\(\int e^{c (a+b x)} \tan ^2(d+e x) \, dx\) [20]

Optimal result

Integrand size = 18, antiderivative size = 130 \[ \int e^{c (a+b x)} \tan ^2(d+e x) \, dx=-\frac {e^{c (a+b x)}}{b c}+\frac {4 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c}{2 e},1-\frac {i b c}{2 e},-e^{2 i (d+e x)}\right )}{b c}-\frac {4 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,-\frac {i b c}{2 e},1-\frac {i b c}{2 e},-e^{2 i (d+e x)}\right )}{b c} \]

[Out]

-exp(c*(b*x+a))/b/c+4*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-4*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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4527, 2225, 2283} \[ \int e^{c (a+b x)} \tan ^2(d+e x) \, dx=\frac {4 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c}{2 e},1-\frac {i b c}{2 e},-e^{2 i (d+e x)}\right )}{b c}-\frac {4 e^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,-\frac {i b c}{2 e},1-\frac {i b c}{2 e},-e^{2 i (d+e x)}\right )}{b c}-\frac {e^{c (a+b x)}}{b c} \]

[In]

Int[E^(c*(a + b*x))*Tan[d + e*x]^2,x]

[Out]

-(E^(c*(a + b*x))/(b*c)) + (4*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) - (4*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)

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2283

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Simp
[a^p*(G^(h*(f + g*x))/(g*h*Log[G]))*Hypergeometric2F1[-p, g*h*(Log[G]/(d*e*Log[F])), g*h*(Log[G]/(d*e*Log[F]))
+ 1, Simplify[(-b/a)*F^(e*(c + d*x))]], x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x] && (ILtQ[p, 0] || G
tQ[a, 0])

Rule 4527

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Tan[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Dist[I^n, Int[ExpandIntegran
d[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]

Rubi steps \begin{align*} \text {integral}& = -\int \left (e^{c (a+b x)}+\frac {4 e^{c (a+b x)}}{\left (1+e^{2 i (d+e x)}\right )^2}-\frac {4 e^{c (a+b x)}}{1+e^{2 i (d+e x)}}\right ) \, dx \\ & = -\left (4 \int \frac {e^{c (a+b x)}}{\left (1+e^{2 i (d+e x)}\right )^2} \, dx\right )+4 \int \frac {e^{c (a+b x)}}{1+e^{2 i (d+e x)}} \, dx-\int e^{c (a+b x)} \, dx \\ & = -\frac {e^{c (a+b x)}}{b c}+\frac {4 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 {4 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} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.15 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.35 \[ \int e^{c (a+b x)} \tan ^2(d+e x) \, dx=e^{c (a+b x)} \left (-\frac {1}{b c}+\frac {2 e^{2 i d} \left (i b c e^{2 i e x} \operatorname {Hypergeometric2F1}\left (1,1-\frac {i b c}{2 e},2-\frac {i b c}{2 e},-e^{2 i (d+e x)}\right )+(-i b c+2 e) \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c}{2 e},1-\frac {i b c}{2 e},-e^{2 i (d+e x)}\right )\right )}{(b c+2 i e) e \left (1+e^{2 i d}\right )}+\frac {\sec (d) \sec (d+e x) \sin (e x)}{e}\right ) \]

[In]

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

[Out]

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

Maple [F]

\[\int {\mathrm e}^{c \left (x b +a \right )} \tan \left (e x +d \right )^{2}d x\]

[In]

int(exp(c*(b*x+a))*tan(e*x+d)^2,x)

[Out]

int(exp(c*(b*x+a))*tan(e*x+d)^2,x)

Fricas [F]

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

[In]

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

[Out]

integral(e^(b*c*x + a*c)*tan(e*x + d)^2, x)

Sympy [F]

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

[In]

integrate(exp(c*(b*x+a))*tan(e*x+d)**2,x)

[Out]

exp(a*c)*Integral(exp(b*c*x)*tan(d + e*x)**2, x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(exp(c*(a + b*x))*tan(d + e*x)^2,x)

[Out]

int(exp(c*(a + b*x))*tan(d + e*x)^2, x)