3.715 \(\int \frac{(d+e x)^m}{\left (a+c x^2\right )^3} \, dx\)

Optimal. Leaf size=472 \[ \frac{(d+e x)^{m+1} \left (c d x \left (a e^2 (5-2 m)+3 c d^2\right )+a e \left (a e^2 (3-m)+c d^2 (m+1)\right )\right )}{8 a^2 \left (a+c x^2\right ) \left (a e^2+c d^2\right )^2}+\frac{(d+e x)^{m+1} \left (a \sqrt{c} d e m \left (a e^2 (5-2 m)+3 c d^2\right )-\sqrt{-a} \left (a^2 e^4 \left (m^2-4 m+3\right )+a c d^2 e^2 \left (-m^2-2 m+6\right )+3 c^2 d^4\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{16 a^3 (m+1) \left (\sqrt{c} d-\sqrt{-a} e\right ) \left (a e^2+c d^2\right )^2}+\frac{(d+e x)^{m+1} \left (\sqrt{-a} \left (a^2 e^4 \left (m^2-4 m+3\right )+a c d^2 e^2 \left (-m^2-2 m+6\right )+3 c^2 d^4\right )+a \sqrt{c} d e m \left (a e^2 (5-2 m)+3 c d^2\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{16 a^3 (m+1) \left (\sqrt{-a} e+\sqrt{c} d\right ) \left (a e^2+c d^2\right )^2}+\frac{(d+e x)^{m+1} (a e+c d x)}{4 a \left (a+c x^2\right )^2 \left (a e^2+c d^2\right )} \]

[Out]

((a*e + c*d*x)*(d + e*x)^(1 + m))/(4*a*(c*d^2 + a*e^2)*(a + c*x^2)^2) + ((d + e*
x)^(1 + m)*(a*e*(a*e^2*(3 - m) + c*d^2*(1 + m)) + c*d*(3*c*d^2 + a*e^2*(5 - 2*m)
)*x))/(8*a^2*(c*d^2 + a*e^2)^2*(a + c*x^2)) + ((a*Sqrt[c]*d*e*(3*c*d^2 + a*e^2*(
5 - 2*m))*m - Sqrt[-a]*(3*c^2*d^4 + a*c*d^2*e^2*(6 - 2*m - m^2) + a^2*e^4*(3 - 4
*m + m^2)))*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (Sqrt[c]*(d + e
*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(16*a^3*(Sqrt[c]*d - Sqrt[-a]*e)*(c*d^2 + a*e^2)
^2*(1 + m)) + ((a*Sqrt[c]*d*e*(3*c*d^2 + a*e^2*(5 - 2*m))*m + Sqrt[-a]*(3*c^2*d^
4 + a*c*d^2*e^2*(6 - 2*m - m^2) + a^2*e^4*(3 - 4*m + m^2)))*(d + e*x)^(1 + m)*Hy
pergeometric2F1[1, 1 + m, 2 + m, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/
(16*a^3*(Sqrt[c]*d + Sqrt[-a]*e)*(c*d^2 + a*e^2)^2*(1 + m))

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Rubi [A]  time = 1.99662, antiderivative size = 472, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235 \[ \frac{(d+e x)^{m+1} \left (c d x \left (a e^2 (5-2 m)+3 c d^2\right )+a e \left (a e^2 (3-m)+c d^2 (m+1)\right )\right )}{8 a^2 \left (a+c x^2\right ) \left (a e^2+c d^2\right )^2}+\frac{(d+e x)^{m+1} \left (a \sqrt{c} d e m \left (a e^2 (5-2 m)+3 c d^2\right )-\sqrt{-a} \left (a^2 e^4 \left (m^2-4 m+3\right )+a c d^2 e^2 \left (-m^2-2 m+6\right )+3 c^2 d^4\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{16 a^3 (m+1) \left (\sqrt{c} d-\sqrt{-a} e\right ) \left (a e^2+c d^2\right )^2}+\frac{(d+e x)^{m+1} \left (\sqrt{-a} \left (a^2 e^4 \left (m^2-4 m+3\right )+a c d^2 e^2 \left (-m^2-2 m+6\right )+3 c^2 d^4\right )+a \sqrt{c} d e m \left (a e^2 (5-2 m)+3 c d^2\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{16 a^3 (m+1) \left (\sqrt{-a} e+\sqrt{c} d\right ) \left (a e^2+c d^2\right )^2}+\frac{(d+e x)^{m+1} (a e+c d x)}{4 a \left (a+c x^2\right )^2 \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

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

[Out]

((a*e + c*d*x)*(d + e*x)^(1 + m))/(4*a*(c*d^2 + a*e^2)*(a + c*x^2)^2) + ((d + e*
x)^(1 + m)*(a*e*(a*e^2*(3 - m) + c*d^2*(1 + m)) + c*d*(3*c*d^2 + a*e^2*(5 - 2*m)
)*x))/(8*a^2*(c*d^2 + a*e^2)^2*(a + c*x^2)) + ((a*Sqrt[c]*d*e*(3*c*d^2 + a*e^2*(
5 - 2*m))*m - Sqrt[-a]*(3*c^2*d^4 + a*c*d^2*e^2*(6 - 2*m - m^2) + a^2*e^4*(3 - 4
*m + m^2)))*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (Sqrt[c]*(d + e
*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(16*a^3*(Sqrt[c]*d - Sqrt[-a]*e)*(c*d^2 + a*e^2)
^2*(1 + m)) + ((a*Sqrt[c]*d*e*(3*c*d^2 + a*e^2*(5 - 2*m))*m + Sqrt[-a]*(3*c^2*d^
4 + a*c*d^2*e^2*(6 - 2*m - m^2) + a^2*e^4*(3 - 4*m + m^2)))*(d + e*x)^(1 + m)*Hy
pergeometric2F1[1, 1 + m, 2 + m, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/
(16*a^3*(Sqrt[c]*d + Sqrt[-a]*e)*(c*d^2 + a*e^2)^2*(1 + m))

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Rubi in Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Timed out

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Mathematica [A]  time = 0.146503, size = 0, normalized size = 0. \[ \int \frac{(d+e x)^m}{\left (a+c x^2\right )^3} \, dx \]

Verification is Not applicable to the result.

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

[Out]

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

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Maple [F]  time = 0.134, size = 0, normalized size = 0. \[ \int{\frac{ \left ( ex+d \right ) ^{m}}{ \left ( c{x}^{2}+a \right ) ^{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((e*x+d)^m/(c*x^2+a)^3,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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