3.2.42 \(\int \frac {F^{c (a+b x)}}{(f+f \cos (d+e x))^2} \, dx\) [142]
Optimal. Leaf size=169 \[ -\frac {2 e^{i (d+e x)} F^{c (a+b x)} \, _2F_1\left (2,1-\frac {i b c \log (F)}{e};2-\frac {i b c \log (F)}{e};-e^{i (d+e x)}\right ) (i e-b c \log (F))}{3 e^2 f^2}-\frac {b c F^{c (a+b x)} \log (F) \sec ^2\left (\frac {d}{2}+\frac {e x}{2}\right )}{6 e^2 f^2}+\frac {F^{c (a+b x)} \sec ^2\left (\frac {d}{2}+\frac {e x}{2}\right ) \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{6 e f^2} \]
[Out]
-2/3*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],-exp(I*(e*x+d)))*(I*e-b*c*l
n(F))/e^2/f^2-1/6*b*c*F^(c*(b*x+a))*ln(F)*sec(1/2*e*x+1/2*d)^2/e^2/f^2+1/6*F^(c*(b*x+a))*sec(1/2*e*x+1/2*d)^2*
tan(1/2*e*x+1/2*d)/e/f^2
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Rubi [A]
time = 0.07, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4542, 4533,
4536} \begin {gather*} -\frac {2 e^{i (d+e x)} F^{c (a+b x)} (-b c \log (F)+i e) \, _2F_1\left (2,1-\frac {i b c \log (F)}{e};2-\frac {i b c \log (F)}{e};-e^{i (d+e x)}\right )}{3 e^2 f^2}-\frac {b c \log (F) \sec ^2\left (\frac {d}{2}+\frac {e x}{2}\right ) F^{c (a+b x)}}{6 e^2 f^2}+\frac {\tan \left (\frac {d}{2}+\frac {e x}{2}\right ) \sec ^2\left (\frac {d}{2}+\frac {e x}{2}\right ) F^{c (a+b x)}}{6 e f^2} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[F^(c*(a + b*x))/(f + f*Cos[d + e*x])^2,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, -E^(I*(d
+ e*x))]*(I*e - b*c*Log[F]))/(3*e^2*f^2) - (b*c*F^(c*(a + b*x))*Log[F]*Sec[d/2 + (e*x)/2]^2)/(6*e^2*f^2) + (F^
(c*(a + b*x))*Sec[d/2 + (e*x)/2]^2*Tan[d/2 + (e*x)/2])/(6*e*f^2)
Rule 4533
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sec[(d_.) + (e_.)*(x_)]^(n_), x_Symbol] :> Simp[(-b)*c*Log[F]*F^(c*(a +
b*x))*(Sec[d + e*x]^(n - 2)/(e^2*(n - 1)*(n - 2))), x] + (Dist[(e^2*(n - 2)^2 + b^2*c^2*Log[F]^2)/(e^2*(n - 1)
*(n - 2)), Int[F^(c*(a + b*x))*Sec[d + e*x]^(n - 2), x], x] + Simp[F^(c*(a + b*x))*Sec[d + e*x]^(n - 1)*(Sin[d
+ e*x]/(e*(n - 1))), x]) /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[b^2*c^2*Log[F]^2 + e^2*(n - 2)^2, 0] && GtQ[n
, 1] && NeQ[n, 2]
Rule 4536
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sec[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Simp[2^n*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*(d + e*x))], x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]
Rule 4542
Int[(Cos[(d_.) + (e_.)*(x_)]*(g_.) + (f_))^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Dist[2^n*f^n,
Int[F^(c*(a + b*x))*Cos[d/2 + e*(x/2)]^(2*n), x], x] /; FreeQ[{F, a, b, c, d, e, f, g}, x] && EqQ[f - g, 0] &
& ILtQ[n, 0]
Rubi steps
\begin {align*} \int \frac {F^{c (a+b x)}}{(f+f \cos (d+e x))^2} \, dx &=\frac {\int F^{c (a+b x)} \sec ^4\left (\frac {d}{2}+\frac {e x}{2}\right ) \, dx}{4 f^2}\\ &=-\frac {b c F^{c (a+b x)} \log (F) \sec ^2\left (\frac {d}{2}+\frac {e x}{2}\right )}{6 e^2 f^2}+\frac {F^{c (a+b x)} \sec ^2\left (\frac {d}{2}+\frac {e x}{2}\right ) \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{6 e f^2}+\frac {\left (1+\frac {b^2 c^2 \log ^2(F)}{e^2}\right ) \int F^{c (a+b x)} \sec ^2\left (\frac {d}{2}+\frac {e x}{2}\right ) \, dx}{6 f^2}\\ &=-\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};-e^{i (d+e x)}\right ) (i e-b c \log (F))}{3 e^2 f^2}-\frac {b c F^{c (a+b x)} \log (F) \sec ^2\left (\frac {d}{2}+\frac {e x}{2}\right )}{6 e^2 f^2}+\frac {F^{c (a+b x)} \sec ^2\left (\frac {d}{2}+\frac {e x}{2}\right ) \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{6 e f^2}\\ \end {align*}
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Mathematica [A]
time = 0.28, size = 145, normalized size = 0.86 \begin {gather*} \frac {2 F^{c (a+b x)} \cos \left (\frac {1}{2} (d+e x)\right ) \left (-b c \cos \left (\frac {1}{2} (d+e x)\right ) \log (F)+4 e^{i (d+e x)} \cos ^3\left (\frac {1}{2} (d+e x)\right ) \, _2F_1\left (2,1-\frac {i b c \log (F)}{e};2-\frac {i b c \log (F)}{e};-e^{i (d+e x)}\right ) (-i e+b c \log (F))+e \sin \left (\frac {1}{2} (d+e x)\right )\right )}{3 e^2 f^2 (1+\cos (d+e x))^2} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[F^(c*(a + b*x))/(f + f*Cos[d + e*x])^2,x]
[Out]
(2*F^(c*(a + b*x))*Cos[(d + e*x)/2]*(-(b*c*Cos[(d + e*x)/2]*Log[F]) + 4*E^(I*(d + e*x))*Cos[(d + e*x)/2]^3*Hyp
ergeometric2F1[2, 1 - (I*b*c*Log[F])/e, 2 - (I*b*c*Log[F])/e, -E^(I*(d + e*x))]*((-I)*e + b*c*Log[F]) + e*Sin[
(d + e*x)/2]))/(3*e^2*f^2*(1 + Cos[d + e*x])^2)
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Maple [F]
time = 0.23, size = 0, normalized size = 0.00 \[\int \frac {F^{c \left (b x +a \right )}}{\left (f +f \cos \left (e x +d \right )\right )^{2}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(F^(c*(b*x+a))/(f+f*cos(e*x+d))^2,x)
[Out]
int(F^(c*(b*x+a))/(f+f*cos(e*x+d))^2,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*cos(e*x+d))^2,x, algorithm="maxima")
[Out]
4*(6*(F^(a*c)*b^5*c^5*log(F)^5 + 25*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 144*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*cos(2
*x*e + 2*d)^2 + 80*(F^(a*c)*b^3*c^3*e^2*log(F)^3 + 16*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*cos(x*e + d)^2 + 6*(F^
(a*c)*b^5*c^5*log(F)^5 + 25*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 144*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*sin(2*x*e + 2
*d)^2 + 80*(F^(a*c)*b^3*c^3*e^2*log(F)^3 + 16*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*sin(x*e + d)^2 - 140*(F^(a*c)*
b^3*c^3*e^2*log(F)^3 - 8*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*cos(x*e + d) + 20*(F^(a*c)*b^4*c^4*e*log(F)^4 - 26*
F^(a*c)*b^2*c^2*e^3*log(F)^2)*F^(b*c*x)*sin(x*e + d) - 40*(F^(a*c)*b^3*c^3*e^2*log(F)^3 - 5*F^(a*c)*b*c*e^4*lo
g(F))*F^(b*c*x) + ((F^(a*c)*b^5*c^5*log(F)^5 + 25*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 144*F^(a*c)*b*c*e^4*log(F))*F
^(b*c*x)*cos(2*x*e + 2*d) + 20*(F^(a*c)*b^3*c^3*e^2*log(F)^3 + 16*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*cos(x*e +
d) - 2*(F^(a*c)*b^4*c^4*e*log(F)^4 + 25*F^(a*c)*b^2*c^2*e^3*log(F)^2 + 144*F^(a*c)*e^5)*F^(b*c*x)*sin(2*x*e +
2*d) + 4*(F^(a*c)*b^4*c^4*e*log(F)^4 + 10*F^(a*c)*b^2*c^2*e^3*log(F)^2 - 96*F^(a*c)*e^5)*F^(b*c*x)*sin(x*e + d
) - 40*(F^(a*c)*b^3*c^3*e^2*log(F)^3 - 5*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x))*cos(4*x*e + 4*d) + 4*((F^(a*c)*b^5
*c^5*log(F)^5 + 25*F^(a*c)*b^3*c^3*e^2*log(F)^3 + 144*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*cos(2*x*e + 2*d) + 20*
(F^(a*c)*b^3*c^3*e^2*log(F)^3 + 16*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*cos(x*e + d) - 2*(F^(a*c)*b^4*c^4*e*log(F
)^4 + 25*F^(a*c)*b^2*c^2*e^3*log(F)^2 + 144*F^(a*c)*e^5)*F^(b*c*x)*sin(2*x*e + 2*d) + 4*(F^(a*c)*b^4*c^4*e*log
(F)^4 + 10*F^(a*c)*b^2*c^2*e^3*log(F)^2 - 96*F^(a*c)*e^5)*F^(b*c*x)*sin(x*e + d) - 40*(F^(a*c)*b^3*c^3*e^2*log
(F)^3 - 5*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x))*cos(3*x*e + 3*d) + (4*(F^(a*c)*b^5*c^5*log(F)^5 + 55*F^(a*c)*b^3*
c^3*e^2*log(F)^3 + 624*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x)*cos(x*e + d) + 8*(4*F^(a*c)*b^4*c^4*e*log(F)^4 + 55*F
^(a*c)*b^2*c^2*e^3*log(F)^2 - 144*F^(a*c)*e^5)*F^(b*c*x)*sin(x*e + d) + (F^(a*c)*b^5*c^5*log(F)^5 - 215*F^(a*c
)*b^3*c^3*e^2*log(F)^3 + 1344*F^(a*c)*b*c*e^4*log(F))*F^(b*c*x))*cos(2*x*e + 2*d) + 4*((F^(a*c)*b^7*c^7*e^2*lo
g(F)^7 + 29*F^(a*c)*b^5*c^5*e^4*log(F)^5 + 244*F^(a*c)*b^3*c^3*e^6*log(F)^3 + 576*F^(a*c)*b*c*e^8*log(F))*f^2*
cos(4*x*e + 4*d)^2 + 16*(F^(a*c)*b^7*c^7*e^2*log(F)^7 + 29*F^(a*c)*b^5*c^5*e^4*log(F)^5 + 244*F^(a*c)*b^3*c^3*
e^6*log(F)^3 + 576*F^(a*c)*b*c*e^8*log(F))*f^2*cos(3*x*e + 3*d)^2 + 36*(F^(a*c)*b^7*c^7*e^2*log(F)^7 + 29*F^(a
*c)*b^5*c^5*e^4*log(F)^5 + 244*F^(a*c)*b^3*c^3*e^6*log(F)^3 + 576*F^(a*c)*b*c*e^8*log(F))*f^2*cos(2*x*e + 2*d)
^2 + 16*(F^(a*c)*b^7*c^7*e^2*log(F)^7 + 29*F^(a*c)*b^5*c^5*e^4*log(F)^5 + 244*F^(a*c)*b^3*c^3*e^6*log(F)^3 + 5
76*F^(a*c)*b*c*e^8*log(F))*f^2*cos(x*e + d)^2 + (F^(a*c)*b^7*c^7*e^2*log(F)^7 + 29*F^(a*c)*b^5*c^5*e^4*log(F)^
5 + 244*F^(a*c)*b^3*c^3*e^6*log(F)^3 + 576*F^(a*c)*b*c*e^8*log(F))*f^2*sin(4*x*e + 4*d)^2 + 16*(F^(a*c)*b^7*c^
7*e^2*log(F)^7 + 29*F^(a*c)*b^5*c^5*e^4*log(F)^5 + 244*F^(a*c)*b^3*c^3*e^6*log(F)^3 + 576*F^(a*c)*b*c*e^8*log(
F))*f^2*sin(3*x*e + 3*d)^2 + 36*(F^(a*c)*b^7*c^7*e^2*log(F)^7 + 29*F^(a*c)*b^5*c^5*e^4*log(F)^5 + 244*F^(a*c)*
b^3*c^3*e^6*log(F)^3 + 576*F^(a*c)*b*c*e^8*log(F))*f^2*sin(2*x*e + 2*d)^2 + 48*(F^(a*c)*b^7*c^7*e^2*log(F)^7 +
29*F^(a*c)*b^5*c^5*e^4*log(F)^5 + 244*F^(a*c)*b^3*c^3*e^6*log(F)^3 + 576*F^(a*c)*b*c*e^8*log(F))*f^2*sin(2*x*
e + 2*d)*sin(x*e + d) + 16*(F^(a*c)*b^7*c^7*e^2*log(F)^7 + 29*F^(a*c)*b^5*c^5*e^4*log(F)^5 + 244*F^(a*c)*b^3*c
^3*e^6*log(F)^3 + 576*F^(a*c)*b*c*e^8*log(F))*f^2*sin(x*e + d)^2 + 8*(F^(a*c)*b^7*c^7*e^2*log(F)^7 + 29*F^(a*c
)*b^5*c^5*e^4*log(F)^5 + 244*F^(a*c)*b^3*c^3*e^6*log(F)^3 + 576*F^(a*c)*b*c*e^8*log(F))*f^2*cos(x*e + d) + (F^
(a*c)*b^7*c^7*e^2*log(F)^7 + 29*F^(a*c)*b^5*c^5*e^4*log(F)^5 + 244*F^(a*c)*b^3*c^3*e^6*log(F)^3 + 576*F^(a*c)*
b*c*e^8*log(F))*f^2 + 2*(4*(F^(a*c)*b^7*c^7*e^2*log(F)^7 + 29*F^(a*c)*b^5*c^5*e^4*log(F)^5 + 244*F^(a*c)*b^3*c
^3*e^6*log(F)^3 + 576*F^(a*c)*b*c*e^8*log(F))*f^2*cos(3*x*e + 3*d) + 6*(F^(a*c)*b^7*c^7*e^2*log(F)^7 + 29*F^(a
*c)*b^5*c^5*e^4*log(F)^5 + 244*F^(a*c)*b^3*c^3*e^6*log(F)^3 + 576*F^(a*c)*b*c*e^8*log(F))*f^2*cos(2*x*e + 2*d)
+ 4*(F^(a*c)*b^7*c^7*e^2*log(F)^7 + 29*F^(a*c)*b^5*c^5*e^4*log(F)^5 + 244*F^(a*c)*b^3*c^3*e^6*log(F)^3 + 576*
F^(a*c)*b*c*e^8*log(F))*f^2*cos(x*e + d) + (F^(a*c)*b^7*c^7*e^2*log(F)^7 + 29*F^(a*c)*b^5*c^5*e^4*log(F)^5 + 2
44*F^(a*c)*b^3*c^3*e^6*log(F)^3 + 576*F^(a*c)*b*c*e^8*log(F))*f^2)*cos(4*x*e + 4*d) + 8*(6*(F^(a*c)*b^7*c^7*e^
2*log(F)^7 + 29*F^(a*c)*b^5*c^5*e^4*log(F)^5 + 244*F^(a*c)*b^3*c^3*e^6*log(F)^3 + 576*F^(a*c)*b*c*e^8*log(F))*
f^2*cos(2*x*e + 2*d) + 4*(F^(a*c)*b^7*c^7*e^2*log(F)^7 + 29*F^(a*c)*b^5*c^5*e^4*log(F)^5 + 244*F^(a*c)*b^3*c^3
*e^6*log(F)^3 + 576*F^(a*c)*b*c*e^8*log(F))*f^2*cos(x*e + d) + (F^(a*c)*b^7*c^7*e^2*log(F)^7 + 29*F^(a*c)*b^5*
c^5*e^4*log(F)^5 + 244*F^(a*c)*b^3*c^3*e^6*log(F)^3 + 576*F^(a*c)*b*c*e^8*log(F))*f^2)*cos(3*x*e + 3*d) + 12*(
4*(F^(a*c)*b^7*c^7*e^2*log(F)^7 + 29*F^(a*c)*b^5*c^5*e^4*log(F)^5 + 244*F^(a*c)*b^3*c^3*e^6*log(F)^3 + 576*F^(
a*c)*b*c*e^8*log(F))*f^2*cos(x*e + d) + (F^(a*c...
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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*cos(e*x+d))^2,x, algorithm="fricas")
[Out]
integral(F^(b*c*x + a*c)/(f^2*cos(x*e + d)^2 + 2*f^2*cos(x*e + d) + f^2), 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}}{\cos ^{2}{\left (d + e x \right )} + 2 \cos {\left (d + e x \right )} + 1}\, dx}{f^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(F**(c*(b*x+a))/(f+f*cos(e*x+d))**2,x)
[Out]
Integral(F**(a*c)*F**(b*c*x)/(cos(d + e*x)**2 + 2*cos(d + e*x) + 1), x)/f**2
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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*cos(e*x+d))^2,x, algorithm="giac")
[Out]
integrate(F^((b*x + a)*c)/(f*cos(e*x + d) + f)^2, 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 )}}{{\left (f+f\,\cos \left (d+e\,x\right )\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(F^(c*(a + b*x))/(f + f*cos(d + e*x))^2,x)
[Out]
int(F^(c*(a + b*x))/(f + f*cos(d + e*x))^2, x)
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