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

Optimal. Leaf size=78 \[ \frac{2 i e^{c (a+b x)} \text{Hypergeometric2F1}\left (1,-\frac{i b c}{2 e},1-\frac{i b c}{2 e},-e^{2 i (d+e x)}\right )}{b c}-\frac{i e^{c (a+b x)}}{b c} \]

[Out]

((-I)*E^(c*(a + b*x)))/(b*c) + ((2*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)

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Rubi [A]  time = 0.0756806, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {4442, 2194, 2251} \[ \frac{2 i e^{c (a+b x)} \, _2F_1\left (1,-\frac{i b c}{2 e};1-\frac{i b c}{2 e};-e^{2 i (d+e x)}\right )}{b c}-\frac{i e^{c (a+b x)}}{b c} \]

Antiderivative was successfully verified.

[In]

Int[E^(c*(a + b*x))*Tan[d + e*x],x]

[Out]

((-I)*E^(c*(a + b*x)))/(b*c) + ((2*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)

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

Mathematica [B]  time = 0.46909, size = 166, normalized size = 2.13 \[ \frac{e^{c (a+b x)} \left (2 b c e^{2 i (d+e x)} \text{Hypergeometric2F1}\left (1,1-\frac{i b c}{2 e},2-\frac{i b c}{2 e},-e^{2 i (d+e x)}\right )-(b c+2 i e) \left (2 e^{2 i d} \text{Hypergeometric2F1}\left (1,-\frac{i b c}{2 e},1-\frac{i b c}{2 e},-e^{2 i (d+e x)}\right )-e^{2 i d}+1\right )\right )}{b c \left (1+e^{2 i d}\right ) (-2 e+i b c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(c*(a + b*x))*(2*b*c*E^((2*I)*(d + 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)*(1 - E^((2*I)*d) + 2*E^((2*I)*d)*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)*(1 + E^((2*I)*d)))

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Maple [F]  time = 0.046, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{c \left ( bx+a \right ) }}\tan \left ( ex+d \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [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 )\,{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),x, algorithm="maxima")

[Out]

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

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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 ), 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),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )\,{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),x, algorithm="giac")

[Out]

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