3.1211 \(\int \frac {x (a+b \tan ^{-1}(c x))}{(d+e x^2)^{3/2}} \, dx\)

Optimal. Leaf size=71 \[ \frac {b c \tan ^{-1}\left (\frac {x \sqrt {c^2 d-e}}{\sqrt {d+e x^2}}\right )}{e \sqrt {c^2 d-e}}-\frac {a+b \tan ^{-1}(c x)}{e \sqrt {d+e x^2}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4974, 377, 203} \[ \frac {b c \tan ^{-1}\left (\frac {x \sqrt {c^2 d-e}}{\sqrt {d+e x^2}}\right )}{e \sqrt {c^2 d-e}}-\frac {a+b \tan ^{-1}(c x)}{e \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [C]  time = 0.40, size = 210, normalized size = 2.96 \[ -\frac {\frac {2 a}{\sqrt {d+e x^2}}+\frac {i b c \log \left (-\frac {4 i e \left (\sqrt {c^2 d-e} \sqrt {d+e x^2}+c d-i e x\right )}{b (c x+i) \sqrt {c^2 d-e}}\right )}{\sqrt {c^2 d-e}}-\frac {i b c \log \left (\frac {4 i e \left (\sqrt {c^2 d-e} \sqrt {d+e x^2}+c d+i e x\right )}{b (c x-i) \sqrt {c^2 d-e}}\right )}{\sqrt {c^2 d-e}}+\frac {2 b \tan ^{-1}(c x)}{\sqrt {d+e x^2}}}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((2*a)/Sqrt[d + e*x^2] + (2*b*ArcTan[c*x])/Sqrt[d + e*x^2] + (I*b*c*Log[((-4*I)*e*(c*d - I*e*x + Sqrt[c^2
*d - e]*Sqrt[d + e*x^2]))/(b*Sqrt[c^2*d - e]*(I + c*x))])/Sqrt[c^2*d - e] - (I*b*c*Log[((4*I)*e*(c*d + I*e*x +
Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*Sqrt[c^2*d - e]*(-I + c*x))])/Sqrt[c^2*d - e])/e

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fricas [B]  time = 0.53, size = 379, normalized size = 5.34 \[ \left [-\frac {{\left (b c e x^{2} + b c d\right )} \sqrt {-c^{2} d + e} \log \left (\frac {{\left (c^{4} d^{2} - 8 \, c^{2} d e + 8 \, e^{2}\right )} x^{4} - 2 \, {\left (3 \, c^{2} d^{2} - 4 \, d e\right )} x^{2} - 4 \, {\left ({\left (c^{2} d - 2 \, e\right )} x^{3} - d x\right )} \sqrt {-c^{2} d + e} \sqrt {e x^{2} + d} + d^{2}}{c^{4} x^{4} + 2 \, c^{2} x^{2} + 1}\right ) + 4 \, {\left (a c^{2} d - a e + {\left (b c^{2} d - b e\right )} \arctan \left (c x\right )\right )} \sqrt {e x^{2} + d}}{4 \, {\left (c^{2} d^{2} e - d e^{2} + {\left (c^{2} d e^{2} - e^{3}\right )} x^{2}\right )}}, \frac {{\left (b c e x^{2} + b c d\right )} \sqrt {c^{2} d - e} \arctan \left (\frac {\sqrt {c^{2} d - e} {\left ({\left (c^{2} d - 2 \, e\right )} x^{2} - d\right )} \sqrt {e x^{2} + d}}{2 \, {\left ({\left (c^{2} d e - e^{2}\right )} x^{3} + {\left (c^{2} d^{2} - d e\right )} x\right )}}\right ) - 2 \, {\left (a c^{2} d - a e + {\left (b c^{2} d - b e\right )} \arctan \left (c x\right )\right )} \sqrt {e x^{2} + d}}{2 \, {\left (c^{2} d^{2} e - d e^{2} + {\left (c^{2} d e^{2} - e^{3}\right )} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

sage0*x

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maple [F]  time = 0.99, size = 0, normalized size = 0.00 \[ \int \frac {x \left (a +b \arctan \left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x*(a+b*arctan(c*x))/(e*x^2+d)^(3/2),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*(a+b*arctan(c*x))/(e*x^2+d)^(3/2),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 positive, negative or zero?

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

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 \[ \int \frac {x \left (a + b \operatorname {atan}{\left (c x \right )}\right )}{\left (d + e x^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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