3.2.37 \(\int \frac {F^{c (a+b x)}}{f+f \sin (d+e x)} \, dx\) [137]
Optimal. Leaf size=80 \[ -\frac {2 e^{i (d+e x)} F^{c (a+b x)} \, _2F_1\left (2,1-\frac {i b c \log (F)}{e};2-\frac {i b c \log (F)}{e};i e^{i (d+e x)}\right )}{f (e-i b c \log (F))} \]
[Out]
-2*exp(I*(e*x+d))*F^(c*(b*x+a))*hypergeom([2, 1-I*b*c*ln(F)/e],[2-I*b*c*ln(F)/e],I*exp(I*(e*x+d)))/f/(e-I*b*c*
ln(F))
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Rubi [A]
time = 0.05, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps
used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4541, 4535}
\begin {gather*} -\frac {2 e^{i (d+e x)} F^{c (a+b x)} \, _2F_1\left (2,1-\frac {i b c \log (F)}{e};2-\frac {i b c \log (F)}{e};i e^{i (d+e x)}\right )}{f (e-i b c \log (F))} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[F^(c*(a + b*x))/(f + f*Sin[d + e*x]),x]
[Out]
(-2*E^(I*(d + e*x))*F^(c*(a + b*x))*Hypergeometric2F1[2, 1 - (I*b*c*Log[F])/e, 2 - (I*b*c*Log[F])/e, I*E^(I*(d
+ e*x))])/(f*(e - I*b*c*Log[F]))
Rule 4535
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sec[(d_.) + Pi*(k_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Simp[2^n*E^(I*k*n
*Pi)*E^(I*n*(d + e*x))*(F^(c*(a + b*x))/(I*e*n + b*c*Log[F]))*Hypergeometric2F1[n, n/2 - I*b*c*(Log[F]/(2*e)),
1 + n/2 - I*b*c*(Log[F]/(2*e)), (-E^(2*I*k*Pi))*E^(2*I*(d + e*x))], x] /; FreeQ[{F, a, b, c, d, e}, x] && Int
egerQ[4*k] && IntegerQ[n]
Rule 4541
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_) + (g_.)*Sin[(d_.) + (e_.)*(x_)])^(n_.), x_Symbol] :> Dist[2^n*f^n,
Int[F^(c*(a + b*x))*Cos[d/2 - f*(Pi/(4*g)) + e*(x/2)]^(2*n), x], x] /; FreeQ[{F, a, b, c, d, e, f, g}, x] &&
EqQ[f^2 - g^2, 0] && ILtQ[n, 0]
Rubi steps
\begin {align*} \int \frac {F^{c (a+b x)}}{f+f \sin (d+e x)} \, dx &=\frac {\int F^{c (a+b x)} \sec ^2\left (\frac {d}{2}-\frac {\pi }{4}+\frac {e x}{2}\right ) \, dx}{2 f}\\ &=-\frac {2 e^{i (d+e x)} F^{c (a+b x)} \, _2F_1\left (2,1-\frac {i b c \log (F)}{e};2-\frac {i b c \log (F)}{e};i e^{i (d+e x)}\right )}{f (e-i b c \log (F))}\\ \end {align*}
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Mathematica [A]
time = 1.06, size = 128, normalized size = 1.60 \begin {gather*} \frac {2 F^{c (a+b x)} \left (-i \, _2F_1\left (1,-\frac {i b c \log (F)}{e};1-\frac {i b c \log (F)}{e};i \cos (d+e x)-\sin (d+e x)\right )-\frac {1}{\cos (d)+i (1+\sin (d))}+\frac {\sin \left (\frac {e x}{2}\right )}{\left (\cos \left (\frac {d}{2}\right )+\sin \left (\frac {d}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )}\right )}{e f} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[F^(c*(a + b*x))/(f + f*Sin[d + e*x]),x]
[Out]
(2*F^(c*(a + b*x))*((-I)*Hypergeometric2F1[1, ((-I)*b*c*Log[F])/e, 1 - (I*b*c*Log[F])/e, I*Cos[d + e*x] - Sin[
d + e*x]] - (Cos[d] + I*(1 + Sin[d]))^(-1) + Sin[(e*x)/2]/((Cos[d/2] + Sin[d/2])*(Cos[(d + e*x)/2] + Sin[(d +
e*x)/2]))))/(e*f)
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Maple [F]
time = 0.10, size = 0, normalized size = 0.00 \[\int \frac {F^{c \left (b x +a \right )}}{f +f \sin \left (e x +d \right )}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(F^(c*(b*x+a))/(f+f*sin(e*x+d)),x)
[Out]
int(F^(c*(b*x+a))/(f+f*sin(e*x+d)),x)
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(F^(c*(b*x+a))/(f+f*sin(e*x+d)),x, algorithm="maxima")
[Out]
2*(6*F^(a*c)*b*c*e^(b*c*x*log(F) + 2)*log(F) + 2*(F^(a*c)*b^3*c^3*log(F)^3 + 4*F^(a*c)*b*c*e^2*log(F))*F^(b*c*
x)*cos(x*e + d)^2 + 2*(F^(a*c)*b^3*c^3*log(F)^3 + 4*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*sin(x*e + d)^2 + (5*F^(a
*c)*b^2*c^2*e*log(F)^2 - 4*F^(a*c)*e^3)*F^(b*c*x)*cos(x*e + d) + (F^(a*c)*b^3*c^3*log(F)^3 + 16*F^(a*c)*b*c*e^
2*log(F))*F^(b*c*x)*sin(x*e + d) - (6*F^(a*c)*b*c*e^(b*c*x*log(F) + 2)*log(F) + (F^(a*c)*b^2*c^2*e*log(F)^2 +
4*F^(a*c)*e^3)*F^(b*c*x)*cos(x*e + d) + (F^(a*c)*b^3*c^3*log(F)^3 + 4*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*sin(x*
e + d))*cos(2*x*e + 2*d) + 2*((F^(a*c)*b^5*c^5*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 4*F^(a*c)*b*c*e^4*l
og(F))*f*cos(2*x*e + 2*d)^2 + 4*(F^(a*c)*b^5*c^5*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 4*F^(a*c)*b*c*e^4
*log(F))*f*cos(x*e + d)^2 + 4*(F^(a*c)*b^5*c^5*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 4*F^(a*c)*b*c*e^4*l
og(F))*f*cos(x*e + d)*sin(2*x*e + 2*d) + (F^(a*c)*b^5*c^5*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 4*F^(a*c
)*b*c*e^4*log(F))*f*sin(2*x*e + 2*d)^2 + 4*(F^(a*c)*b^5*c^5*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 4*F^(a
*c)*b*c*e^4*log(F))*f*sin(x*e + d)^2 + 4*(F^(a*c)*b^5*c^5*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 4*F^(a*c
)*b*c*e^4*log(F))*f*sin(x*e + d) + (F^(a*c)*b^5*c^5*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 4*F^(a*c)*b*c*
e^4*log(F))*f - 2*(2*(F^(a*c)*b^5*c^5*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 4*F^(a*c)*b*c*e^4*log(F))*f*
sin(x*e + d) + (F^(a*c)*b^5*c^5*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 4*F^(a*c)*b*c*e^4*log(F))*f)*cos(2
*x*e + 2*d))*integrate(-(3*b*c*cos(3*x*e + 3*d)*e^(b*c*x*log(F) + 2)*log(F) - 9*b*c*cos(x*e + d)*e^(b*c*x*log(
F) + 2)*log(F) - 9*b*c*e^(b*c*x*log(F) + 2)*log(F)*sin(2*x*e + 2*d) - 3*(b^2*c^2*e*log(F)^2 - 2*e^3)*F^(b*c*x)
*cos(2*x*e + 2*d) - (b^2*c^2*e*log(F)^2 - 2*e^3)*F^(b*c*x)*sin(3*x*e + 3*d) + 3*(b^2*c^2*e*log(F)^2 - 2*e^3)*F
^(b*c*x)*sin(x*e + d) + (b^2*c^2*e*log(F)^2 - 2*e^3)*F^(b*c*x))/((b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 +
4*e^4)*f*cos(3*x*e + 3*d)^2 + 9*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*cos(2*x*e + 2*d)^2 + 9*(
b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*cos(x*e + d)^2 + (b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)
^2 + 4*e^4)*f*sin(3*x*e + 3*d)^2 + 18*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*cos(x*e + d)*sin(2
*x*e + 2*d) + 9*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(2*x*e + 2*d)^2 + 9*(b^4*c^4*log(F)^4
+ 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(x*e + d)^2 + 6*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*
sin(x*e + d) + (b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f - 6*((b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*lo
g(F)^2 + 4*e^4)*f*cos(x*e + d) + (b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(2*x*e + 2*d))*cos(3
*x*e + 3*d) - 6*(3*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(x*e + d) + (b^4*c^4*log(F)^4 + 5*
b^2*c^2*e^2*log(F)^2 + 4*e^4)*f)*cos(2*x*e + 2*d) + 2*(3*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f
*cos(2*x*e + 2*d) - 3*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(x*e + d) - (b^4*c^4*log(F)^4 +
5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f)*sin(3*x*e + 3*d)), x) + ((F^(a*c)*b^3*c^3*log(F)^3 + 4*F^(a*c)*b*c*e^2*log
(F))*F^(b*c*x)*cos(x*e + d) - (F^(a*c)*b^2*c^2*e*log(F)^2 + 4*F^(a*c)*e^3)*F^(b*c*x)*sin(x*e + d) + 2*(F^(a*c)
*b^2*c^2*e*log(F)^2 - 2*F^(a*c)*e^3)*F^(b*c*x))*sin(2*x*e + 2*d))/((b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2
+ 4*e^4)*f*cos(2*x*e + 2*d)^2 + 4*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*cos(x*e + d)^2 + 4*(b^
4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*cos(x*e + d)*sin(2*x*e + 2*d) + (b^4*c^4*log(F)^4 + 5*b^2*c
^2*e^2*log(F)^2 + 4*e^4)*f*sin(2*x*e + 2*d)^2 + 4*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(x*
e + d)^2 + 4*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(x*e + d) + (b^4*c^4*log(F)^4 + 5*b^2*c^
2*e^2*log(F)^2 + 4*e^4)*f - 2*(2*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(x*e + d) + (b^4*c^4
*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f)*cos(2*x*e + 2*d))
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(F^(c*(b*x+a))/(f+f*sin(e*x+d)),x, algorithm="fricas")
[Out]
integral(F^(b*c*x + a*c)/(f*sin(x*e + d) + f), x)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {F^{a c} F^{b c x}}{\sin {\left (d + e x \right )} + 1}\, dx}{f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(F**(c*(b*x+a))/(f+f*sin(e*x+d)),x)
[Out]
Integral(F**(a*c)*F**(b*c*x)/(sin(d + e*x) + 1), x)/f
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(F^(c*(b*x+a))/(f+f*sin(e*x+d)),x, algorithm="giac")
[Out]
integrate(F^((b*x + a)*c)/(f*sin(e*x + d) + f), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {F^{c\,\left (a+b\,x\right )}}{f+f\,\sin \left (d+e\,x\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(F^(c*(a + b*x))/(f + f*sin(d + e*x)),x)
[Out]
int(F^(c*(a + b*x))/(f + f*sin(d + e*x)), x)
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