3.25 \(\int F^{a+b x} \tan (\frac{\pi }{4}+\frac{1}{2} (-c-d x)) \, dx\)

Optimal. Leaf size=76 \[ \frac{i F^{a+b x}}{b \log (F)}-\frac{2 i F^{a+b x} \text{Hypergeometric2F1}\left (1,-\frac{i b \log (F)}{d},1-\frac{i b \log (F)}{d},i e^{i (c+d x)}\right )}{b \log (F)} \]

[Out]

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

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Rubi [A]  time = 0.0935517, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {4464, 4442, 2194, 2251} \[ \frac{i F^{a+b x}}{b \log (F)}-\frac{2 i F^{a+b x} \, _2F_1\left (1,-\frac{i b \log (F)}{d};1-\frac{i b \log (F)}{d};i e^{i (c+d x)}\right )}{b \log (F)} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b*x)*Tan[Pi/4 + (-c - d*x)/2],x]

[Out]

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

Rule 4464

Int[(F_)^((c_.)*(u_))*(G_)[v_]^(n_.), x_Symbol] :> Int[F^(c*ExpandToSum[u, x])*G[ExpandToSum[v, x]]^n, x] /; F
reeQ[{F, c, n}, x] && TrigQ[G] && LinearQ[{u, v}, x] &&  !LinearMatchQ[{u, v}, x]

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 F^{a+b x} \tan \left (\frac{\pi }{4}+\frac{1}{2} (-c-d x)\right ) \, dx &=-\int F^{a+b x} \tan \left (\frac{c}{2}-\frac{\pi }{4}+\frac{d x}{2}\right ) \, dx\\ &=-\left (i \int \left (-F^{a+b x}+\frac{2 F^{a+b x}}{1+e^{2 i \left (\frac{c}{2}-\frac{\pi }{4}+\frac{d x}{2}\right )}}\right ) \, dx\right )\\ &=i \int F^{a+b x} \, dx-2 i \int \frac{F^{a+b x}}{1+e^{2 i \left (\frac{c}{2}-\frac{\pi }{4}+\frac{d x}{2}\right )}} \, dx\\ &=\frac{i F^{a+b x}}{b \log (F)}-\frac{2 i F^{a+b x} \, _2F_1\left (1,-\frac{i b \log (F)}{d};1-\frac{i b \log (F)}{d};i e^{i (c+d x)}\right )}{b \log (F)}\\ \end{align*}

Mathematica [A]  time = 0.317999, size = 133, normalized size = 1.75 \[ \frac{F^{a+b x} \left (b \log (F) e^{i (c+d x)} \text{Hypergeometric2F1}\left (1,1-\frac{i b \log (F)}{d},2-\frac{i b \log (F)}{d},i e^{i (c+d x)}\right )+(d-i b \log (F)) \text{Hypergeometric2F1}\left (1,-\frac{i b \log (F)}{d},1-\frac{i b \log (F)}{d},i e^{i (c+d x)}\right )\right )}{b \log (F) (b \log (F)+i d)} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b*x)*Tan[Pi/4 + (-c - d*x)/2],x]

[Out]

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

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Maple [F]  time = 0.087, size = 0, normalized size = 0. \begin{align*} \int{F}^{bx+a}\cot \left ({\frac{\pi }{4}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(b*x+a)*cot(1/4*Pi+1/2*d*x+1/2*c),x)

[Out]

int(F^(b*x+a)*cot(1/4*Pi+1/2*d*x+1/2*c),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int F^{b x + a} \cot \left (\frac{1}{4} \, \pi + \frac{1}{2} \, d x + \frac{1}{2} \, c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(b*x+a)*cot(1/2*c+1/4*pi+1/2*d*x),x, algorithm="maxima")

[Out]

integrate(F^(b*x + a)*cot(1/4*pi + 1/2*d*x + 1/2*c), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (F^{b x + a} \cot \left (\frac{1}{4} \, \pi + \frac{1}{2} \, d x + \frac{1}{2} \, c\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(b*x+a)*cot(1/2*c+1/4*pi+1/2*d*x),x, algorithm="fricas")

[Out]

integral(F^(b*x + a)*cot(1/4*pi + 1/2*d*x + 1/2*c), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int F^{a + b x} \cot{\left (\frac{c}{2} + \frac{d x}{2} + \frac{\pi }{4} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(b*x+a)*cot(1/2*c+1/4*pi+1/2*d*x),x)

[Out]

Integral(F**(a + b*x)*cot(c/2 + d*x/2 + pi/4), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int F^{b x + a} \cot \left (\frac{1}{4} \, \pi + \frac{1}{2} \, d x + \frac{1}{2} \, c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(b*x+a)*cot(1/2*c+1/4*pi+1/2*d*x),x, algorithm="giac")

[Out]

integrate(F^(b*x + a)*cot(1/4*pi + 1/2*d*x + 1/2*c), x)