3.1731 \(\int \frac{(d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx\)

Optimal. Leaf size=79 \[ -\frac{e^4 (a+b x) (d+e x)^{m+1} \, _2F_1\left (5,m+1;m+2;\frac{b (d+e x)}{b d-a e}\right )}{(m+1) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^5} \]

[Out]

-((e^4*(a + b*x)*(d + e*x)^(1 + m)*Hypergeometric2F1[5, 1 + m, 2 + m, (b*(d + e*
x))/(b*d - a*e)])/((b*d - a*e)^5*(1 + m)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]))

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Rubi [A]  time = 0.133378, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071 \[ -\frac{e^4 (a+b x) (d+e x)^{m+1} \, _2F_1\left (5,m+1;m+2;\frac{b (d+e x)}{b d-a e}\right )}{(m+1) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^5} \]

Antiderivative was successfully verified.

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

[Out]

-((e^4*(a + b*x)*(d + e*x)^(1 + m)*Hypergeometric2F1[5, 1 + m, 2 + m, (b*(d + e*
x))/(b*d - a*e)])/((b*d - a*e)^5*(1 + m)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]))

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Rubi in Sympy [A]  time = 20.0671, size = 68, normalized size = 0.86 \[ \frac{e^{4} \left (d + e x\right )^{m + 1} \sqrt{a^{2} + 2 a b x + b^{2} x^{2}}{{}_{2}F_{1}\left (\begin{matrix} 5, m + 1 \\ m + 2 \end{matrix}\middle |{\frac{b \left (- d - e x\right )}{a e - b d}} \right )}}{\left (a + b x\right ) \left (m + 1\right ) \left (a e - b d\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

e**4*(d + e*x)**(m + 1)*sqrt(a**2 + 2*a*b*x + b**2*x**2)*hyper((5, m + 1), (m +
2,), b*(-d - e*x)/(a*e - b*d))/((a + b*x)*(m + 1)*(a*e - b*d)**5)

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Mathematica [A]  time = 0.103084, size = 0, normalized size = 0. \[ \int \frac{(d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx \]

Verification is Not applicable to the result.

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

[Out]

Integrate[(d + e*x)^m/(a^2 + 2*a*b*x + b^2*x^2)^(5/2), x]

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Maple [F]  time = 0.138, size = 0, normalized size = 0. \[ \int{ \left ( ex+d \right ) ^{m} \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2} \right ) ^{-{\frac{5}{2}}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

int((e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integrate((e*x + d)^m/(b^2*x^2 + 2*a*b*x + a^2)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (e x + d\right )}^{m}}{{\left (b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((e*x + d)^m/((b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b*x + a^4)*
sqrt(b^2*x^2 + 2*a*b*x + a^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((e*x+d)**m/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integrate((e*x + d)^m/(b^2*x^2 + 2*a*b*x + a^2)^(5/2), x)