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

Optimal. Leaf size=76 \[ \frac {i F^{a+b x}}{b \log (F)}-\frac {2 i F^{a+b x} \, _2F_1\left (1,-\frac {i b \log (F)}{d};1-\frac {i b \log (F)}{d};i e^{i (c+d x)}\right )}{b \log (F)} \]

[Out]

I*F^(b*x+a)/b/ln(F)-2*I*F^(b*x+a)*hypergeom([1, -I*b*ln(F)/d],[1-I*b*ln(F)/d],I*exp(I*(d*x+c)))/b/ln(F)

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Rubi [A]  time = 0.09, antiderivative size = 76, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.31, size = 133, normalized size = 1.75 \[ \frac {F^{a+b x} \left (b \log (F) e^{i (c+d x)} \, _2F_1\left (1,1-\frac {i b \log (F)}{d};2-\frac {i b \log (F)}{d};i e^{i (c+d x)}\right )+(d-i b \log (F)) \, _2F_1\left (1,-\frac {i b \log (F)}{d};1-\frac {i b \log (F)}{d};i e^{i (c+d x)}\right )\right )}{b \log (F) (b \log (F)+i d)} \]

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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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (F^{b x + a} \cot \left (\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ), x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int F^{b x + a} \cot \left (\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\,{d x} \]

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)

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maple [F]  time = 0.22, size = 0, normalized size = 0.00 \[ \int F^{b x +a} \cot \left (\frac {\pi }{4}+\frac {d x}{2}+\frac {c}{2}\right )\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int F^{b x + a} \cot \left (\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int F^{a+b\,x}\,\mathrm {cot}\left (\frac {\Pi }{4}+\frac {c}{2}+\frac {d\,x}{2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int F^{a + b x} \cot {\left (\frac {c}{2} + \frac {d x}{2} + \frac {\pi }{4} \right )}\, dx \]

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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