\(\int e^{c (a+b x)} \tan ^3(d+e x) \, dx\) [19]

Optimal result

Integrand size = 18, antiderivative size = 194 \[ \int e^{c (a+b x)} \tan ^3(d+e x) \, dx=\frac {i e^{c (a+b x)}}{b c}-\frac {6 i e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c}{2 e},1-\frac {i b c}{2 e},-e^{2 i (d+e x)}\right )}{b c}+\frac {12 i e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,-\frac {i b c}{2 e},1-\frac {i b c}{2 e},-e^{2 i (d+e x)}\right )}{b c}-\frac {8 i e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (3,-\frac {i b c}{2 e},1-\frac {i b c}{2 e},-e^{2 i (d+e x)}\right )}{b c} \] Output:

I*exp(c*(b*x+a))/b/c-6*I*exp(c*(b*x+a))*hypergeom([1, -1/2*I*b*c/e],[1-1/2 
*I*b*c/e],-exp(2*I*(e*x+d)))/b/c+12*I*exp(c*(b*x+a))*hypergeom([2, -1/2*I* 
b*c/e],[1-1/2*I*b*c/e],-exp(2*I*(e*x+d)))/b/c-8*I*exp(c*(b*x+a))*hypergeom 
([3, -1/2*I*b*c/e],[1-1/2*I*b*c/e],-exp(2*I*(e*x+d)))/b/c
 

Mathematica [A] (verified)

Time = 1.44 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.09 \[ \int e^{c (a+b x)} \tan ^3(d+e x) \, dx=\frac {1}{2} e^{c (a+b x)} \left (\frac {2 \left (b^2 c^2-2 e^2\right ) e^{2 i d} \left (b c e^{2 i e x} \operatorname {Hypergeometric2F1}\left (1,1-\frac {i b c}{2 e},2-\frac {i b c}{2 e},-e^{2 i (d+e x)}\right )-(b c+2 i e) \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c}{2 e},1-\frac {i b c}{2 e},-e^{2 i (d+e x)}\right )\right )}{b c (i b c-2 e) e^2 \left (1+e^{2 i d}\right )}+\frac {\sec ^2(d+e x)}{e}-\frac {b c \sec (d) \sec (d+e x) \sin (e x)}{e^2}-\frac {2 \tan (d)}{b c}\right ) \] Input:

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

Output:

(E^(c*(a + b*x))*((2*(b^2*c^2 - 2*e^2)*E^((2*I)*d)*(b*c*E^((2*I)*e*x)*Hype 
rgeometric2F1[1, 1 - ((I/2)*b*c)/e, 2 - ((I/2)*b*c)/e, -E^((2*I)*(d + e*x) 
)] - (b*c + (2*I)*e)*Hypergeometric2F1[1, ((-1/2*I)*b*c)/e, 1 - ((I/2)*b*c 
)/e, -E^((2*I)*(d + e*x))]))/(b*c*(I*b*c - 2*e)*e^2*(1 + E^((2*I)*d))) + S 
ec[d + e*x]^2/e - (b*c*Sec[d]*Sec[d + e*x]*Sin[e*x])/e^2 - (2*Tan[d])/(b*c 
)))/2
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.98, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4942, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int e^{c (a+b x)} \tan ^3(d+e x) \, dx\)

\(\Big \downarrow \) 4942

\(\displaystyle -i \int \left (-e^{c (a+b x)}+\frac {6 e^{c (a+b x)}}{1+e^{2 i (d+e x)}}-\frac {12 e^{c (a+b x)}}{\left (1+e^{2 i (d+e x)}\right )^2}+\frac {8 e^{c (a+b x)}}{\left (1+e^{2 i (d+e x)}\right )^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -i \left (\frac {6 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c}{2 e},1-\frac {i b c}{2 e},-e^{2 i (d+e x)}\right )}{b c}-\frac {12 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,-\frac {i b c}{2 e},1-\frac {i b c}{2 e},-e^{2 i (d+e x)}\right )}{b c}+\frac {8 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (3,-\frac {i b c}{2 e},1-\frac {i b c}{2 e},-e^{2 i (d+e x)}\right )}{b c}-\frac {e^{c (a+b x)}}{b c}\right )\)

Input:

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

Output:

(-I)*(-(E^(c*(a + b*x))/(b*c)) + (6*E^(c*(a + b*x))*Hypergeometric2F1[1, ( 
(-1/2*I)*b*c)/e, 1 - ((I/2)*b*c)/e, -E^((2*I)*(d + e*x))])/(b*c) - (12*E^( 
c*(a + b*x))*Hypergeometric2F1[2, ((-1/2*I)*b*c)/e, 1 - ((I/2)*b*c)/e, -E^ 
((2*I)*(d + e*x))])/(b*c) + (8*E^(c*(a + b*x))*Hypergeometric2F1[3, ((-1/2 
*I)*b*c)/e, 1 - ((I/2)*b*c)/e, -E^((2*I)*(d + e*x))])/(b*c))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4942

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Tan[(d_.) + (e_.)*(x_)]^(n_.), x_Symb 
ol] :> Simp[I^n   Int[ExpandIntegrand[F^(c*(a + b*x))*((1 - E^(2*I*(d + e*x 
)))^n/(1 + E^(2*I*(d + e*x)))^n), x], x], x] /; FreeQ[{F, a, b, c, d, e}, x 
] && IntegerQ[n]
 

Maple [F]

\[\int {\mathrm e}^{c \left (b x +a \right )} \tan \left (e x +d \right )^{3}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

(4*e*cos(2*e*x + 2*d)^2*e^(b*c*x + a*c) - b*c*e^(b*c*x + a*c)*sin(2*e*x + 
2*d) + 4*e*e^(b*c*x + a*c)*sin(2*e*x + 2*d)^2 + 2*e*cos(2*e*x + 2*d)*e^(b* 
c*x + a*c) + (b*c*e^(b*c*x + a*c)*sin(2*e*x + 2*d) + 2*e*cos(2*e*x + 2*d)* 
e^(b*c*x + a*c))*cos(4*e*x + 4*d) + (b^2*c^2*e^4*e^(a*c) - 2*e^6*e^(a*c) + 
 (b^2*c^2*e^4*e^(a*c) - 2*e^6*e^(a*c))*cos(4*e*x + 4*d)^2 + 4*(b^2*c^2*e^4 
*e^(a*c) - 2*e^6*e^(a*c))*cos(2*e*x + 2*d)^2 + (b^2*c^2*e^4*e^(a*c) - 2*e^ 
6*e^(a*c))*sin(4*e*x + 4*d)^2 + 4*(b^2*c^2*e^4*e^(a*c) - 2*e^6*e^(a*c))*si 
n(4*e*x + 4*d)*sin(2*e*x + 2*d) + 4*(b^2*c^2*e^4*e^(a*c) - 2*e^6*e^(a*c))* 
sin(2*e*x + 2*d)^2 + 2*(b^2*c^2*e^4*e^(a*c) - 2*e^6*e^(a*c) + 2*(b^2*c^2*e 
^4*e^(a*c) - 2*e^6*e^(a*c))*cos(2*e*x + 2*d))*cos(4*e*x + 4*d) + 4*(b^2*c^ 
2*e^4*e^(a*c) - 2*e^6*e^(a*c))*cos(2*e*x + 2*d))*integrate(e^(b*c*x)*sin(2 
*e*x + 2*d)/(e^4*cos(2*e*x + 2*d)^2 + e^4*sin(2*e*x + 2*d)^2 + 2*e^4*cos(2 
*e*x + 2*d) + e^4), x) - (b*c*cos(2*e*x + 2*d)*e^(b*c*x + a*c) + b*c*e^(b* 
c*x + a*c) - 2*e*e^(b*c*x + a*c)*sin(2*e*x + 2*d))*sin(4*e*x + 4*d))/(e^2* 
cos(4*e*x + 4*d)^2 + 4*e^2*cos(2*e*x + 2*d)^2 + e^2*sin(4*e*x + 4*d)^2 + 4 
*e^2*sin(4*e*x + 4*d)*sin(2*e*x + 2*d) + 4*e^2*sin(2*e*x + 2*d)^2 + 4*e^2* 
cos(2*e*x + 2*d) + e^2 + 2*(2*e^2*cos(2*e*x + 2*d) + e^2)*cos(4*e*x + 4*d) 
)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int e^{c (a+b x)} \tan ^3(d+e x) \, dx=\int {\mathrm {e}}^{c\,\left (a+b\,x\right )}\,{\mathrm {tan}\left (d+e\,x\right )}^3 \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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