3.1175 \(\int x \sqrt{d+e x^2} (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=140 \[ \frac{\left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right )}{3 e}-\frac{b \left (c^2 d-e\right )^{3/2} \tan ^{-1}\left (\frac{x \sqrt{c^2 d-e}}{\sqrt{d+e x^2}}\right )}{3 c^3 e}-\frac{b \left (3 c^2 d-2 e\right ) \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{6 c^3 \sqrt{e}}-\frac{b x \sqrt{d+e x^2}}{6 c} \]

[Out]

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

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Rubi [A]  time = 0.142284, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4974, 416, 523, 217, 206, 377, 203} \[ \frac{\left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right )}{3 e}-\frac{b \left (c^2 d-e\right )^{3/2} \tan ^{-1}\left (\frac{x \sqrt{c^2 d-e}}{\sqrt{d+e x^2}}\right )}{3 c^3 e}-\frac{b \left (3 c^2 d-2 e\right ) \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{6 c^3 \sqrt{e}}-\frac{b x \sqrt{d+e x^2}}{6 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*x*Sqrt[d + e*x^2])/(6*c) + ((d + e*x^2)^(3/2)*(a + b*ArcTan[c*x]))/(3*e) - (b*(c^2*d - e)^(3/2)*ArcTan[(Sq
rt[c^2*d - e]*x)/Sqrt[d + e*x^2]])/(3*c^3*e) - (b*(3*c^2*d - 2*e)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(6*c^3
*Sqrt[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 416

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

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

Mathematica [C]  time = 0.517287, size = 279, normalized size = 1.99 \[ \frac{c^2 \sqrt{d+e x^2} \left (2 a c \left (d+e x^2\right )-b e x\right )-i b \left (c^2 d-e\right )^{3/2} \log \left (\frac{12 c^4 e \left (-i \sqrt{c^2 d-e} \sqrt{d+e x^2}-i c d+e x\right )}{b (c x-i) \left (c^2 d-e\right )^{5/2}}\right )+i b \left (c^2 d-e\right )^{3/2} \log \left (\frac{12 c^4 e \left (i \sqrt{c^2 d-e} \sqrt{d+e x^2}+i c d+e x\right )}{b (c x+i) \left (c^2 d-e\right )^{5/2}}\right )+b \sqrt{e} \left (2 e-3 c^2 d\right ) \log \left (\sqrt{e} \sqrt{d+e x^2}+e x\right )+2 b c^3 \tan ^{-1}(c x) \left (d+e x^2\right )^{3/2}}{6 c^3 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*Sqrt[d + e*x^2]*(-(b*e*x) + 2*a*c*(d + e*x^2)) + 2*b*c^3*(d + e*x^2)^(3/2)*ArcTan[c*x] - I*b*(c^2*d - e)^
(3/2)*Log[(12*c^4*e*((-I)*c*d + e*x - I*Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*(c^2*d - e)^(5/2)*(-I + c*x))] +
I*b*(c^2*d - e)^(3/2)*Log[(12*c^4*e*(I*c*d + e*x + I*Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*(c^2*d - e)^(5/2)*(I
+ c*x))] + b*Sqrt[e]*(-3*c^2*d + 2*e)*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2]])/(6*c^3*e)

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Maple [F]  time = 0.806, size = 0, normalized size = 0. \begin{align*} \int x\sqrt{e{x}^{2}+d} \left ( a+b\arctan \left ( cx \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 6.76812, size = 1947, normalized size = 13.91 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x^2+d)^(1/2)*(a+b*arctan(c*x)),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) + (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 + 2*a*c^3*d - b*c^2*e*x + 2*
(b*c^3*e*x^2 + b*c^3*d)*arctan(c*x))*sqrt(e*x^2 + d))/(c^3*e), -1/12*(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 + 2*a*c^3*d - b*c^2
*e*x + 2*(b*c^3*e*x^2 + b*c^3*d)*arctan(c*x))*sqrt(e*x^2 + d))/(c^3*e), 1/12*(2*(3*b*c^2*d - 2*b*e)*sqrt(-e)*a
rctan(sqrt(-e)*x/sqrt(e*x^2 + d)) - (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 + 2*a*c^3*d - b*c^2*e*x + 2*(b*c^3*e*x^2 + b*c^3*d)*arctan(c*x))*sqrt(e*x^2 +
d))/(c^3*e), -1/6*((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(sqrt(-e)*x/sqrt(e*x^2
+ d)) - (2*a*c^3*e*x^2 + 2*a*c^3*d - b*c^2*e*x + 2*(b*c^3*e*x^2 + b*c^3*d)*arctan(c*x))*sqrt(e*x^2 + d))/(c^3
*e)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \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*(e*x**2+d)**(1/2)*(a+b*atan(c*x)),x)

[Out]

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

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Giac [A]  time = 1.29327, size = 251, normalized size = 1.79 \begin{align*} \frac{1}{3} \,{\left (x^{2} e + d\right )}^{\frac{3}{2}} a e^{\left (-1\right )} + \frac{1}{12} \,{\left (4 \,{\left (x^{2} e + d\right )}^{\frac{3}{2}} \arctan \left (c x\right ) e^{\left (-1\right )} - c{\left (\frac{2 \, \sqrt{x^{2} e + d} x}{c^{2}} - \frac{{\left (3 \, c^{2} d - 2 \, e\right )} e^{\left (-\frac{1}{2}\right )} \log \left ({\left (x e^{\frac{1}{2}} - \sqrt{x^{2} e + d}\right )}^{2}\right )}{c^{4}} - \frac{4 \,{\left (c^{4} d^{2} e^{\frac{1}{2}} - 2 \, c^{2} d e^{\frac{3}{2}} + e^{\frac{5}{2}}\right )} \arctan \left (\frac{{\left ({\left (x e^{\frac{1}{2}} - \sqrt{x^{2} e + d}\right )}^{2} c^{2} - c^{2} d + 2 \, e\right )} e^{\left (-\frac{1}{2}\right )}}{2 \, \sqrt{c^{2} d - e}}\right ) e^{\left (-\frac{3}{2}\right )}}{\sqrt{c^{2} d - e} c^{4}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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