3.496 \(\int \frac{(e x)^m (A+B x)}{\left (a+c x^2\right )^{3/2}} \, dx\)

Optimal. Leaf size=145 \[ \frac{A \sqrt{\frac{c x^2}{a}+1} (e x)^{m+1} \, _2F_1\left (\frac{3}{2},\frac{m+1}{2};\frac{m+3}{2};-\frac{c x^2}{a}\right )}{a e (m+1) \sqrt{a+c x^2}}+\frac{B \sqrt{\frac{c x^2}{a}+1} (e x)^{m+2} \, _2F_1\left (\frac{3}{2},\frac{m+2}{2};\frac{m+4}{2};-\frac{c x^2}{a}\right )}{a e^2 (m+2) \sqrt{a+c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.210771, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136 \[ \frac{A \sqrt{\frac{c x^2}{a}+1} (e x)^{m+1} \, _2F_1\left (\frac{3}{2},\frac{m+1}{2};\frac{m+3}{2};-\frac{c x^2}{a}\right )}{a e (m+1) \sqrt{a+c x^2}}+\frac{B \sqrt{\frac{c x^2}{a}+1} (e x)^{m+2} \, _2F_1\left (\frac{3}{2},\frac{m+2}{2};\frac{m+4}{2};-\frac{c x^2}{a}\right )}{a e^2 (m+2) \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 21.2893, size = 119, normalized size = 0.82 \[ \frac{A \left (e x\right )^{m + 1} \sqrt{a + c x^{2}}{{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, \frac{m}{2} + \frac{1}{2} \\ \frac{m}{2} + \frac{3}{2} \end{matrix}\middle |{- \frac{c x^{2}}{a}} \right )}}{a^{2} e \sqrt{1 + \frac{c x^{2}}{a}} \left (m + 1\right )} + \frac{B \left (e x\right )^{m + 2} \sqrt{a + c x^{2}}{{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, \frac{m}{2} + 1 \\ \frac{m}{2} + 2 \end{matrix}\middle |{- \frac{c x^{2}}{a}} \right )}}{a^{2} e^{2} \sqrt{1 + \frac{c x^{2}}{a}} \left (m + 2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

A*(e*x)**(m + 1)*sqrt(a + c*x**2)*hyper((3/2, m/2 + 1/2), (m/2 + 3/2,), -c*x**2/
a)/(a**2*e*sqrt(1 + c*x**2/a)*(m + 1)) + B*(e*x)**(m + 2)*sqrt(a + c*x**2)*hyper
((3/2, m/2 + 1), (m/2 + 2,), -c*x**2/a)/(a**2*e**2*sqrt(1 + c*x**2/a)*(m + 2))

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Mathematica [A]  time = 0.147166, size = 111, normalized size = 0.77 \[ \frac{x \sqrt{\frac{c x^2}{a}+1} (e x)^m \left (A (m+2) \, _2F_1\left (\frac{3}{2},\frac{m+1}{2};\frac{m+3}{2};-\frac{c x^2}{a}\right )+B (m+1) x \, _2F_1\left (\frac{3}{2},\frac{m}{2}+1;\frac{m}{2}+2;-\frac{c x^2}{a}\right )\right )}{a (m+1) (m+2) \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

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

[Out]

(x*(e*x)^m*Sqrt[1 + (c*x^2)/a]*(B*(1 + m)*x*Hypergeometric2F1[3/2, 1 + m/2, 2 +
m/2, -((c*x^2)/a)] + A*(2 + m)*Hypergeometric2F1[3/2, (1 + m)/2, (3 + m)/2, -((c
*x^2)/a)]))/(a*(1 + m)*(2 + m)*Sqrt[a + c*x^2])

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((B*x + A)*(e*x)^m/(c*x^2 + a)^(3/2), x)

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Sympy [A]  time = 52.8038, size = 112, normalized size = 0.77 \[ \frac{A e^{m} x x^{m} \Gamma \left (\frac{m}{2} + \frac{1}{2}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, \frac{m}{2} + \frac{1}{2} \\ \frac{m}{2} + \frac{3}{2} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{3}{2}} \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )} + \frac{B e^{m} x^{2} x^{m} \Gamma \left (\frac{m}{2} + 1\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, \frac{m}{2} + 1 \\ \frac{m}{2} + 2 \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{3}{2}} \Gamma \left (\frac{m}{2} + 2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x)**m*(B*x+A)/(c*x**2+a)**(3/2),x)

[Out]

A*e**m*x*x**m*gamma(m/2 + 1/2)*hyper((3/2, m/2 + 1/2), (m/2 + 3/2,), c*x**2*exp_
polar(I*pi)/a)/(2*a**(3/2)*gamma(m/2 + 3/2)) + B*e**m*x**2*x**m*gamma(m/2 + 1)*h
yper((3/2, m/2 + 1), (m/2 + 2,), c*x**2*exp_polar(I*pi)/a)/(2*a**(3/2)*gamma(m/2
+ 2))

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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