3.1220 \(\int \frac{x (a+b \tan ^{-1}(c x))}{(d+e x^2)^{5/2}} \, dx\)

Optimal. Leaf size=110 \[ -\frac{a+b \tan ^{-1}(c x)}{3 e \left (d+e x^2\right )^{3/2}}-\frac{b c x}{3 d \left (c^2 d-e\right ) \sqrt{d+e x^2}}+\frac{b c^3 \tan ^{-1}\left (\frac{x \sqrt{c^2 d-e}}{\sqrt{d+e x^2}}\right )}{3 e \left (c^2 d-e\right )^{3/2}} \]

[Out]

-(b*c*x)/(3*d*(c^2*d - e)*Sqrt[d + e*x^2]) - (a + b*ArcTan[c*x])/(3*e*(d + e*x^2)^(3/2)) + (b*c^3*ArcTan[(Sqrt
[c^2*d - e]*x)/Sqrt[d + e*x^2]])/(3*(c^2*d - e)^(3/2)*e)

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Rubi [A]  time = 0.093573, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {4974, 382, 377, 203} \[ -\frac{a+b \tan ^{-1}(c x)}{3 e \left (d+e x^2\right )^{3/2}}-\frac{b c x}{3 d \left (c^2 d-e\right ) \sqrt{d+e x^2}}+\frac{b c^3 \tan ^{-1}\left (\frac{x \sqrt{c^2 d-e}}{\sqrt{d+e x^2}}\right )}{3 e \left (c^2 d-e\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*x)/(3*d*(c^2*d - e)*Sqrt[d + e*x^2]) - (a + b*ArcTan[c*x])/(3*e*(d + e*x^2)^(3/2)) + (b*c^3*ArcTan[(Sqrt
[c^2*d - e]*x)/Sqrt[d + e*x^2]])/(3*(c^2*d - e)^(3/2)*e)

Rule 4974

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*(x_)*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[((d + e*x^2)^(q +
1)*(a + b*ArcTan[c*x]))/(2*e*(q + 1)), x] - Dist[(b*c)/(2*e*(q + 1)), Int[(d + e*x^2)^(q + 1)/(1 + c^2*x^2), x
], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[q, -1]

Rule 382

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [C]  time = 0.725135, size = 259, normalized size = 2.35 \[ \frac{1}{6} \left (-\frac{2 a}{e \left (d+e x^2\right )^{3/2}}-\frac{2 b c x}{\left (c^2 d^2-d e\right ) \sqrt{d+e x^2}}-\frac{i b c^3 \log \left (-\frac{12 i e \sqrt{c^2 d-e} \left (\sqrt{c^2 d-e} \sqrt{d+e x^2}+c d-i e x\right )}{b c^2 (c x+i)}\right )}{e \left (c^2 d-e\right )^{3/2}}+\frac{i b c^3 \log \left (\frac{12 i e \sqrt{c^2 d-e} \left (\sqrt{c^2 d-e} \sqrt{d+e x^2}+c d+i e x\right )}{b c^2 (c x-i)}\right )}{e \left (c^2 d-e\right )^{3/2}}-\frac{2 b \tan ^{-1}(c x)}{e \left (d+e x^2\right )^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*a)/(e*(d + e*x^2)^(3/2)) - (2*b*c*x)/((c^2*d^2 - d*e)*Sqrt[d + e*x^2]) - (2*b*ArcTan[c*x])/(e*(d + e*x^2)
^(3/2)) - (I*b*c^3*Log[((-12*I)*Sqrt[c^2*d - e]*e*(c*d - I*e*x + Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*c^2*(I +
c*x))])/((c^2*d - e)^(3/2)*e) + (I*b*c^3*Log[((12*I)*Sqrt[c^2*d - e]*e*(c*d + I*e*x + Sqrt[c^2*d - e]*Sqrt[d
+ e*x^2]))/(b*c^2*(-I + c*x))])/((c^2*d - e)^(3/2)*e))/6

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Maple [F]  time = 0.607, size = 0, normalized size = 0. \begin{align*} \int{x \left ( a+b\arctan \left ( cx \right ) \right ) \left ( e{x}^{2}+d \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 5.24779, size = 1385, normalized size = 12.59 \begin{align*} \left [\frac{{\left (b c^{3} d e^{2} x^{4} + 2 \, b c^{3} d^{2} e x^{2} + b c^{3} d^{3}\right )} \sqrt{-c^{2} d + e} \log \left (\frac{{\left (c^{4} d^{2} - 8 \, c^{2} d e + 8 \, e^{2}\right )} x^{4} - 2 \,{\left (3 \, c^{2} d^{2} - 4 \, d e\right )} x^{2} + 4 \,{\left ({\left (c^{2} d - 2 \, e\right )} x^{3} - d x\right )} \sqrt{-c^{2} d + e} \sqrt{e x^{2} + d} + d^{2}}{c^{4} x^{4} + 2 \, c^{2} x^{2} + 1}\right ) - 4 \,{\left (a c^{4} d^{3} - 2 \, a c^{2} d^{2} e + a d e^{2} +{\left (b c^{3} d e^{2} - b c e^{3}\right )} x^{3} +{\left (b c^{3} d^{2} e - b c d e^{2}\right )} x +{\left (b c^{4} d^{3} - 2 \, b c^{2} d^{2} e + b d e^{2}\right )} \arctan \left (c x\right )\right )} \sqrt{e x^{2} + d}}{12 \,{\left (c^{4} d^{5} e - 2 \, c^{2} d^{4} e^{2} + d^{3} e^{3} +{\left (c^{4} d^{3} e^{3} - 2 \, c^{2} d^{2} e^{4} + d e^{5}\right )} x^{4} + 2 \,{\left (c^{4} d^{4} e^{2} - 2 \, c^{2} d^{3} e^{3} + d^{2} e^{4}\right )} x^{2}\right )}}, \frac{{\left (b c^{3} d e^{2} x^{4} + 2 \, b c^{3} d^{2} e x^{2} + b c^{3} d^{3}\right )} \sqrt{c^{2} d - e} \arctan \left (\frac{\sqrt{c^{2} d - e}{\left ({\left (c^{2} d - 2 \, e\right )} x^{2} - d\right )} \sqrt{e x^{2} + d}}{2 \,{\left ({\left (c^{2} d e - e^{2}\right )} x^{3} +{\left (c^{2} d^{2} - d e\right )} x\right )}}\right ) - 2 \,{\left (a c^{4} d^{3} - 2 \, a c^{2} d^{2} e + a d e^{2} +{\left (b c^{3} d e^{2} - b c e^{3}\right )} x^{3} +{\left (b c^{3} d^{2} e - b c d e^{2}\right )} x +{\left (b c^{4} d^{3} - 2 \, b c^{2} d^{2} e + b d e^{2}\right )} \arctan \left (c x\right )\right )} \sqrt{e x^{2} + d}}{6 \,{\left (c^{4} d^{5} e - 2 \, c^{2} d^{4} e^{2} + d^{3} e^{3} +{\left (c^{4} d^{3} e^{3} - 2 \, c^{2} d^{2} e^{4} + d e^{5}\right )} x^{4} + 2 \,{\left (c^{4} d^{4} e^{2} - 2 \, c^{2} d^{3} e^{3} + d^{2} e^{4}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*((b*c^3*d*e^2*x^4 + 2*b*c^3*d^2*e*x^2 + b*c^3*d^3)*sqrt(-c^2*d + e)*log(((c^4*d^2 - 8*c^2*d*e + 8*e^2)*x
^4 - 2*(3*c^2*d^2 - 4*d*e)*x^2 + 4*((c^2*d - 2*e)*x^3 - d*x)*sqrt(-c^2*d + e)*sqrt(e*x^2 + d) + d^2)/(c^4*x^4
+ 2*c^2*x^2 + 1)) - 4*(a*c^4*d^3 - 2*a*c^2*d^2*e + a*d*e^2 + (b*c^3*d*e^2 - b*c*e^3)*x^3 + (b*c^3*d^2*e - b*c*
d*e^2)*x + (b*c^4*d^3 - 2*b*c^2*d^2*e + b*d*e^2)*arctan(c*x))*sqrt(e*x^2 + d))/(c^4*d^5*e - 2*c^2*d^4*e^2 + d^
3*e^3 + (c^4*d^3*e^3 - 2*c^2*d^2*e^4 + d*e^5)*x^4 + 2*(c^4*d^4*e^2 - 2*c^2*d^3*e^3 + d^2*e^4)*x^2), 1/6*((b*c^
3*d*e^2*x^4 + 2*b*c^3*d^2*e*x^2 + b*c^3*d^3)*sqrt(c^2*d - e)*arctan(1/2*sqrt(c^2*d - e)*((c^2*d - 2*e)*x^2 - d
)*sqrt(e*x^2 + d)/((c^2*d*e - e^2)*x^3 + (c^2*d^2 - d*e)*x)) - 2*(a*c^4*d^3 - 2*a*c^2*d^2*e + a*d*e^2 + (b*c^3
*d*e^2 - b*c*e^3)*x^3 + (b*c^3*d^2*e - b*c*d*e^2)*x + (b*c^4*d^3 - 2*b*c^2*d^2*e + b*d*e^2)*arctan(c*x))*sqrt(
e*x^2 + d))/(c^4*d^5*e - 2*c^2*d^4*e^2 + d^3*e^3 + (c^4*d^3*e^3 - 2*c^2*d^2*e^4 + d*e^5)*x^4 + 2*(c^4*d^4*e^2
- 2*c^2*d^3*e^3 + d^2*e^4)*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \left (a + b \operatorname{atan}{\left (c x \right )}\right )}{\left (d + e x^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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