3.1233 \(\int x^m (d+e x^2)^{5/2} (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=75 \[ \frac{a x^{m+1} \left (d+e x^2\right )^{7/2} \text{Hypergeometric2F1}\left (1,\frac{m+8}{2},\frac{m+3}{2},-\frac{e x^2}{d}\right )}{d (m+1)}+b \text{Unintegrable}\left (x^m \tan ^{-1}(c x) \left (d+e x^2\right )^{5/2},x\right ) \]

[Out]

(a*x^(1 + m)*(d + e*x^2)^(7/2)*Hypergeometric2F1[1, (8 + m)/2, (3 + m)/2, -((e*x^2)/d)])/(d*(1 + m)) + b*Unint
egrable[x^m*(d + e*x^2)^(5/2)*ArcTan[c*x], x]

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Rubi [A]  time = 0.176079, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int x^m \left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

(a*d^2*x^(1 + m)*Sqrt[d + e*x^2]*Hypergeometric2F1[-5/2, (1 + m)/2, (3 + m)/2, -((e*x^2)/d)])/((1 + m)*Sqrt[1
+ (e*x^2)/d]) + b*Defer[Int][x^m*(d + e*x^2)^(5/2)*ArcTan[c*x], x]

Rubi steps

\begin{align*} \int x^m \left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right ) \, dx &=a \int x^m \left (d+e x^2\right )^{5/2} \, dx+b \int x^m \left (d+e x^2\right )^{5/2} \tan ^{-1}(c x) \, dx\\ &=b \int x^m \left (d+e x^2\right )^{5/2} \tan ^{-1}(c x) \, dx+\frac{\left (a d^2 \sqrt{d+e x^2}\right ) \int x^m \left (1+\frac{e x^2}{d}\right )^{5/2} \, dx}{\sqrt{1+\frac{e x^2}{d}}}\\ &=\frac{a d^2 x^{1+m} \sqrt{d+e x^2} \, _2F_1\left (-\frac{5}{2},\frac{1+m}{2};\frac{3+m}{2};-\frac{e x^2}{d}\right )}{(1+m) \sqrt{1+\frac{e x^2}{d}}}+b \int x^m \left (d+e x^2\right )^{5/2} \tan ^{-1}(c x) \, dx\\ \end{align*}

Mathematica [A]  time = 3.94408, size = 0, normalized size = 0. \[ \int x^m \left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [A]  time = 0.615, size = 0, normalized size = 0. \begin{align*} \int{x}^{m} \left ( e{x}^{2}+d \right ) ^{{\frac{5}{2}}} \left ( a+b\arctan \left ( cx \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x^{2} + d\right )}^{\frac{5}{2}}{\left (b \arctan \left (c x\right ) + a\right )} x^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a e^{2} x^{4} + 2 \, a d e x^{2} + a d^{2} +{\left (b e^{2} x^{4} + 2 \, b d e x^{2} + b d^{2}\right )} \arctan \left (c x\right )\right )} \sqrt{e x^{2} + d} x^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*e^2*x^4 + 2*a*d*e*x^2 + a*d^2 + (b*e^2*x^4 + 2*b*d*e*x^2 + b*d^2)*arctan(c*x))*sqrt(e*x^2 + d)*x^m
, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x^{2} + d\right )}^{\frac{5}{2}}{\left (b \arctan \left (c x\right ) + a\right )} x^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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