3.1201 \(\int \frac{x^3 (a+b \tan ^{-1}(c x))}{\sqrt{d+e x^2}} \, dx\)

Optimal. Leaf size=176 \[ \frac{\left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right )}{3 e^2}-\frac{d \sqrt{d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{e^2}+\frac{b \sqrt{c^2 d-e} \left (2 c^2 d+e\right ) \tan ^{-1}\left (\frac{x \sqrt{c^2 d-e}}{\sqrt{d+e x^2}}\right )}{3 c^3 e^2}+\frac{b \left (3 c^2 d+2 e\right ) \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{6 c^3 e^{3/2}}-\frac{b x \sqrt{d+e x^2}}{6 c e} \]

[Out]

-(b*x*Sqrt[d + e*x^2])/(6*c*e) - (d*Sqrt[d + e*x^2]*(a + b*ArcTan[c*x]))/e^2 + ((d + e*x^2)^(3/2)*(a + b*ArcTa
n[c*x]))/(3*e^2) + (b*Sqrt[c^2*d - e]*(2*c^2*d + e)*ArcTan[(Sqrt[c^2*d - e]*x)/Sqrt[d + e*x^2]])/(3*c^3*e^2) +
(b*(3*c^2*d + 2*e)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(6*c^3*e^(3/2))

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Rubi [A]  time = 0.247594, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 10, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.435, Rules used = {266, 43, 4976, 12, 528, 523, 217, 206, 377, 203} \[ \frac{\left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right )}{3 e^2}-\frac{d \sqrt{d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{e^2}+\frac{b \sqrt{c^2 d-e} \left (2 c^2 d+e\right ) \tan ^{-1}\left (\frac{x \sqrt{c^2 d-e}}{\sqrt{d+e x^2}}\right )}{3 c^3 e^2}+\frac{b \left (3 c^2 d+2 e\right ) \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{6 c^3 e^{3/2}}-\frac{b x \sqrt{d+e x^2}}{6 c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*x*Sqrt[d + e*x^2])/(6*c*e) - (d*Sqrt[d + e*x^2]*(a + b*ArcTan[c*x]))/e^2 + ((d + e*x^2)^(3/2)*(a + b*ArcTa
n[c*x]))/(3*e^2) + (b*Sqrt[c^2*d - e]*(2*c^2*d + e)*ArcTan[(Sqrt[c^2*d - e]*x)/Sqrt[d + e*x^2]])/(3*c^3*e^2) +
(b*(3*c^2*d + 2*e)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(6*c^3*e^(3/2))

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 4976

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^q, x]}, Dist[a + b*ArcTan[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/(1 + c^
2*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && ((IGtQ[q, 0] &&  !(ILtQ[(m - 1)/2, 0] && GtQ[m +
2*q + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[q, 0] && GtQ[m + 2*q + 3, 0])) || (ILtQ[(m + 2*q + 1)/2, 0] &&
!ILtQ[(m - 1)/2, 0]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 528

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

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

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

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^3 \left (a+b \tan ^{-1}(c x)\right )}{\sqrt{d+e x^2}} \, dx &=-\frac{d \sqrt{d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{e^2}+\frac{\left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right )}{3 e^2}-(b c) \int \frac{\left (-2 d+e x^2\right ) \sqrt{d+e x^2}}{3 e^2 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{d \sqrt{d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{e^2}+\frac{\left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right )}{3 e^2}-\frac{(b c) \int \frac{\left (-2 d+e x^2\right ) \sqrt{d+e x^2}}{1+c^2 x^2} \, dx}{3 e^2}\\ &=-\frac{b x \sqrt{d+e x^2}}{6 c e}-\frac{d \sqrt{d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{e^2}+\frac{\left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right )}{3 e^2}-\frac{b \int \frac{-d \left (4 c^2 d+e\right )-e \left (3 c^2 d+2 e\right ) x^2}{\left (1+c^2 x^2\right ) \sqrt{d+e x^2}} \, dx}{6 c e^2}\\ &=-\frac{b x \sqrt{d+e x^2}}{6 c e}-\frac{d \sqrt{d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{e^2}+\frac{\left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right )}{3 e^2}+\frac{\left (b \left (c^2 d-e\right ) \left (2 c^2 d+e\right )\right ) \int \frac{1}{\left (1+c^2 x^2\right ) \sqrt{d+e x^2}} \, dx}{3 c^3 e^2}+\frac{\left (b \left (3 c^2 d+2 e\right )\right ) \int \frac{1}{\sqrt{d+e x^2}} \, dx}{6 c^3 e}\\ &=-\frac{b x \sqrt{d+e x^2}}{6 c e}-\frac{d \sqrt{d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{e^2}+\frac{\left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right )}{3 e^2}+\frac{\left (b \left (c^2 d-e\right ) \left (2 c^2 d+e\right )\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 c^3 e^2}+\frac{\left (b \left (3 c^2 d+2 e\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-e x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{6 c^3 e}\\ &=-\frac{b x \sqrt{d+e x^2}}{6 c e}-\frac{d \sqrt{d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{e^2}+\frac{\left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right )}{3 e^2}+\frac{b \sqrt{c^2 d-e} \left (2 c^2 d+e\right ) \tan ^{-1}\left (\frac{\sqrt{c^2 d-e} x}{\sqrt{d+e x^2}}\right )}{3 c^3 e^2}+\frac{b \left (3 c^2 d+2 e\right ) \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{6 c^3 e^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.484655, size = 377, normalized size = 2.14 \[ \frac{-\frac{\sqrt{d+e x^2} \left (a c \left (4 d-2 e x^2\right )+b e x\right )}{c}-\frac{i b \left (2 c^4 d^2-c^2 d e-e^2\right ) \log \left (\frac{12 i c^4 e^2 \left (\sqrt{c^2 d-e} \sqrt{d+e x^2}+c d-i e x\right )}{b (c x+i) \sqrt{c^2 d-e} \left (-2 c^4 d^2+c^2 d e+e^2\right )}\right )}{c^3 \sqrt{c^2 d-e}}+\frac{i b \left (2 c^4 d^2-c^2 d e-e^2\right ) \log \left (-\frac{12 i c^4 e^2 \left (\sqrt{c^2 d-e} \sqrt{d+e x^2}+c d+i e x\right )}{b (c x-i) \sqrt{c^2 d-e} \left (-2 c^4 d^2+c^2 d e+e^2\right )}\right )}{c^3 \sqrt{c^2 d-e}}+\frac{b \sqrt{e} \left (3 c^2 d+2 e\right ) \log \left (\sqrt{e} \sqrt{d+e x^2}+e x\right )}{c^3}+2 b \tan ^{-1}(c x) \left (e x^2-2 d\right ) \sqrt{d+e x^2}}{6 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((Sqrt[d + e*x^2]*(b*e*x + a*c*(4*d - 2*e*x^2)))/c) + 2*b*(-2*d + e*x^2)*Sqrt[d + e*x^2]*ArcTan[c*x] - (I*b*
(2*c^4*d^2 - c^2*d*e - e^2)*Log[((12*I)*c^4*e^2*(c*d - I*e*x + Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*Sqrt[c^2*d
- e]*(-2*c^4*d^2 + c^2*d*e + e^2)*(I + c*x))])/(c^3*Sqrt[c^2*d - e]) + (I*b*(2*c^4*d^2 - c^2*d*e - e^2)*Log[(
(-12*I)*c^4*e^2*(c*d + I*e*x + Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*Sqrt[c^2*d - e]*(-2*c^4*d^2 + c^2*d*e + e^
2)*(-I + c*x))])/(c^3*Sqrt[c^2*d - e]) + (b*Sqrt[e]*(3*c^2*d + 2*e)*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2]])/c^3)/(
6*e^2)

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Maple [F]  time = 0.831, size = 0, normalized size = 0. \begin{align*} \int{{x}^{3} \left ( a+b\arctan \left ( cx \right ) \right ){\frac{1}{\sqrt{e{x}^{2}+d}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arctan(c*x))/(e*x^2+d)^(1/2),x)

[Out]

int(x^3*(a+b*arctan(c*x))/(e*x^2+d)^(1/2),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^3*(a+b*arctan(c*x))/(e*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 7.63185, size = 1976, normalized size = 11.23 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctan(c*x))/(e*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \left (a + b \operatorname{atan}{\left (c x \right )}\right )}{\sqrt{d + e x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*atan(c*x))/(e*x**2+d)**(1/2),x)

[Out]

Integral(x**3*(a + b*atan(c*x))/sqrt(d + e*x**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^{3}}{\sqrt{e x^{2} + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctan(c*x))/(e*x^2+d)^(1/2),x, algorithm="giac")

[Out]

integrate((b*arctan(c*x) + a)*x^3/sqrt(e*x^2 + d), x)