3.13 \(\int \frac {x^3}{a+b \tan (c+d x^2)} \, dx\)

Optimal. Leaf size=122 \[ -\frac {i b \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (d x^2+c\right )}}{(a+i b)^2}\right )}{4 d^2 \left (a^2+b^2\right )}+\frac {b x^2 \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d x^2\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}+\frac {x^4}{4 (a+i b)} \]

[Out]

1/4*x^4/(a+I*b)+1/2*b*x^2*ln(1+(a^2+b^2)*exp(2*I*(d*x^2+c))/(a+I*b)^2)/(a^2+b^2)/d-1/4*I*b*polylog(2,-(a^2+b^2
)*exp(2*I*(d*x^2+c))/(a+I*b)^2)/(a^2+b^2)/d^2

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Rubi [A]  time = 0.21, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {3747, 3732, 2190, 2279, 2391} \[ -\frac {i b \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (d x^2+c\right )}}{(a+i b)^2}\right )}{4 d^2 \left (a^2+b^2\right )}+\frac {b x^2 \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d x^2\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}+\frac {x^4}{4 (a+i b)} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a + b*Tan[c + d*x^2]),x]

[Out]

x^4/(4*(a + I*b)) + (b*x^2*Log[1 + ((a^2 + b^2)*E^((2*I)*(c + d*x^2)))/(a + I*b)^2])/(2*(a^2 + b^2)*d) - ((I/4
)*b*PolyLog[2, -(((a^2 + b^2)*E^((2*I)*(c + d*x^2)))/(a + I*b)^2)])/((a^2 + b^2)*d^2)

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3732

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + d*x)^(m + 1)/(d*
(m + 1)*(a + I*b)), x] + Dist[2*I*b, Int[((c + d*x)^m*E^Simp[2*I*(e + f*x), x])/((a + I*b)^2 + (a^2 + b^2)*E^S
imp[2*I*(e + f*x), x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 + b^2, 0] && IGtQ[m, 0]

Rule 3747

Int[(x_)^(m_.)*((a_.) + (b_.)*Tan[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Tan[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[
(m + 1)/n], 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {x^3}{a+b \tan \left (c+d x^2\right )} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x}{a+b \tan (c+d x)} \, dx,x,x^2\right )\\ &=\frac {x^4}{4 (a+i b)}+(i b) \operatorname {Subst}\left (\int \frac {e^{2 i (c+d x)} x}{(a+i b)^2+\left (a^2+b^2\right ) e^{2 i (c+d x)}} \, dx,x,x^2\right )\\ &=\frac {x^4}{4 (a+i b)}+\frac {b x^2 \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d x^2\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d}-\frac {b \operatorname {Subst}\left (\int \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,x^2\right )}{2 \left (a^2+b^2\right ) d}\\ &=\frac {x^4}{4 (a+i b)}+\frac {b x^2 \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d x^2\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d}+\frac {(i b) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {\left (a^2+b^2\right ) x}{(a+i b)^2}\right )}{x} \, dx,x,e^{2 i \left (c+d x^2\right )}\right )}{4 \left (a^2+b^2\right ) d^2}\\ &=\frac {x^4}{4 (a+i b)}+\frac {b x^2 \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d x^2\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d}-\frac {i b \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d x^2\right )}}{(a+i b)^2}\right )}{4 \left (a^2+b^2\right ) d^2}\\ \end {align*}

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Mathematica [A]  time = 1.83, size = 110, normalized size = 0.90 \[ \frac {i b \text {Li}_2\left (\frac {(-a-i b) e^{-2 i \left (d x^2+c\right )}}{a-i b}\right )+d x^2 \left (2 b \log \left (1+\frac {(a+i b) e^{-2 i \left (c+d x^2\right )}}{a-i b}\right )+d x^2 (a+i b)\right )}{4 d^2 \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*x^2*((a + I*b)*d*x^2 + 2*b*Log[1 + (a + I*b)/((a - I*b)*E^((2*I)*(c + d*x^2)))]) + I*b*PolyLog[2, (-a - I*b
)/((a - I*b)*E^((2*I)*(c + d*x^2)))])/(4*(a^2 + b^2)*d^2)

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fricas [B]  time = 0.72, size = 542, normalized size = 4.44 \[ \frac {2 \, a d^{2} x^{4} - 2 \, b c \log \left (\frac {{\left (i \, a b + b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} - a^{2} + i \, a b + {\left (i \, a^{2} + i \, b^{2}\right )} \tan \left (d x^{2} + c\right )}{\tan \left (d x^{2} + c\right )^{2} + 1}\right ) - 2 \, b c \log \left (\frac {{\left (i \, a b - b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} + a^{2} + i \, a b + {\left (i \, a^{2} + i \, b^{2}\right )} \tan \left (d x^{2} + c\right )}{\tan \left (d x^{2} + c\right )^{2} + 1}\right ) + i \, b {\rm Li}_2\left (\frac {2 \, {\left (i \, a b - b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} - 2 \, a^{2} - 2 i \, a b - {\left (-2 i \, a^{2} + 4 \, a b + 2 i \, b^{2}\right )} \tan \left (d x^{2} + c\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} + a^{2} + b^{2}} + 1\right ) - i \, b {\rm Li}_2\left (\frac {2 \, {\left (-i \, a b - b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} - 2 \, a^{2} + 2 i \, a b - {\left (2 i \, a^{2} + 4 \, a b - 2 i \, b^{2}\right )} \tan \left (d x^{2} + c\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} + a^{2} + b^{2}} + 1\right ) + 2 \, {\left (b d x^{2} + b c\right )} \log \left (-\frac {2 \, {\left (i \, a b - b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} - 2 \, a^{2} - 2 i \, a b - {\left (-2 i \, a^{2} + 4 \, a b + 2 i \, b^{2}\right )} \tan \left (d x^{2} + c\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} + a^{2} + b^{2}}\right ) + 2 \, {\left (b d x^{2} + b c\right )} \log \left (-\frac {2 \, {\left (-i \, a b - b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} - 2 \, a^{2} + 2 i \, a b - {\left (2 i \, a^{2} + 4 \, a b - 2 i \, b^{2}\right )} \tan \left (d x^{2} + c\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} + a^{2} + b^{2}}\right )}{8 \, {\left (a^{2} + b^{2}\right )} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*a*d^2*x^4 - 2*b*c*log(((I*a*b + b^2)*tan(d*x^2 + c)^2 - a^2 + I*a*b + (I*a^2 + I*b^2)*tan(d*x^2 + c))/(
tan(d*x^2 + c)^2 + 1)) - 2*b*c*log(((I*a*b - b^2)*tan(d*x^2 + c)^2 + a^2 + I*a*b + (I*a^2 + I*b^2)*tan(d*x^2 +
c))/(tan(d*x^2 + c)^2 + 1)) + I*b*dilog((2*(I*a*b - b^2)*tan(d*x^2 + c)^2 - 2*a^2 - 2*I*a*b - (-2*I*a^2 + 4*a
*b + 2*I*b^2)*tan(d*x^2 + c))/((a^2 + b^2)*tan(d*x^2 + c)^2 + a^2 + b^2) + 1) - I*b*dilog((2*(-I*a*b - b^2)*ta
n(d*x^2 + c)^2 - 2*a^2 + 2*I*a*b - (2*I*a^2 + 4*a*b - 2*I*b^2)*tan(d*x^2 + c))/((a^2 + b^2)*tan(d*x^2 + c)^2 +
a^2 + b^2) + 1) + 2*(b*d*x^2 + b*c)*log(-(2*(I*a*b - b^2)*tan(d*x^2 + c)^2 - 2*a^2 - 2*I*a*b - (-2*I*a^2 + 4*
a*b + 2*I*b^2)*tan(d*x^2 + c))/((a^2 + b^2)*tan(d*x^2 + c)^2 + a^2 + b^2)) + 2*(b*d*x^2 + b*c)*log(-(2*(-I*a*b
- b^2)*tan(d*x^2 + c)^2 - 2*a^2 + 2*I*a*b - (2*I*a^2 + 4*a*b - 2*I*b^2)*tan(d*x^2 + c))/((a^2 + b^2)*tan(d*x^
2 + c)^2 + a^2 + b^2)))/((a^2 + b^2)*d^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3/(b*tan(d*x^2 + c) + a), x)

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maple [F]  time = 0.91, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{a +b \tan \left (d \,x^{2}+c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.82, size = 267, normalized size = 2.19 \[ \frac {{\left (a - i \, b\right )} d^{2} x^{4} - 2 i \, b d x^{2} \arctan \left (\frac {2 \, a b \cos \left (2 \, d x^{2} + 2 \, c\right ) - {\left (a^{2} - b^{2}\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )}{a^{2} + b^{2}}, \frac {2 \, a b \sin \left (2 \, d x^{2} + 2 \, c\right ) + a^{2} + b^{2} + {\left (a^{2} - b^{2}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right )}{a^{2} + b^{2}}\right ) + b d x^{2} \log \left (\frac {{\left (a^{2} + b^{2}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + 4 \, a b \sin \left (2 \, d x^{2} + 2 \, c\right ) + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} + a^{2} + b^{2} + 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right )}{a^{2} + b^{2}}\right ) - i \, b {\rm Li}_2\left (\frac {{\left (i \, a + b\right )} e^{\left (2 i \, d x^{2} + 2 i \, c\right )}}{-i \, a + b}\right )}{4 \, {\left (a^{2} + b^{2}\right )} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((a - I*b)*d^2*x^4 - 2*I*b*d*x^2*arctan2((2*a*b*cos(2*d*x^2 + 2*c) - (a^2 - b^2)*sin(2*d*x^2 + 2*c))/(a^2
+ b^2), (2*a*b*sin(2*d*x^2 + 2*c) + a^2 + b^2 + (a^2 - b^2)*cos(2*d*x^2 + 2*c))/(a^2 + b^2)) + b*d*x^2*log(((a
^2 + b^2)*cos(2*d*x^2 + 2*c)^2 + 4*a*b*sin(2*d*x^2 + 2*c) + (a^2 + b^2)*sin(2*d*x^2 + 2*c)^2 + a^2 + b^2 + 2*(
a^2 - b^2)*cos(2*d*x^2 + 2*c))/(a^2 + b^2)) - I*b*dilog((I*a + b)*e^(2*I*d*x^2 + 2*I*c)/(-I*a + b)))/((a^2 + b
^2)*d^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a + b*tan(c + d*x^2)),x)

[Out]

int(x^3/(a + b*tan(c + d*x^2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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