\(\int (c+d x)^3 \tan (a+b x) \, dx\) [210]

Optimal result

Integrand size = 14, antiderivative size = 132 \[ \int (c+d x)^3 \tan (a+b x) \, dx=\frac {i (c+d x)^4}{4 d}-\frac {(c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac {3 i d^3 \operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3800, 2221, 2611, 6744, 2320, 6724} \[ \int (c+d x)^3 \tan (a+b x) \, dx=-\frac {3 i d^3 \operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4}-\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {(c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {i (c+d x)^4}{4 d} \]

[In]

Int[(c + d*x)^3*Tan[a + b*x],x]

[Out]

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

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 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 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 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]

Rule 6744

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.95 \[ \int (c+d x)^3 \tan (a+b x) \, dx=\frac {1}{4} i \left (\frac {(c+d x)^4}{d}+\frac {4 i (c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {3 d \left (2 b^2 (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )+d \left (2 i b (c+d x) \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )-d \operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )\right )\right )}{b^4}\right ) \]

[In]

Integrate[(c + d*x)^3*Tan[a + b*x],x]

[Out]

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 431 vs. \(2 (113 ) = 226\).

Time = 1.22 (sec) , antiderivative size = 432, normalized size of antiderivative = 3.27

method	result	size
risch	\(\frac {6 i d \,c^{2} x a}{b}+\frac {3 i c \,d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (x b +a \right )}\right ) x}{b^{2}}-\frac {6 i c \,d^{2} a^{2} x}{b^{2}}-i c^{3} x -\frac {i c^{4}}{4 d}-\frac {c^{3} \ln \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )}{b}+\frac {2 c^{3} \ln \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b}+\frac {6 c \,d^{2} a^{2} \ln \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b^{3}}-\frac {6 c^{2} d a \ln \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b^{2}}-\frac {3 d \,c^{2} \ln \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right ) x}{b}-\frac {3 c \,d^{2} \ln \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right ) x^{2}}{b}+\frac {3 i d \,c^{2} a^{2}}{b^{2}}+\frac {3 i d^{3} \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (x b +a \right )}\right ) x^{2}}{2 b^{2}}-\frac {4 i c \,d^{2} a^{3}}{b^{3}}+\frac {3 i d \,c^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (x b +a \right )}\right )}{2 b^{2}}+\frac {2 i d^{3} a^{3} x}{b^{3}}+i d^{2} c \,x^{3}+\frac {3 i d \,c^{2} x^{2}}{2}-\frac {3 c \,d^{2} \operatorname {polylog}\left (3, -{\mathrm e}^{2 i \left (x b +a \right )}\right )}{2 b^{3}}-\frac {d^{3} \ln \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right ) x^{3}}{b}-\frac {3 d^{3} \operatorname {polylog}\left (3, -{\mathrm e}^{2 i \left (x b +a \right )}\right ) x}{2 b^{3}}-\frac {3 i d^{3} \operatorname {polylog}\left (4, -{\mathrm e}^{2 i \left (x b +a \right )}\right )}{4 b^{4}}-\frac {2 d^{3} a^{3} \ln \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b^{4}}+\frac {3 i d^{3} a^{4}}{2 b^{4}}+\frac {i d^{3} x^{4}}{4}\)	\(432\)

[In]

int((d*x+c)^3*sec(b*x+a)*sin(b*x+a),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 974 vs. \(2 (109) = 218\).

Time = 0.29 (sec) , antiderivative size = 974, normalized size of antiderivative = 7.38 \[ \int (c+d x)^3 \tan (a+b x) \, dx=\text {Too large to display} \]

[In]

integrate((d*x+c)^3*sec(b*x+a)*sin(b*x+a),x, algorithm="fricas")

[Out]

1/2*(6*I*d^3*polylog(4, I*cos(b*x + a) + sin(b*x + a)) - 6*I*d^3*polylog(4, I*cos(b*x + a) - sin(b*x + a)) - 6
*I*d^3*polylog(4, -I*cos(b*x + a) + sin(b*x + a)) + 6*I*d^3*polylog(4, -I*cos(b*x + a) - sin(b*x + a)) - 3*(I*
b^2*d^3*x^2 + 2*I*b^2*c*d^2*x + I*b^2*c^2*d)*dilog(I*cos(b*x + a) + sin(b*x + a)) - 3*(-I*b^2*d^3*x^2 - 2*I*b^
2*c*d^2*x - I*b^2*c^2*d)*dilog(I*cos(b*x + a) - sin(b*x + a)) - 3*(-I*b^2*d^3*x^2 - 2*I*b^2*c*d^2*x - I*b^2*c^
2*d)*dilog(-I*cos(b*x + a) + sin(b*x + a)) - 3*(I*b^2*d^3*x^2 + 2*I*b^2*c*d^2*x + I*b^2*c^2*d)*dilog(-I*cos(b*
x + a) - sin(b*x + a)) - (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*log(cos(b*x + a) + I*sin(b*x + a)
+ I) - (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*log(cos(b*x + a) - I*sin(b*x + a) + I) - (b^3*d^3*
x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^3)*log(I*cos(b*x + a) + sin(b*x
+ a) + 1) - (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^3)*log(I*co
s(b*x + a) - sin(b*x + a) + 1) - (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3*a*b^2*c^2*d - 3*a^2*b*c*d^
2 + a^3*d^3)*log(-I*cos(b*x + a) + sin(b*x + a) + 1) - (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3*a*b^
2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^3)*log(-I*cos(b*x + a) - sin(b*x + a) + 1) - (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*
b*c*d^2 - a^3*d^3)*log(-cos(b*x + a) + I*sin(b*x + a) + I) - (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^
3)*log(-cos(b*x + a) - I*sin(b*x + a) + I) - 6*(b*d^3*x + b*c*d^2)*polylog(3, I*cos(b*x + a) + sin(b*x + a)) -
6*(b*d^3*x + b*c*d^2)*polylog(3, I*cos(b*x + a) - sin(b*x + a)) - 6*(b*d^3*x + b*c*d^2)*polylog(3, -I*cos(b*x
+ a) + sin(b*x + a)) - 6*(b*d^3*x + b*c*d^2)*polylog(3, -I*cos(b*x + a) - sin(b*x + a)))/b^4

Sympy [F]

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

[In]

integrate((d*x+c)**3*sec(b*x+a)*sin(b*x+a),x)

[Out]

Integral((c + d*x)**3*sin(a + b*x)*sec(a + b*x), x)

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 497 vs. \(2 (109) = 218\).

Time = 0.38 (sec) , antiderivative size = 497, normalized size of antiderivative = 3.77 \[ \int (c+d x)^3 \tan (a+b x) \, dx=-\frac {6 \, c^{3} \log \left (-\sin \left (b x + a\right )^{2} + 1\right ) - \frac {18 \, a c^{2} d \log \left (-\sin \left (b x + a\right )^{2} + 1\right )}{b} + \frac {18 \, a^{2} c d^{2} \log \left (-\sin \left (b x + a\right )^{2} + 1\right )}{b^{2}} - \frac {6 \, a^{3} d^{3} \log \left (-\sin \left (b x + a\right )^{2} + 1\right )}{b^{3}} + \frac {-3 i \, {\left (b x + a\right )}^{4} d^{3} - 12 \, {\left (i \, b c d^{2} - i \, a d^{3}\right )} {\left (b x + a\right )}^{3} + 12 i \, d^{3} {\rm Li}_{4}(-e^{\left (2 i \, b x + 2 i \, a\right )}) - 18 \, {\left (i \, b^{2} c^{2} d - 2 i \, a b c d^{2} + i \, a^{2} d^{3}\right )} {\left (b x + a\right )}^{2} - 4 \, {\left (-4 i \, {\left (b x + a\right )}^{3} d^{3} + 9 \, {\left (-i \, b c d^{2} + i \, a d^{3}\right )} {\left (b x + a\right )}^{2} + 9 \, {\left (-i \, b^{2} c^{2} d + 2 i \, a b c d^{2} - i \, a^{2} d^{3}\right )} {\left (b x + a\right )}\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - 6 \, {\left (3 i \, b^{2} c^{2} d - 6 i \, a b c d^{2} + 4 i \, {\left (b x + a\right )}^{2} d^{3} + 3 i \, a^{2} d^{3} + 6 \, {\left (i \, b c d^{2} - i \, a d^{3}\right )} {\left (b x + a\right )}\right )} {\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 2 \, {\left (4 \, {\left (b x + a\right )}^{3} d^{3} + 9 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}^{2} + 9 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} {\left (b x + a\right )}\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 6 \, {\left (3 \, b c d^{2} + 4 \, {\left (b x + a\right )} d^{3} - 3 \, a d^{3}\right )} {\rm Li}_{3}(-e^{\left (2 i \, b x + 2 i \, a\right )})}{b^{3}}}{12 \, b} \]

[In]

integrate((d*x+c)^3*sec(b*x+a)*sin(b*x+a),x, algorithm="maxima")

[Out]

-1/12*(6*c^3*log(-sin(b*x + a)^2 + 1) - 18*a*c^2*d*log(-sin(b*x + a)^2 + 1)/b + 18*a^2*c*d^2*log(-sin(b*x + a)
^2 + 1)/b^2 - 6*a^3*d^3*log(-sin(b*x + a)^2 + 1)/b^3 + (-3*I*(b*x + a)^4*d^3 - 12*(I*b*c*d^2 - I*a*d^3)*(b*x +
a)^3 + 12*I*d^3*polylog(4, -e^(2*I*b*x + 2*I*a)) - 18*(I*b^2*c^2*d - 2*I*a*b*c*d^2 + I*a^2*d^3)*(b*x + a)^2 -
4*(-4*I*(b*x + a)^3*d^3 + 9*(-I*b*c*d^2 + I*a*d^3)*(b*x + a)^2 + 9*(-I*b^2*c^2*d + 2*I*a*b*c*d^2 - I*a^2*d^3)
*(b*x + a))*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) - 6*(3*I*b^2*c^2*d - 6*I*a*b*c*d^2 + 4*I*(b*x + a)
^2*d^3 + 3*I*a^2*d^3 + 6*(I*b*c*d^2 - I*a*d^3)*(b*x + a))*dilog(-e^(2*I*b*x + 2*I*a)) + 2*(4*(b*x + a)^3*d^3 +
9*(b*c*d^2 - a*d^3)*(b*x + a)^2 + 9*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*(b*x + a))*log(cos(2*b*x + 2*a)^2 + s
in(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) + 6*(3*b*c*d^2 + 4*(b*x + a)*d^3 - 3*a*d^3)*polylog(3, -e^(2*I*b*x
+ 2*I*a)))/b^3)/b

Giac [F]

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

[In]

integrate((d*x+c)^3*sec(b*x+a)*sin(b*x+a),x, algorithm="giac")

[Out]

integrate((d*x + c)^3*sec(b*x + a)*sin(b*x + a), x)

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^3 \tan (a+b x) \, dx=\int \frac {\sin \left (a+b\,x\right )\,{\left (c+d\,x\right )}^3}{\cos \left (a+b\,x\right )} \,d x \]

[In]

int((sin(a + b*x)*(c + d*x)^3)/cos(a + b*x),x)

[Out]

int((sin(a + b*x)*(c + d*x)^3)/cos(a + b*x), x)