3.1185 \(\int x (d+e x^2)^{3/2} (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=181 \[ \frac {\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e}-\frac {b \left (c^2 d-e\right )^{5/2} \tan ^{-1}\left (\frac {x \sqrt {c^2 d-e}}{\sqrt {d+e x^2}}\right )}{5 c^5 e}-\frac {b x \left (7 c^2 d-4 e\right ) \sqrt {d+e x^2}}{40 c^3}-\frac {b \left (15 c^4 d^2-20 c^2 d e+8 e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{40 c^5 \sqrt {e}}-\frac {b x \left (d+e x^2\right )^{3/2}}{20 c} \]

[Out]

-1/20*b*x*(e*x^2+d)^(3/2)/c+1/5*(e*x^2+d)^(5/2)*(a+b*arctan(c*x))/e-1/5*b*(c^2*d-e)^(5/2)*arctan(x*(c^2*d-e)^(
1/2)/(e*x^2+d)^(1/2))/c^5/e-1/40*b*(15*c^4*d^2-20*c^2*d*e+8*e^2)*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/c^5/e^(1/2
)-1/40*b*(7*c^2*d-4*e)*x*(e*x^2+d)^(1/2)/c^3

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Rubi [A]  time = 0.23, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {4974, 416, 528, 523, 217, 206, 377, 203} \[ \frac {\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e}-\frac {b \left (15 c^4 d^2-20 c^2 d e+8 e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{40 c^5 \sqrt {e}}-\frac {b x \left (7 c^2 d-4 e\right ) \sqrt {d+e x^2}}{40 c^3}-\frac {b \left (c^2 d-e\right )^{5/2} \tan ^{-1}\left (\frac {x \sqrt {c^2 d-e}}{\sqrt {d+e x^2}}\right )}{5 c^5 e}-\frac {b x \left (d+e x^2\right )^{3/2}}{20 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(7*c^2*d - 4*e)*x*Sqrt[d + e*x^2])/(40*c^3) - (b*x*(d + e*x^2)^(3/2))/(20*c) + ((d + e*x^2)^(5/2)*(a + b*A
rcTan[c*x]))/(5*e) - (b*(c^2*d - e)^(5/2)*ArcTan[(Sqrt[c^2*d - e]*x)/Sqrt[d + e*x^2]])/(5*c^5*e) - (b*(15*c^4*
d^2 - 20*c^2*d*e + 8*e^2)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(40*c^5*Sqrt[e])

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])

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 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 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 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 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 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]

Rubi steps

\begin {align*} \int x \left (d+e x^2\right )^{3/2} \left (a+b \tan ^{-1}(c x)\right ) \, dx &=\frac {\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e}-\frac {(b c) \int \frac {\left (d+e x^2\right )^{5/2}}{1+c^2 x^2} \, dx}{5 e}\\ &=-\frac {b x \left (d+e x^2\right )^{3/2}}{20 c}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e}-\frac {b \int \frac {\sqrt {d+e x^2} \left (d \left (4 c^2 d-e\right )+\left (7 c^2 d-4 e\right ) e x^2\right )}{1+c^2 x^2} \, dx}{20 c e}\\ &=-\frac {b \left (7 c^2 d-4 e\right ) x \sqrt {d+e x^2}}{40 c^3}-\frac {b x \left (d+e x^2\right )^{3/2}}{20 c}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e}-\frac {b \int \frac {d \left (8 c^4 d^2-9 c^2 d e+4 e^2\right )+e \left (15 c^4 d^2-20 c^2 d e+8 e^2\right ) x^2}{\left (1+c^2 x^2\right ) \sqrt {d+e x^2}} \, dx}{40 c^3 e}\\ &=-\frac {b \left (7 c^2 d-4 e\right ) x \sqrt {d+e x^2}}{40 c^3}-\frac {b x \left (d+e x^2\right )^{3/2}}{20 c}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e}-\frac {\left (b \left (c^2 d-e\right )^3\right ) \int \frac {1}{\left (1+c^2 x^2\right ) \sqrt {d+e x^2}} \, dx}{5 c^5 e}-\frac {\left (b \left (15 c^4 d^2-20 c^2 d e+8 e^2\right )\right ) \int \frac {1}{\sqrt {d+e x^2}} \, dx}{40 c^5}\\ &=-\frac {b \left (7 c^2 d-4 e\right ) x \sqrt {d+e x^2}}{40 c^3}-\frac {b x \left (d+e x^2\right )^{3/2}}{20 c}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e}-\frac {\left (b \left (c^2 d-e\right )^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 )}{5 c^5 e}-\frac {\left (b \left (15 c^4 d^2-20 c^2 d e+8 e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-e x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{40 c^5}\\ &=-\frac {b \left (7 c^2 d-4 e\right ) x \sqrt {d+e x^2}}{40 c^3}-\frac {b x \left (d+e x^2\right )^{3/2}}{20 c}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 e}-\frac {b \left (c^2 d-e\right )^{5/2} \tan ^{-1}\left (\frac {\sqrt {c^2 d-e} x}{\sqrt {d+e x^2}}\right )}{5 c^5 e}-\frac {b \left (15 c^4 d^2-20 c^2 d e+8 e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{40 c^5 \sqrt {e}}\\ \end {align*}

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Mathematica [C]  time = 0.46, size = 313, normalized size = 1.73 \[ \frac {c^2 \sqrt {d+e x^2} \left (8 a c^3 \left (d+e x^2\right )^2+b e x \left (4 e-c^2 \left (9 d+2 e x^2\right )\right )\right )+8 b c^5 \tan ^{-1}(c x) \left (d+e x^2\right )^{5/2}-4 i b \left (c^2 d-e\right )^{5/2} \log \left (\frac {20 c^6 e \left (-i \sqrt {c^2 d-e} \sqrt {d+e x^2}-i c d+e x\right )}{b (c x-i) \left (c^2 d-e\right )^{7/2}}\right )+4 i b \left (c^2 d-e\right )^{5/2} \log \left (\frac {20 c^6 e \left (i \sqrt {c^2 d-e} \sqrt {d+e x^2}+i c d+e x\right )}{b (c x+i) \left (c^2 d-e\right )^{7/2}}\right )-b \sqrt {e} \left (15 c^4 d^2-20 c^2 d e+8 e^2\right ) \log \left (\sqrt {e} \sqrt {d+e x^2}+e x\right )}{40 c^5 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*Sqrt[d + e*x^2]*(8*a*c^3*(d + e*x^2)^2 + b*e*x*(4*e - c^2*(9*d + 2*e*x^2))) + 8*b*c^5*(d + e*x^2)^(5/2)*A
rcTan[c*x] - (4*I)*b*(c^2*d - e)^(5/2)*Log[(20*c^6*e*((-I)*c*d + e*x - I*Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*
(c^2*d - e)^(7/2)*(-I + c*x))] + (4*I)*b*(c^2*d - e)^(5/2)*Log[(20*c^6*e*(I*c*d + e*x + I*Sqrt[c^2*d - e]*Sqrt
[d + e*x^2]))/(b*(c^2*d - e)^(7/2)*(I + c*x))] - b*Sqrt[e]*(15*c^4*d^2 - 20*c^2*d*e + 8*e^2)*Log[e*x + Sqrt[e]
*Sqrt[d + e*x^2]])/(40*c^5*e)

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fricas [A]  time = 3.63, size = 1192, normalized size = 6.59 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/80*((15*b*c^4*d^2 - 20*b*c^2*d*e + 8*b*e^2)*sqrt(e)*log(-2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) + 4*(b*
c^4*d^2 - 2*b*c^2*d*e + b*e^2)*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*(8*a
*c^5*e^2*x^4 + 16*a*c^5*d*e*x^2 - 2*b*c^4*e^2*x^3 + 8*a*c^5*d^2 - (9*b*c^4*d*e - 4*b*c^2*e^2)*x + 8*(b*c^5*e^2
*x^4 + 2*b*c^5*d*e*x^2 + b*c^5*d^2)*arctan(c*x))*sqrt(e*x^2 + d))/(c^5*e), -1/80*(8*(b*c^4*d^2 - 2*b*c^2*d*e +
b*e^2)*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)) - (15*b*c^4*d^2 - 20*b*c^2*d*e + 8*b*e^2)*sqrt(e)*log(-2*e*x^2 + 2*sqrt(e*x^2 + d)*sqr
t(e)*x - d) - 2*(8*a*c^5*e^2*x^4 + 16*a*c^5*d*e*x^2 - 2*b*c^4*e^2*x^3 + 8*a*c^5*d^2 - (9*b*c^4*d*e - 4*b*c^2*e
^2)*x + 8*(b*c^5*e^2*x^4 + 2*b*c^5*d*e*x^2 + b*c^5*d^2)*arctan(c*x))*sqrt(e*x^2 + d))/(c^5*e), 1/40*((15*b*c^4
*d^2 - 20*b*c^2*d*e + 8*b*e^2)*sqrt(-e)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) + 2*(b*c^4*d^2 - 2*b*c^2*d*e + b*e^
2)*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)) + (8*a*c^5*e^2*x^4 + 16*a*c^5*d*e*x^
2 - 2*b*c^4*e^2*x^3 + 8*a*c^5*d^2 - (9*b*c^4*d*e - 4*b*c^2*e^2)*x + 8*(b*c^5*e^2*x^4 + 2*b*c^5*d*e*x^2 + b*c^5
*d^2)*arctan(c*x))*sqrt(e*x^2 + d))/(c^5*e), -1/40*(4*(b*c^4*d^2 - 2*b*c^2*d*e + b*e^2)*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)) - (15*
b*c^4*d^2 - 20*b*c^2*d*e + 8*b*e^2)*sqrt(-e)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) - (8*a*c^5*e^2*x^4 + 16*a*c^5*
d*e*x^2 - 2*b*c^4*e^2*x^3 + 8*a*c^5*d^2 - (9*b*c^4*d*e - 4*b*c^2*e^2)*x + 8*(b*c^5*e^2*x^4 + 2*b*c^5*d*e*x^2 +
b*c^5*d^2)*arctan(c*x))*sqrt(e*x^2 + d))/(c^5*e)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [F]  time = 0.97, size = 0, normalized size = 0.00 \[ \int x \left (e \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \arctan \left (c x \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e-c^2*d>0)', see `assume?` for
more details)Is e-c^2*d zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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