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

Optimal. Leaf size=194 \[ -\frac{6 i e^{c (a+b x)} \text{Hypergeometric2F1}\left (1,-\frac{i b c}{2 e},1-\frac{i b c}{2 e},-e^{2 i (d+e x)}\right )}{b c}+\frac{12 i e^{c (a+b x)} \text{Hypergeometric2F1}\left (2,-\frac{i b c}{2 e},1-\frac{i b c}{2 e},-e^{2 i (d+e x)}\right )}{b c}-\frac{8 i e^{c (a+b x)} \text{Hypergeometric2F1}\left (3,-\frac{i b c}{2 e},1-\frac{i b c}{2 e},-e^{2 i (d+e x)}\right )}{b c}+\frac{i e^{c (a+b x)}}{b c} \]

[Out]

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

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Rubi [A]  time = 0.198565, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {4442, 2194, 2251} \[ -\frac{6 i e^{c (a+b x)} \, _2F_1\left (1,-\frac{i b c}{2 e};1-\frac{i b c}{2 e};-e^{2 i (d+e x)}\right )}{b c}+\frac{12 i e^{c (a+b x)} \, _2F_1\left (2,-\frac{i b c}{2 e};1-\frac{i b c}{2 e};-e^{2 i (d+e x)}\right )}{b c}-\frac{8 i e^{c (a+b x)} \, _2F_1\left (3,-\frac{i b c}{2 e};1-\frac{i b c}{2 e};-e^{2 i (d+e x)}\right )}{b c}+\frac{i e^{c (a+b x)}}{b c} \]

Antiderivative was successfully verified.

[In]

Int[E^(c*(a + b*x))*Tan[d + e*x]^3,x]

[Out]

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

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]

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])

Rubi steps

\begin{align*} \int e^{c (a+b x)} \tan ^3(d+e x) \, dx &=-\left (i \int \left (-e^{c (a+b x)}+\frac{8 e^{c (a+b x)}}{\left (1+e^{2 i (d+e x)}\right )^3}-\frac{12 e^{c (a+b x)}}{\left (1+e^{2 i (d+e x)}\right )^2}+\frac{6 e^{c (a+b x)}}{1+e^{2 i (d+e x)}}\right ) \, dx\right )\\ &=i \int e^{c (a+b x)} \, dx-6 i \int \frac{e^{c (a+b x)}}{1+e^{2 i (d+e x)}} \, dx-8 i \int \frac{e^{c (a+b x)}}{\left (1+e^{2 i (d+e x)}\right )^3} \, dx+12 i \int \frac{e^{c (a+b x)}}{\left (1+e^{2 i (d+e x)}\right )^2} \, dx\\ &=\frac{i e^{c (a+b x)}}{b c}-\frac{6 i 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{12 i 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{8 i e^{c (a+b x)} \, _2F_1\left (3,-\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]  time = 2.12599, size = 212, normalized size = 1.09 \[ \frac{1}{2} e^{c (a+b x)} \left (\frac{2 e^{2 i d} \left (b^2 c^2-2 e^2\right ) \left (b c e^{2 i e x} \text{Hypergeometric2F1}\left (1,1-\frac{i b c}{2 e},2-\frac{i b c}{2 e},-e^{2 i (d+e x)}\right )-(b c+2 i e) \text{Hypergeometric2F1}\left (1,-\frac{i b c}{2 e},1-\frac{i b c}{2 e},-e^{2 i (d+e x)}\right )\right )}{b c \left (1+e^{2 i d}\right ) e^2 (-2 e+i b c)}-\frac{b c \sec (d) \sin (e x) \sec (d+e x)}{e^2}-\frac{2 \tan (d)}{b c}+\frac{\sec ^2(d+e x)}{e}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.102, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{c \left ( bx+a \right ) }} \left ( \tan \left ( ex+d \right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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))*sin(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))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (e^{\left (b c x + a c\right )} \tan \left (e x + d\right )^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int e^{\left ({\left (b x + a\right )} c\right )} \tan \left (e x + d\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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