\(\int (a+b \tan (c+d \sqrt [3]{x}))^2 \, dx\) [54]

Optimal result

Integrand size = 16, antiderivative size = 206 \[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx=-\frac {3 i b^2 x^{2/3}}{d}+a^2 x+2 i a b x-b^2 x+\frac {6 b^2 \sqrt [3]{x} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac {3 i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {6 i a b \sqrt [3]{x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {3 a b \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {3 b^2 x^{2/3} \tan \left (c+d \sqrt [3]{x}\right )}{d} \]

[Out]

-3*I*b^2*x^(2/3)/d+a^2*x+2*I*a*b*x-b^2*x+6*b^2*x^(1/3)*ln(1+exp(2*I*(c+d*x^(1/3))))/d^2-6*a*b*x^(2/3)*ln(1+exp
(2*I*(c+d*x^(1/3))))/d-3*I*b^2*polylog(2,-exp(2*I*(c+d*x^(1/3))))/d^3+6*I*a*b*x^(1/3)*polylog(2,-exp(2*I*(c+d*
x^(1/3))))/d^2-3*a*b*polylog(3,-exp(2*I*(c+d*x^(1/3))))/d^3+3*b^2*x^(2/3)*tan(c+d*x^(1/3))/d

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.688, Rules used = {3824, 3803, 3800, 2221, 2611, 2320, 6724, 3801, 2317, 2438, 30} \[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx=a^2 x-\frac {3 a b \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {6 i a b \sqrt [3]{x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+2 i a b x-\frac {3 i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {6 b^2 \sqrt [3]{x} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {3 b^2 x^{2/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-\frac {3 i b^2 x^{2/3}}{d}-b^2 x \]

[In]

Int[(a + b*Tan[c + d*x^(1/3)])^2,x]

[Out]

((-3*I)*b^2*x^(2/3))/d + a^2*x + (2*I)*a*b*x - b^2*x + (6*b^2*x^(1/3)*Log[1 + E^((2*I)*(c + d*x^(1/3)))])/d^2
- (6*a*b*x^(2/3)*Log[1 + E^((2*I)*(c + d*x^(1/3)))])/d - ((3*I)*b^2*PolyLog[2, -E^((2*I)*(c + d*x^(1/3)))])/d^
3 + ((6*I)*a*b*x^(1/3)*PolyLog[2, -E^((2*I)*(c + d*x^(1/3)))])/d^2 - (3*a*b*PolyLog[3, -E^((2*I)*(c + d*x^(1/3
)))])/d^3 + (3*b^2*x^(2/3)*Tan[c + d*x^(1/3)])/d

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2438

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 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 3800

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

Rule 3801

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(c + d*x)^m*((b*Tan[e
+ f*x])^(n - 1)/(f*(n - 1))), x] + (-Dist[b*d*(m/(f*(n - 1))), Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1)
, x], x] - Dist[b^2, Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n,
1] && GtQ[m, 0]

Rule 3803

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 3824

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

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps \begin{align*} \text {integral}& = 3 \text {Subst}\left (\int x^2 (a+b \tan (c+d x))^2 \, dx,x,\sqrt [3]{x}\right ) \\ & = 3 \text {Subst}\left (\int \left (a^2 x^2+2 a b x^2 \tan (c+d x)+b^2 x^2 \tan ^2(c+d x)\right ) \, dx,x,\sqrt [3]{x}\right ) \\ & = a^2 x+(6 a b) \text {Subst}\left (\int x^2 \tan (c+d x) \, dx,x,\sqrt [3]{x}\right )+\left (3 b^2\right ) \text {Subst}\left (\int x^2 \tan ^2(c+d x) \, dx,x,\sqrt [3]{x}\right ) \\ & = a^2 x+2 i a b x+\frac {3 b^2 x^{2/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-(12 i a b) \text {Subst}\left (\int \frac {e^{2 i (c+d x)} x^2}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt [3]{x}\right )-\left (3 b^2\right ) \text {Subst}\left (\int x^2 \, dx,x,\sqrt [3]{x}\right )-\frac {\left (6 b^2\right ) \text {Subst}\left (\int x \tan (c+d x) \, dx,x,\sqrt [3]{x}\right )}{d} \\ & = -\frac {3 i b^2 x^{2/3}}{d}+a^2 x+2 i a b x-b^2 x-\frac {6 a b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {3 b^2 x^{2/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}+\frac {(12 a b) \text {Subst}\left (\int x \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d}+\frac {\left (12 i b^2\right ) \text {Subst}\left (\int \frac {e^{2 i (c+d x)} x}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt [3]{x}\right )}{d} \\ & = -\frac {3 i b^2 x^{2/3}}{d}+a^2 x+2 i a b x-b^2 x+\frac {6 b^2 \sqrt [3]{x} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {6 i a b \sqrt [3]{x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {3 b^2 x^{2/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-\frac {(6 i a b) \text {Subst}\left (\int \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^2}-\frac {\left (6 b^2\right ) \text {Subst}\left (\int \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^2} \\ & = -\frac {3 i b^2 x^{2/3}}{d}+a^2 x+2 i a b x-b^2 x+\frac {6 b^2 \sqrt [3]{x} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {6 i a b \sqrt [3]{x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {3 b^2 x^{2/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-\frac {(3 a b) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {\left (3 i b^2\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3} \\ & = -\frac {3 i b^2 x^{2/3}}{d}+a^2 x+2 i a b x-b^2 x+\frac {6 b^2 \sqrt [3]{x} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac {3 i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {6 i a b \sqrt [3]{x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {3 a b \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {3 b^2 x^{2/3} \tan \left (c+d \sqrt [3]{x}\right )}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.60 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.90 \[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx=\frac {b \left (\frac {6 i b d^2 x^{2/3}-4 i a d^3 x}{1+e^{2 i c}}+6 d \left (b-a d \sqrt [3]{x}\right ) \sqrt [3]{x} \log \left (1+e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )+3 i \left (b-2 a d \sqrt [3]{x}\right ) \operatorname {PolyLog}\left (2,-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )-3 a \operatorname {PolyLog}\left (3,-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )\right )}{d^3}+\frac {3 b^2 x^{2/3} \sec (c) \sec \left (c+d \sqrt [3]{x}\right ) \sin \left (d \sqrt [3]{x}\right )}{d}+x \left (a^2-b^2+2 a b \tan (c)\right ) \]

[In]

Integrate[(a + b*Tan[c + d*x^(1/3)])^2,x]

[Out]

(b*(((6*I)*b*d^2*x^(2/3) - (4*I)*a*d^3*x)/(1 + E^((2*I)*c)) + 6*d*(b - a*d*x^(1/3))*x^(1/3)*Log[1 + E^((-2*I)*
(c + d*x^(1/3)))] + (3*I)*(b - 2*a*d*x^(1/3))*PolyLog[2, -E^((-2*I)*(c + d*x^(1/3)))] - 3*a*PolyLog[3, -E^((-2
*I)*(c + d*x^(1/3)))]))/d^3 + (3*b^2*x^(2/3)*Sec[c]*Sec[c + d*x^(1/3)]*Sin[d*x^(1/3)])/d + x*(a^2 - b^2 + 2*a*
b*Tan[c])

Maple [F]

\[\int {\left (a +b \tan \left (c +d \,x^{\frac {1}{3}}\right )\right )}^{2}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.55 \[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx=\frac {6 \, b^{2} d^{2} x^{\frac {2}{3}} \tan \left (d x^{\frac {1}{3}} + c\right ) + 2 \, {\left (a^{2} - b^{2}\right )} d^{3} x - 3 \, a b {\rm polylog}\left (3, \frac {\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 2 i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1}\right ) - 3 \, a b {\rm polylog}\left (3, \frac {\tan \left (d x^{\frac {1}{3}} + c\right )^{2} - 2 i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1}\right ) - 3 \, {\left (2 i \, a b d x^{\frac {1}{3}} - i \, b^{2}\right )} {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1} + 1\right ) - 3 \, {\left (-2 i \, a b d x^{\frac {1}{3}} + i \, b^{2}\right )} {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1} + 1\right ) - 6 \, {\left (a b d^{2} x^{\frac {2}{3}} - b^{2} d x^{\frac {1}{3}}\right )} \log \left (-\frac {2 \, {\left (i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1}\right ) - 6 \, {\left (a b d^{2} x^{\frac {2}{3}} - b^{2} d x^{\frac {1}{3}}\right )} \log \left (-\frac {2 \, {\left (-i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1}\right )}{2 \, d^{3}} \]

[In]

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

[Out]

1/2*(6*b^2*d^2*x^(2/3)*tan(d*x^(1/3) + c) + 2*(a^2 - b^2)*d^3*x - 3*a*b*polylog(3, (tan(d*x^(1/3) + c)^2 + 2*I
*tan(d*x^(1/3) + c) - 1)/(tan(d*x^(1/3) + c)^2 + 1)) - 3*a*b*polylog(3, (tan(d*x^(1/3) + c)^2 - 2*I*tan(d*x^(1
/3) + c) - 1)/(tan(d*x^(1/3) + c)^2 + 1)) - 3*(2*I*a*b*d*x^(1/3) - I*b^2)*dilog(2*(I*tan(d*x^(1/3) + c) - 1)/(
tan(d*x^(1/3) + c)^2 + 1) + 1) - 3*(-2*I*a*b*d*x^(1/3) + I*b^2)*dilog(2*(-I*tan(d*x^(1/3) + c) - 1)/(tan(d*x^(
1/3) + c)^2 + 1) + 1) - 6*(a*b*d^2*x^(2/3) - b^2*d*x^(1/3))*log(-2*(I*tan(d*x^(1/3) + c) - 1)/(tan(d*x^(1/3) +
c)^2 + 1)) - 6*(a*b*d^2*x^(2/3) - b^2*d*x^(1/3))*log(-2*(-I*tan(d*x^(1/3) + c) - 1)/(tan(d*x^(1/3) + c)^2 + 1
)))/d^3

Sympy [F]

\[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx=\int \left (a + b \tan {\left (c + d \sqrt [3]{x} \right )}\right )^{2}\, dx \]

[In]

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

[Out]

Integral((a + b*tan(c + d*x**(1/3)))**2, x)

Maxima [F]

\[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx=\int { {\left (b \tan \left (d x^{\frac {1}{3}} + c\right ) + a\right )}^{2} \,d x } \]

[In]

integrate((a+b*tan(c+d*x^(1/3)))^2,x, algorithm="maxima")

[Out]

a^2*x + (6*b^2*x^(2/3)*sin(2*d*x^(1/3) + 2*c) - (b^2*d*cos(2*d*x^(1/3) + 2*c)^2 + b^2*d*sin(2*d*x^(1/3) + 2*c)
^2 + 2*b^2*d*cos(2*d*x^(1/3) + 2*c) + b^2*d)*x - (d*cos(2*d*x^(1/3) + 2*c)^2 + d*sin(2*d*x^(1/3) + 2*c)^2 + 2*
d*cos(2*d*x^(1/3) + 2*c) + d)*integrate(-4*(a*b*d*x*sin(2*d*x^(1/3) + 2*c) - b^2*x^(2/3)*sin(2*d*x^(1/3) + 2*c
))/((d*cos(2*d*x^(1/3) + 2*c)^2 + d*sin(2*d*x^(1/3) + 2*c)^2 + 2*d*cos(2*d*x^(1/3) + 2*c) + d)*x), x))/(d*cos(
2*d*x^(1/3) + 2*c)^2 + d*sin(2*d*x^(1/3) + 2*c)^2 + 2*d*cos(2*d*x^(1/3) + 2*c) + d)

Giac [F]

\[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx=\int { {\left (b \tan \left (d x^{\frac {1}{3}} + c\right ) + a\right )}^{2} \,d x } \]

[In]

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

[Out]

integrate((b*tan(d*x^(1/3) + c) + a)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx=\int {\left (a+b\,\mathrm {tan}\left (c+d\,x^{1/3}\right )\right )}^2 \,d x \]

[In]

int((a + b*tan(c + d*x^(1/3)))^2,x)

[Out]

int((a + b*tan(c + d*x^(1/3)))^2, x)