b e \left (2 m + 23\right )} + \frac{a^{2} \left (e x\right )^{m + 1} \sqrt{a + b x^{3}} \left (A b \left (2 m + 23\right ) - 2 B a \left (m + 1\right )\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{2}, \frac{m}{3} + \frac{1}{3} \\ \frac{m}{3} + \frac{4}{3} \end{matrix}\middle |{- \frac{b x^{3}}{a}} \right )}}{b e \sqrt{1 + \frac{b x^{3}}{a}} \left (m + 1\right ) \left (2 m + 23\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((e*x)**m*(b*x**3+a)**(5/2)*(B*x**3+A),x)

[Out]

2*B*(e*x)**(m + 1)*(a + b*x**3)**(7/2)/(b*e*(2*m + 23)) + a**2*(e*x)**(m + 1)*sq
rt(a + b*x**3)*(A*b*(2*m + 23) - 2*B*a*(m + 1))*hyper((-5/2, m/3 + 1/3), (m/3 +
4/3,), -b*x**3/a)/(b*e*sqrt(1 + b*x**3/a)*(m + 1)*(2*m + 23))

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Mathematica [A]  time = 0.506887, size = 200, normalized size = 1.49 \[ \frac{x \sqrt{a+b x^3} (e x)^m \left (\frac{a^2 A \, _2F_1\left (-\frac{1}{2},\frac{m+1}{3};\frac{m+4}{3};-\frac{b x^3}{a}\right )}{m+1}+\frac{a x^3 (a B+2 A b) \, _2F_1\left (-\frac{1}{2},\frac{m+4}{3};\frac{m+7}{3};-\frac{b x^3}{a}\right )}{m+4}+b x^6 \left (\frac{(2 a B+A b) \, _2F_1\left (-\frac{1}{2},\frac{m+7}{3};\frac{m+10}{3};-\frac{b x^3}{a}\right )}{m+7}+\frac{b B x^3 \, _2F_1\left (-\frac{1}{2},\frac{m+10}{3};\frac{m+13}{3};-\frac{b x^3}{a}\right )}{m+10}\right )\right )}{\sqrt{\frac{b x^3}{a}+1}} \]

Antiderivative was successfully verified.

[In]  Integrate[(e*x)^m*(a + b*x^3)^(5/2)*(A + B*x^3),x]

[Out]

(x*(e*x)^m*Sqrt[a + b*x^3]*((a^2*A*Hypergeometric2F1[-1/2, (1 + m)/3, (4 + m)/3,
-((b*x^3)/a)])/(1 + m) + (a*(2*A*b + a*B)*x^3*Hypergeometric2F1[-1/2, (4 + m)/3
, (7 + m)/3, -((b*x^3)/a)])/(4 + m) + b*x^6*(((A*b + 2*a*B)*Hypergeometric2F1[-1
/2, (7 + m)/3, (10 + m)/3, -((b*x^3)/a)])/(7 + m) + (b*B*x^3*Hypergeometric2F1[-
1/2, (10 + m)/3, (13 + m)/3, -((b*x^3)/a)])/(10 + m))))/Sqrt[1 + (b*x^3)/a]

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Maple [F]  time = 0.034, size = 0, normalized size = 0. \[ \int \left ( ex \right ) ^{m} \left ( b{x}^{3}+a \right ) ^{{\frac{5}{2}}} \left ( B{x}^{3}+A \right ) \, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((e*x)^m*(b*x^3+a)^(5/2)*(B*x^3+A),x)

[Out]

int((e*x)^m*(b*x^3+a)^(5/2)*(B*x^3+A),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int{\left (B x^{3} + A\right )}{\left (b x^{3} + a\right )}^{\frac{5}{2}} \left (e x\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^3 + A)*(b*x^3 + a)^(5/2)*(e*x)^m,x, algorithm="maxima")

[Out]

integrate((B*x^3 + A)*(b*x^3 + a)^(5/2)*(e*x)^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left ({\left (B b^{2} x^{9} +{\left (2 \, B a b + A b^{2}\right )} x^{6} +{\left (B a^{2} + 2 \, A a b\right )} x^{3} + A a^{2}\right )} \sqrt{b x^{3} + a} \left (e x\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^3 + A)*(b*x^3 + a)^(5/2)*(e*x)^m,x, algorithm="fricas")

[Out]

integral((B*b^2*x^9 + (2*B*a*b + A*b^2)*x^6 + (B*a^2 + 2*A*a*b)*x^3 + A*a^2)*sqr
t(b*x^3 + a)*(e*x)^m, 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((e*x)**m*(b*x**3+a)**(5/2)*(B*x**3+A),x)

[Out]

Timed out

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int{\left (B x^{3} + A\right )}{\left (b x^{3} + a\right )}^{\frac{5}{2}} \left (e x\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x^3 + A)*(b*x^3 + a)^(5/2)*(e*x)^m,x, algorithm="giac")

[Out]

integrate((B*x^3 + A)*(b*x^3 + a)^(5/2)*(e*x)^m, x)