3.39 \(\int (c+d x)^3 (a+b \tan (e+f x)) \, dx\)

Optimal. Leaf size=152 \[ \frac {a (c+d x)^4}{4 d}-\frac {3 b d^2 (c+d x) \text {Li}_3\left (-e^{2 i (e+f x)}\right )}{2 f^3}+\frac {3 i b d (c+d x)^2 \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{2 f^2}-\frac {b (c+d x)^3 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {i b (c+d x)^4}{4 d}-\frac {3 i b d^3 \text {Li}_4\left (-e^{2 i (e+f x)}\right )}{4 f^4} \]

[Out]

1/4*a*(d*x+c)^4/d+1/4*I*b*(d*x+c)^4/d-b*(d*x+c)^3*ln(1+exp(2*I*(f*x+e)))/f+3/2*I*b*d*(d*x+c)^2*polylog(2,-exp(
2*I*(f*x+e)))/f^2-3/2*b*d^2*(d*x+c)*polylog(3,-exp(2*I*(f*x+e)))/f^3-3/4*I*b*d^3*polylog(4,-exp(2*I*(f*x+e)))/
f^4

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Rubi [A]  time = 0.25, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3722, 3719, 2190, 2531, 6609, 2282, 6589} \[ \frac {a (c+d x)^4}{4 d}-\frac {3 b d^2 (c+d x) \text {Li}_3\left (-e^{2 i (e+f x)}\right )}{2 f^3}+\frac {3 i b d (c+d x)^2 \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{2 f^2}-\frac {b (c+d x)^3 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {i b (c+d x)^4}{4 d}-\frac {3 i b d^3 \text {Li}_4\left (-e^{2 i (e+f x)}\right )}{4 f^4} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*(a + b*Tan[e + f*x]),x]

[Out]

(a*(c + d*x)^4)/(4*d) + ((I/4)*b*(c + d*x)^4)/d - (b*(c + d*x)^3*Log[1 + E^((2*I)*(e + f*x))])/f + (((3*I)/2)*
b*d*(c + d*x)^2*PolyLog[2, -E^((2*I)*(e + f*x))])/f^2 - (3*b*d^2*(c + d*x)*PolyLog[3, -E^((2*I)*(e + f*x))])/(
2*f^3) - (((3*I)/4)*b*d^3*PolyLog[4, -E^((2*I)*(e + f*x))])/f^4

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 2282

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 2531

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 3719

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 3722

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 6589

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]

Rule 6609

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

Rubi steps

\begin {align*} \int (c+d x)^3 (a+b \tan (e+f x)) \, dx &=\int \left (a (c+d x)^3+b (c+d x)^3 \tan (e+f x)\right ) \, dx\\ &=\frac {a (c+d x)^4}{4 d}+b \int (c+d x)^3 \tan (e+f x) \, dx\\ &=\frac {a (c+d x)^4}{4 d}+\frac {i b (c+d x)^4}{4 d}-(2 i b) \int \frac {e^{2 i (e+f x)} (c+d x)^3}{1+e^{2 i (e+f x)}} \, dx\\ &=\frac {a (c+d x)^4}{4 d}+\frac {i b (c+d x)^4}{4 d}-\frac {b (c+d x)^3 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {(3 b d) \int (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right ) \, dx}{f}\\ &=\frac {a (c+d x)^4}{4 d}+\frac {i b (c+d x)^4}{4 d}-\frac {b (c+d x)^3 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {3 i b d (c+d x)^2 \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{2 f^2}-\frac {\left (3 i b d^2\right ) \int (c+d x) \text {Li}_2\left (-e^{2 i (e+f x)}\right ) \, dx}{f^2}\\ &=\frac {a (c+d x)^4}{4 d}+\frac {i b (c+d x)^4}{4 d}-\frac {b (c+d x)^3 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {3 i b d (c+d x)^2 \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{2 f^2}-\frac {3 b d^2 (c+d x) \text {Li}_3\left (-e^{2 i (e+f x)}\right )}{2 f^3}+\frac {\left (3 b d^3\right ) \int \text {Li}_3\left (-e^{2 i (e+f x)}\right ) \, dx}{2 f^3}\\ &=\frac {a (c+d x)^4}{4 d}+\frac {i b (c+d x)^4}{4 d}-\frac {b (c+d x)^3 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {3 i b d (c+d x)^2 \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{2 f^2}-\frac {3 b d^2 (c+d x) \text {Li}_3\left (-e^{2 i (e+f x)}\right )}{2 f^3}-\frac {\left (3 i b d^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{2 i (e+f x)}\right )}{4 f^4}\\ &=\frac {a (c+d x)^4}{4 d}+\frac {i b (c+d x)^4}{4 d}-\frac {b (c+d x)^3 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {3 i b d (c+d x)^2 \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{2 f^2}-\frac {3 b d^2 (c+d x) \text {Li}_3\left (-e^{2 i (e+f x)}\right )}{2 f^3}-\frac {3 i b d^3 \text {Li}_4\left (-e^{2 i (e+f x)}\right )}{4 f^4}\\ \end {align*}

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Mathematica [A]  time = 0.36, size = 255, normalized size = 1.68 \[ \frac {1}{4} \left (4 a c^3 x+6 a c^2 d x^2+4 a c d^2 x^3+a d^3 x^4-\frac {4 b c^3 \log (\cos (e+f x))}{f}-\frac {12 b c^2 d x \log \left (1+e^{2 i (e+f x)}\right )}{f}+6 i b c^2 d x^2-\frac {6 b d^2 (c+d x) \text {Li}_3\left (-e^{2 i (e+f x)}\right )}{f^3}-\frac {12 b c d^2 x^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+4 i b c d^2 x^3+\frac {6 i b d (c+d x)^2 \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{f^2}-\frac {3 i b d^3 \text {Li}_4\left (-e^{2 i (e+f x)}\right )}{f^4}-\frac {4 b d^3 x^3 \log \left (1+e^{2 i (e+f x)}\right )}{f}+i b d^3 x^4\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*c^3*x + 6*a*c^2*d*x^2 + (6*I)*b*c^2*d*x^2 + 4*a*c*d^2*x^3 + (4*I)*b*c*d^2*x^3 + a*d^3*x^4 + I*b*d^3*x^4 -
(12*b*c^2*d*x*Log[1 + E^((2*I)*(e + f*x))])/f - (12*b*c*d^2*x^2*Log[1 + E^((2*I)*(e + f*x))])/f - (4*b*d^3*x^
3*Log[1 + E^((2*I)*(e + f*x))])/f - (4*b*c^3*Log[Cos[e + f*x]])/f + ((6*I)*b*d*(c + d*x)^2*PolyLog[2, -E^((2*I
)*(e + f*x))])/f^2 - (6*b*d^2*(c + d*x)*PolyLog[3, -E^((2*I)*(e + f*x))])/f^3 - ((3*I)*b*d^3*PolyLog[4, -E^((2
*I)*(e + f*x))])/f^4)/4

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fricas [C]  time = 0.70, size = 498, normalized size = 3.28 \[ \frac {2 \, a d^{3} f^{4} x^{4} + 8 \, a c d^{2} f^{4} x^{3} + 12 \, a c^{2} d f^{4} x^{2} + 8 \, a c^{3} f^{4} x + 3 i \, b d^{3} {\rm polylog}\left (4, \frac {\tan \left (f x + e\right )^{2} + 2 i \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right ) - 3 i \, b d^{3} {\rm polylog}\left (4, \frac {\tan \left (f x + e\right )^{2} - 2 i \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right ) + {\left (-6 i \, b d^{3} f^{2} x^{2} - 12 i \, b c d^{2} f^{2} x - 6 i \, b c^{2} d f^{2}\right )} {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) + {\left (6 i \, b d^{3} f^{2} x^{2} + 12 i \, b c d^{2} f^{2} x + 6 i \, b c^{2} d f^{2}\right )} {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) - 4 \, {\left (b d^{3} f^{3} x^{3} + 3 \, b c d^{2} f^{3} x^{2} + 3 \, b c^{2} d f^{3} x + b c^{3} f^{3}\right )} \log \left (-\frac {2 \, {\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right ) - 4 \, {\left (b d^{3} f^{3} x^{3} + 3 \, b c d^{2} f^{3} x^{2} + 3 \, b c^{2} d f^{3} x + b c^{3} f^{3}\right )} \log \left (-\frac {2 \, {\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right ) - 6 \, {\left (b d^{3} f x + b c d^{2} f\right )} {\rm polylog}\left (3, \frac {\tan \left (f x + e\right )^{2} + 2 i \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right ) - 6 \, {\left (b d^{3} f x + b c d^{2} f\right )} {\rm polylog}\left (3, \frac {\tan \left (f x + e\right )^{2} - 2 i \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right )}{8 \, f^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*a*d^3*f^4*x^4 + 8*a*c*d^2*f^4*x^3 + 12*a*c^2*d*f^4*x^2 + 8*a*c^3*f^4*x + 3*I*b*d^3*polylog(4, (tan(f*x
+ e)^2 + 2*I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1)) - 3*I*b*d^3*polylog(4, (tan(f*x + e)^2 - 2*I*tan(f*x + e)
- 1)/(tan(f*x + e)^2 + 1)) + (-6*I*b*d^3*f^2*x^2 - 12*I*b*c*d^2*f^2*x - 6*I*b*c^2*d*f^2)*dilog(2*(I*tan(f*x +
e) - 1)/(tan(f*x + e)^2 + 1) + 1) + (6*I*b*d^3*f^2*x^2 + 12*I*b*c*d^2*f^2*x + 6*I*b*c^2*d*f^2)*dilog(2*(-I*ta
n(f*x + e) - 1)/(tan(f*x + e)^2 + 1) + 1) - 4*(b*d^3*f^3*x^3 + 3*b*c*d^2*f^3*x^2 + 3*b*c^2*d*f^3*x + b*c^3*f^3
)*log(-2*(I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1)) - 4*(b*d^3*f^3*x^3 + 3*b*c*d^2*f^3*x^2 + 3*b*c^2*d*f^3*x +
b*c^3*f^3)*log(-2*(-I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1)) - 6*(b*d^3*f*x + b*c*d^2*f)*polylog(3, (tan(f*x
+ e)^2 + 2*I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1)) - 6*(b*d^3*f*x + b*c*d^2*f)*polylog(3, (tan(f*x + e)^2 -
2*I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1)))/f^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)^3*(b*tan(f*x + e) + a), x)

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maple [B]  time = 0.52, size = 481, normalized size = 3.16 \[ \frac {3 a \,c^{2} d \,x^{2}}{2}+a c \,d^{2} x^{3}-i b \,c^{3} x +\frac {i b \,d^{3} x^{4}}{4}+\frac {2 b \,c^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f}-\frac {b \,c^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f}+\frac {6 i b \,c^{2} d e x}{f}+\frac {3 i b c \,d^{2} \polylog \left (2, -{\mathrm e}^{2 i \left (f x +e \right )}\right ) x}{f^{2}}-\frac {6 i b c \,d^{2} e^{2} x}{f^{2}}+a \,c^{3} x +\frac {a \,d^{3} x^{4}}{4}-\frac {3 b c \,d^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) x^{2}}{f}-\frac {3 b \,c^{2} d \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) x}{f}-\frac {6 b \,c^{2} d e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}+\frac {6 b c \,d^{2} e^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}+\frac {3 i b \,c^{2} d \,e^{2}}{f^{2}}-\frac {4 i b c \,d^{2} e^{3}}{f^{3}}+\frac {2 i b \,d^{3} e^{3} x}{f^{3}}+\frac {3 i b \,d^{3} \polylog \left (2, -{\mathrm e}^{2 i \left (f x +e \right )}\right ) x^{2}}{2 f^{2}}+\frac {3 i b \,c^{2} d \polylog \left (2, -{\mathrm e}^{2 i \left (f x +e \right )}\right )}{2 f^{2}}-\frac {2 b \,d^{3} e^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{4}}-\frac {b \,d^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) x^{3}}{f}+\frac {3 i b \,c^{2} d \,x^{2}}{2}+\frac {3 i b \,e^{4} d^{3}}{2 f^{4}}-\frac {3 b c \,d^{2} \polylog \left (3, -{\mathrm e}^{2 i \left (f x +e \right )}\right )}{2 f^{3}}-\frac {3 b \,d^{3} \polylog \left (3, -{\mathrm e}^{2 i \left (f x +e \right )}\right ) x}{2 f^{3}}-\frac {3 i b \,d^{3} \polylog \left (4, -{\mathrm e}^{2 i \left (f x +e \right )}\right )}{4 f^{4}}+i b c \,d^{2} x^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^3*(a+b*tan(f*x+e)),x)

[Out]

-I*b*c^3*x+3/2*a*c^2*d*x^2+I*b*c*d^2*x^3+a*c*d^2*x^3+1/4*I*b*d^3*x^4+2/f*b*c^3*ln(exp(I*(f*x+e)))-1/f*b*c^3*ln
(exp(2*I*(f*x+e))+1)+6*I/f*b*c^2*d*e*x+3*I/f^2*b*c*d^2*polylog(2,-exp(2*I*(f*x+e)))*x-6*I/f^2*b*c*d^2*e^2*x+a*
c^3*x+1/4*a*d^3*x^4-3/2/f^3*b*c*d^2*polylog(3,-exp(2*I*(f*x+e)))-2/f^4*b*d^3*e^3*ln(exp(I*(f*x+e)))-3/f*b*c*d^
2*ln(exp(2*I*(f*x+e))+1)*x^2-3/f*b*c^2*d*ln(exp(2*I*(f*x+e))+1)*x-1/f*b*d^3*ln(exp(2*I*(f*x+e))+1)*x^3-6/f^2*b
*c^2*d*e*ln(exp(I*(f*x+e)))+6/f^3*b*c*d^2*e^2*ln(exp(I*(f*x+e)))+3*I/f^2*b*c^2*d*e^2-4*I/f^3*b*c*d^2*e^3+2*I/f
^3*b*d^3*e^3*x+3/2*I/f^2*b*d^3*polylog(2,-exp(2*I*(f*x+e)))*x^2+3/2*I/f^2*b*c^2*d*polylog(2,-exp(2*I*(f*x+e)))
+3/2*I*b*c^2*d*x^2+3/2*I/f^4*b*e^4*d^3-3/2/f^3*b*d^3*polylog(3,-exp(2*I*(f*x+e)))*x-3/4*I*b*d^3*polylog(4,-exp
(2*I*(f*x+e)))/f^4

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maxima [B]  time = 3.41, size = 665, normalized size = 4.38 \[ \frac {12 \, {\left (f x + e\right )} a c^{3} + \frac {3 \, {\left (f x + e\right )}^{4} a d^{3}}{f^{3}} - \frac {12 \, {\left (f x + e\right )}^{3} a d^{3} e}{f^{3}} + \frac {18 \, {\left (f x + e\right )}^{2} a d^{3} e^{2}}{f^{3}} - \frac {12 \, {\left (f x + e\right )} a d^{3} e^{3}}{f^{3}} + \frac {12 \, {\left (f x + e\right )}^{3} a c d^{2}}{f^{2}} - \frac {36 \, {\left (f x + e\right )}^{2} a c d^{2} e}{f^{2}} + \frac {36 \, {\left (f x + e\right )} a c d^{2} e^{2}}{f^{2}} + \frac {18 \, {\left (f x + e\right )}^{2} a c^{2} d}{f} - \frac {36 \, {\left (f x + e\right )} a c^{2} d e}{f} + 12 \, b c^{3} \log \left (\sec \left (f x + e\right )\right ) - \frac {12 \, b d^{3} e^{3} \log \left (\sec \left (f x + e\right )\right )}{f^{3}} + \frac {36 \, b c d^{2} e^{2} \log \left (\sec \left (f x + e\right )\right )}{f^{2}} - \frac {36 \, b c^{2} d e \log \left (\sec \left (f x + e\right )\right )}{f} - \frac {-3 i \, {\left (f x + e\right )}^{4} b d^{3} + 12 i \, b d^{3} {\rm Li}_{4}(-e^{\left (2 i \, f x + 2 i \, e\right )}) + {\left (12 i \, b d^{3} e - 12 i \, b c d^{2} f\right )} {\left (f x + e\right )}^{3} + {\left (-18 i \, b d^{3} e^{2} + 36 i \, b c d^{2} e f - 18 i \, b c^{2} d f^{2}\right )} {\left (f x + e\right )}^{2} + {\left (16 i \, {\left (f x + e\right )}^{3} b d^{3} + {\left (-36 i \, b d^{3} e + 36 i \, b c d^{2} f\right )} {\left (f x + e\right )}^{2} + {\left (36 i \, b d^{3} e^{2} - 72 i \, b c d^{2} e f + 36 i \, b c^{2} d f^{2}\right )} {\left (f x + e\right )}\right )} \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) + {\left (-24 i \, {\left (f x + e\right )}^{2} b d^{3} - 18 i \, b d^{3} e^{2} + 36 i \, b c d^{2} e f - 18 i \, b c^{2} d f^{2} + {\left (36 i \, b d^{3} e - 36 i \, b c d^{2} f\right )} {\left (f x + e\right )}\right )} {\rm Li}_2\left (-e^{\left (2 i \, f x + 2 i \, e\right )}\right ) + 2 \, {\left (4 \, {\left (f x + e\right )}^{3} b d^{3} - 9 \, {\left (b d^{3} e - b c d^{2} f\right )} {\left (f x + e\right )}^{2} + 9 \, {\left (b d^{3} e^{2} - 2 \, b c d^{2} e f + b c^{2} d f^{2}\right )} {\left (f x + e\right )}\right )} \log \left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) + 6 \, {\left (4 \, {\left (f x + e\right )} b d^{3} - 3 \, b d^{3} e + 3 \, b c d^{2} f\right )} {\rm Li}_{3}(-e^{\left (2 i \, f x + 2 i \, e\right )})}{f^{3}}}{12 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*(f*x + e)*a*c^3 + 3*(f*x + e)^4*a*d^3/f^3 - 12*(f*x + e)^3*a*d^3*e/f^3 + 18*(f*x + e)^2*a*d^3*e^2/f^3
- 12*(f*x + e)*a*d^3*e^3/f^3 + 12*(f*x + e)^3*a*c*d^2/f^2 - 36*(f*x + e)^2*a*c*d^2*e/f^2 + 36*(f*x + e)*a*c*d
^2*e^2/f^2 + 18*(f*x + e)^2*a*c^2*d/f - 36*(f*x + e)*a*c^2*d*e/f + 12*b*c^3*log(sec(f*x + e)) - 12*b*d^3*e^3*l
og(sec(f*x + e))/f^3 + 36*b*c*d^2*e^2*log(sec(f*x + e))/f^2 - 36*b*c^2*d*e*log(sec(f*x + e))/f - (-3*I*(f*x +
e)^4*b*d^3 + 12*I*b*d^3*polylog(4, -e^(2*I*f*x + 2*I*e)) + (12*I*b*d^3*e - 12*I*b*c*d^2*f)*(f*x + e)^3 + (-18*
I*b*d^3*e^2 + 36*I*b*c*d^2*e*f - 18*I*b*c^2*d*f^2)*(f*x + e)^2 + (16*I*(f*x + e)^3*b*d^3 + (-36*I*b*d^3*e + 36
*I*b*c*d^2*f)*(f*x + e)^2 + (36*I*b*d^3*e^2 - 72*I*b*c*d^2*e*f + 36*I*b*c^2*d*f^2)*(f*x + e))*arctan2(sin(2*f*
x + 2*e), cos(2*f*x + 2*e) + 1) + (-24*I*(f*x + e)^2*b*d^3 - 18*I*b*d^3*e^2 + 36*I*b*c*d^2*e*f - 18*I*b*c^2*d*
f^2 + (36*I*b*d^3*e - 36*I*b*c*d^2*f)*(f*x + e))*dilog(-e^(2*I*f*x + 2*I*e)) + 2*(4*(f*x + e)^3*b*d^3 - 9*(b*d
^3*e - b*c*d^2*f)*(f*x + e)^2 + 9*(b*d^3*e^2 - 2*b*c*d^2*e*f + b*c^2*d*f^2)*(f*x + e))*log(cos(2*f*x + 2*e)^2
+ sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1) + 6*(4*(f*x + e)*b*d^3 - 3*b*d^3*e + 3*b*c*d^2*f)*polylog(3, -e
^(2*I*f*x + 2*I*e)))/f^3)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(e + f*x))*(c + d*x)^3,x)

[Out]

int((a + b*tan(e + f*x))*(c + d*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**3*(a+b*tan(f*x+e)),x)

[Out]

Integral((a + b*tan(e + f*x))*(c + d*x)**3, x)

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