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

Optimal. Leaf size=130 \[ \frac {4 e^{c (a+b x)} \, _2F_1\left (1,-\frac {i b c}{2 e};1-\frac {i b c}{2 e};-e^{2 i (d+e x)}\right )}{b c}-\frac {4 e^{c (a+b x)} \, _2F_1\left (2,-\frac {i b c}{2 e};1-\frac {i b c}{2 e};-e^{2 i (d+e x)}\right )}{b c}-\frac {e^{c (a+b x)}}{b c} \]

[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

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Rubi [A]  time = 0.13, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4442, 2194, 2251} \[ \frac {4 e^{c (a+b x)} \, _2F_1\left (1,-\frac {i b c}{2 e};1-\frac {i b c}{2 e};-e^{2 i (d+e x)}\right )}{b c}-\frac {4 e^{c (a+b x)} \, _2F_1\left (2,-\frac {i b c}{2 e};1-\frac {i b c}{2 e};-e^{2 i (d+e x)}\right )}{b c}-\frac {e^{c (a+b x)}}{b c} \]

Antiderivative was successfully verified.

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

Rule 2194

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 2251

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

Rule 4442

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*} \int e^{c (a+b x)} \tan ^2(d+e x) \, dx &=-\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)} \, _2F_1\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)} \, _2F_1\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*}

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Mathematica [A]  time = 1.67, size = 174, normalized size = 1.34 \[ e^{c (a+b x)} \left (\frac {2 i e^{2 i d} \left (b c e^{2 i e x} \, _2F_1\left (1,1-\frac {i b c}{2 e};2-\frac {i b c}{2 e};-e^{2 i (d+e x)}\right )-(b c+2 i e) \, _2F_1\left (1,-\frac {i b c}{2 e};1-\frac {i b c}{2 e};-e^{2 i (d+e x)}\right )\right )}{\left (1+e^{2 i d}\right ) e (b c+2 i e)}-\frac {1}{b c}+\frac {\sec (d) \sin (e x) \sec (d+e x)}{e}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left (e^{\left (b c x + a c\right )} \tan \left (e x + d\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{\left ({\left (b x + a\right )} c\right )} \tan \left (e x + d\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{c \left (b x +a \right )} \left (\tan ^{2}\left (e x +d \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {e \cos \left (2 \, e x + 2 \, d\right )^{2} e^{\left (b c x + a c\right )} - 2 \, b c e^{\left (b c x + a c\right )} \sin \left (2 \, e x + 2 \, d\right ) + e e^{\left (b c x + a c\right )} \sin \left (2 \, e x + 2 \, d\right )^{2} + 2 \, e \cos \left (2 \, e x + 2 \, d\right ) e^{\left (b c x + a c\right )} + e e^{\left (b c x + a c\right )} + \frac {2 \, {\left (b^{2} c^{2} e^{2} \cos \left (2 \, e x + 2 \, d\right )^{2} + b^{2} c^{2} e^{2} \sin \left (2 \, e x + 2 \, d\right )^{2} + 2 \, b^{2} c^{2} e^{2} \cos \left (2 \, e x + 2 \, d\right ) + b^{2} c^{2} e^{2}\right )} e^{\left (a c\right )} \int \frac {e^{\left (b c x\right )} \sin \left (2 \, e x + 2 \, d\right )}{\cos \left (2 \, e x + 2 \, d\right )^{2} + \sin \left (2 \, e x + 2 \, d\right )^{2} + 2 \, \cos \left (2 \, e x + 2 \, d\right ) + 1}\,{d x}}{e^{2}}}{b c e \cos \left (2 \, e x + 2 \, d\right )^{2} + b c e \sin \left (2 \, e x + 2 \, d\right )^{2} + 2 \, b c e \cos \left (2 \, e x + 2 \, d\right ) + b c e} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {e}}^{c\,\left (a+b\,x\right )}\,{\mathrm {tan}\left (d+e\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ e^{a c} \int e^{b c x} \tan ^{2}{\left (d + e x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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