3.1.20 \(\int e^{c (a+b x)} \tan ^2(d+e x) \, dx\) [20]
Optimal. Leaf size=130 \[ -\frac {e^{c (a+b x)}}{b c}+\frac {4 e^{c (a+b x)} \, _2F_1\left (1,-\frac {i b c}{2 e};1-\frac {i b c}{2 e};-e^{2 i (d+e x)}\right )}{b c}-\frac {4 e^{c (a+b x)} \, _2F_1\left (2,-\frac {i b c}{2 e};1-\frac {i b c}{2 e};-e^{2 i (d+e x)}\right )}{b c} \]
[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 = 0.09, 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 = {4527, 2225,
2283} \begin {gather*} \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} \end {gather*}
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, ((-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*} \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 = 1.72, size = 174, normalized size = 1.34 \begin {gather*} e^{c (a+b x)} \left (-\frac {1}{b c}+\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 )}{(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 ) \end {gather*}
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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Maple [F]
time = 0.03, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}
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]
-((b^4*c^4 + 20*b^2*c^2*e^2 + 64*e^4)*cos(4*x*e + 4*d)^2*e^(b*c*x + a*c) - 4*(b^4*c^4 + 12*b^2*c^2*e^2 - 64*e^
4)*cos(2*x*e + 2*d)^2*e^(b*c*x + a*c) + (b^4*c^4 + 20*b^2*c^2*e^2 + 64*e^4)*e^(b*c*x + a*c)*sin(4*x*e + 4*d)^2
- 4*(b^4*c^4 + 12*b^2*c^2*e^2 - 64*e^4)*e^(b*c*x + a*c)*sin(2*x*e + 2*d)^2 - 16*(11*b^2*c^2*e^2 - 16*e^4)*cos
(2*x*e + 2*d)*e^(b*c*x + a*c) + 8*(5*b^3*c^3*e - 16*b*c*e^3)*e^(b*c*x + a*c)*sin(2*x*e + 2*d) + 2*(8*(b^2*c^2*
e^2 + 16*e^4)*cos(2*x*e + 2*d)*e^(b*c*x + a*c) + 4*(b^3*c^3*e + 16*b*c*e^3)*e^(b*c*x + a*c)*sin(2*x*e + 2*d) +
(b^4*c^4 - 28*b^2*c^2*e^2 + 64*e^4)*e^(b*c*x + a*c))*cos(4*x*e + 4*d) + (b^4*c^4 - 76*b^2*c^2*e^2 + 64*e^4)*e
^(b*c*x + a*c) - 16*(b^6*c^6 + 20*b^4*c^4*e^2 + 64*b^2*c^2*e^4 + (b^6*c^6 + 20*b^4*c^4*e^2 + 64*b^2*c^2*e^4)*c
os(4*x*e + 4*d)^2 + 4*(b^6*c^6 + 20*b^4*c^4*e^2 + 64*b^2*c^2*e^4)*cos(2*x*e + 2*d)^2 + (b^6*c^6 + 20*b^4*c^4*e
^2 + 64*b^2*c^2*e^4)*sin(4*x*e + 4*d)^2 + 4*(b^6*c^6 + 20*b^4*c^4*e^2 + 64*b^2*c^2*e^4)*sin(4*x*e + 4*d)*sin(2
*x*e + 2*d) + 4*(b^6*c^6 + 20*b^4*c^4*e^2 + 64*b^2*c^2*e^4)*sin(2*x*e + 2*d)^2 + 2*(b^6*c^6 + 20*b^4*c^4*e^2 +
64*b^2*c^2*e^4 + 2*(b^6*c^6 + 20*b^4*c^4*e^2 + 64*b^2*c^2*e^4)*cos(2*x*e + 2*d))*cos(4*x*e + 4*d) + 4*(b^6*c^
6 + 20*b^4*c^4*e^2 + 64*b^2*c^2*e^4)*cos(2*x*e + 2*d))*integrate(-(6*b*c*cos(6*x*e + 6*d)*e^(b*c*x + a*c + 2)
+ 18*b*c*cos(4*x*e + 4*d)*e^(b*c*x + a*c + 2) + 18*b*c*cos(2*x*e + 2*d)*e^(b*c*x + a*c + 2) + 6*b*c*e^(b*c*x +
a*c + 2) - (b^2*c^2*e - 8*e^3)*e^(b*c*x + a*c)*sin(6*x*e + 6*d) - 3*(b^2*c^2*e - 8*e^3)*e^(b*c*x + a*c)*sin(4
*x*e + 4*d) - 3*(b^2*c^2*e - 8*e^3)*e^(b*c*x + a*c)*sin(2*x*e + 2*d))/(b^4*c^4 + 20*b^2*c^2*e^2 + (b^4*c^4 + 2
0*b^2*c^2*e^2 + 64*e^4)*cos(6*x*e + 6*d)^2 + 9*(b^4*c^4 + 20*b^2*c^2*e^2 + 64*e^4)*cos(4*x*e + 4*d)^2 + 9*(b^4
*c^4 + 20*b^2*c^2*e^2 + 64*e^4)*cos(2*x*e + 2*d)^2 + (b^4*c^4 + 20*b^2*c^2*e^2 + 64*e^4)*sin(6*x*e + 6*d)^2 +
9*(b^4*c^4 + 20*b^2*c^2*e^2 + 64*e^4)*sin(4*x*e + 4*d)^2 + 18*(b^4*c^4 + 20*b^2*c^2*e^2 + 64*e^4)*sin(4*x*e +
4*d)*sin(2*x*e + 2*d) + 9*(b^4*c^4 + 20*b^2*c^2*e^2 + 64*e^4)*sin(2*x*e + 2*d)^2 + 2*(b^4*c^4 + 20*b^2*c^2*e^2
+ 3*(b^4*c^4 + 20*b^2*c^2*e^2 + 64*e^4)*cos(4*x*e + 4*d) + 3*(b^4*c^4 + 20*b^2*c^2*e^2 + 64*e^4)*cos(2*x*e +
2*d) + 64*e^4)*cos(6*x*e + 6*d) + 6*(b^4*c^4 + 20*b^2*c^2*e^2 + 3*(b^4*c^4 + 20*b^2*c^2*e^2 + 64*e^4)*cos(2*x*
e + 2*d) + 64*e^4)*cos(4*x*e + 4*d) + 6*(b^4*c^4 + 20*b^2*c^2*e^2 + 64*e^4)*cos(2*x*e + 2*d) + 6*((b^4*c^4 + 2
0*b^2*c^2*e^2 + 64*e^4)*sin(4*x*e + 4*d) + (b^4*c^4 + 20*b^2*c^2*e^2 + 64*e^4)*sin(2*x*e + 2*d))*sin(6*x*e + 6
*d) + 64*e^4), x) - 8*((b^3*c^3*e + 16*b*c*e^3)*cos(2*x*e + 2*d)*e^(b*c*x + a*c) - 2*(b^2*c^2*e^2 + 16*e^4)*e^
(b*c*x + a*c)*sin(2*x*e + 2*d) - 2*(b^3*c^3*e - 8*b*c*e^3)*e^(b*c*x + a*c))*sin(4*x*e + 4*d))/(b^5*c^5 + 20*b^
3*c^3*e^2 + (b^5*c^5 + 20*b^3*c^3*e^2 + 64*b*c*e^4)*cos(4*x*e + 4*d)^2 + 4*(b^5*c^5 + 20*b^3*c^3*e^2 + 64*b*c*
e^4)*cos(2*x*e + 2*d)^2 + 64*b*c*e^4 + (b^5*c^5 + 20*b^3*c^3*e^2 + 64*b*c*e^4)*sin(4*x*e + 4*d)^2 + 4*(b^5*c^5
+ 20*b^3*c^3*e^2 + 64*b*c*e^4)*sin(4*x*e + 4*d)*sin(2*x*e + 2*d) + 4*(b^5*c^5 + 20*b^3*c^3*e^2 + 64*b*c*e^4)*
sin(2*x*e + 2*d)^2 + 2*(b^5*c^5 + 20*b^3*c^3*e^2 + 64*b*c*e^4 + 2*(b^5*c^5 + 20*b^3*c^3*e^2 + 64*b*c*e^4)*cos(
2*x*e + 2*d))*cos(4*x*e + 4*d) + 4*(b^5*c^5 + 20*b^3*c^3*e^2 + 64*b*c*e^4)*cos(2*x*e + 2*d))
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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(x*e + d)^2, x)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} e^{a c} \int e^{b c x} \tan ^{2}{\left (d + e x \right )}\, dx \end {gather*}
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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {e}}^{c\,\left (a+b\,x\right )}\,{\mathrm {tan}\left (d+e\,x\right )}^2 \,d x \end {gather*}
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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