\(\int F^{a+b x} \tan (\frac {\pi }{4}+\frac {1}{2} (-c-d x)) \, dx\) [25]

Optimal result

Integrand size = 27, antiderivative size = 76 \[ \int F^{a+b x} \tan \left (\frac {\pi }{4}+\frac {1}{2} (-c-d x)\right ) \, dx=\frac {i F^{a+b x}}{b \log (F)}-\frac {2 i F^{a+b x} \operatorname {Hypergeometric2F1}\left (1,-\frac {i b \log (F)}{d},1-\frac {i b \log (F)}{d},i e^{i (c+d x)}\right )}{b \log (F)} \] Output:

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)
 

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.75 \[ \int F^{a+b x} \tan \left (\frac {\pi }{4}+\frac {1}{2} (-c-d x)\right ) \, dx=\frac {F^{a+b x} \left (b e^{i (c+d x)} \operatorname {Hypergeometric2F1}\left (1,1-\frac {i b \log (F)}{d},2-\frac {i b \log (F)}{d},i e^{i (c+d x)}\right ) \log (F)+\operatorname {Hypergeometric2F1}\left (1,-\frac {i b \log (F)}{d},1-\frac {i b \log (F)}{d},i e^{i (c+d x)}\right ) (d-i b \log (F))\right )}{b \log (F) (i d+b \log (F))} \] Input:

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

Output:

(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]))
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {4967, 25, 25, 4943, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 4967

\(\displaystyle \int -F^{a+b x} \tan \left (\frac {c}{2}+\frac {d x}{2}-\frac {\pi }{4}\right )dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int -F^{a+b x} \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )dx\)

\(\Big \downarrow \) 25

\(\displaystyle \int F^{a+b x} \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )dx\)

\(\Big \downarrow \) 4943

\(\displaystyle -i \int \left (\frac {2 F^{a+b x}}{1-e^{\frac {1}{2} i (2 c+2 d x+\pi )}}-F^{a+b x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -i \left (-\frac {F^{a+b x}}{b \log (F)}+\frac {2 F^{a+b x} \operatorname {Hypergeometric2F1}\left (1,-\frac {i b \log (F)}{d},1-\frac {i b \log (F)}{d},e^{\frac {1}{2} i (2 c+2 d x+\pi )}\right )}{b \log (F)}\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4943

Int[Cot[(d_.) + (e_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symb 
ol] :> Simp[(-I)^n   Int[ExpandIntegrand[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 4967

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

Maple [F]

\[\int F^{b x +a} \cot \left (\frac {\pi }{4}+\frac {d x}{2}+\frac {c}{2}\right )d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int F^{a+b x} \tan \left (\frac {\pi }{4}+\frac {1}{2} (-c-d x)\right ) \, dx=\int { F^{b x + a} \cot \left (\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int F^{a+b x} \tan \left (\frac {\pi }{4}+\frac {1}{2} (-c-d x)\right ) \, dx=\int F^{a + b x} \cot {\left (\frac {c}{2} + \frac {d x}{2} + \frac {\pi }{4} \right )}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int F^{a+b x} \tan \left (\frac {\pi }{4}+\frac {1}{2} (-c-d x)\right ) \, dx=\int { F^{b x + a} \cot \left (\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int F^{a+b x} \tan \left (\frac {\pi }{4}+\frac {1}{2} (-c-d x)\right ) \, dx=\int { F^{b x + a} \cot \left (\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int F^{a+b x} \tan \left (\frac {\pi }{4}+\frac {1}{2} (-c-d x)\right ) \, dx=\int F^{a+b\,x}\,\mathrm {cot}\left (\frac {\Pi }{4}+\frac {c}{2}+\frac {d\,x}{2}\right ) \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int F^{a+b x} \tan \left (\frac {\pi }{4}+\frac {1}{2} (-c-d x)\right ) \, dx=f^{a} \left (\int f^{b x} \cot \left (\frac {d x}{2}+\frac {c}{2}+\frac {\pi }{4}\right )d x \right ) \] Input:

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

Output:

f**a*int(f**(b*x)*cot((2*c + 2*d*x + pi)/4),x)