3.7 \(\int F^{c (a+b x)} \csc ^3(d+e x) \, dx\)

Optimal. Leaf size=137 \[ -\frac{e^{i (d+e x)} F^{c (a+b x)} (e+i b c \log (F)) \text{Hypergeometric2F1}\left (1,\frac{e-i b c \log (F)}{2 e},\frac{1}{2} \left (3-\frac{i b c \log (F)}{e}\right ),e^{2 i (d+e x)}\right )}{e^2}-\frac{b c \log (F) \csc (d+e x) F^{c (a+b x)}}{2 e^2}-\frac{\cot (d+e x) \csc (d+e x) F^{c (a+b x)}}{2 e} \]

[Out]

-(F^(c*(a + b*x))*Cot[d + e*x]*Csc[d + e*x])/(2*e) - (b*c*F^(c*(a + b*x))*Csc[d + e*x]*Log[F])/(2*e^2) - (E^(I
*(d + e*x))*F^(c*(a + b*x))*Hypergeometric2F1[1, (e - I*b*c*Log[F])/(2*e), (3 - (I*b*c*Log[F])/e)/2, E^((2*I)*
(d + e*x))]*(e + I*b*c*Log[F]))/e^2

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Rubi [A]  time = 0.0490738, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {4449, 4453} \[ -\frac{e^{i (d+e x)} F^{c (a+b x)} (e+i b c \log (F)) \, _2F_1\left (1,\frac{e-i b c \log (F)}{2 e};\frac{1}{2} \left (3-\frac{i b c \log (F)}{e}\right );e^{2 i (d+e x)}\right )}{e^2}-\frac{b c \log (F) \csc (d+e x) F^{c (a+b x)}}{2 e^2}-\frac{\cot (d+e x) \csc (d+e x) F^{c (a+b x)}}{2 e} \]

Antiderivative was successfully verified.

[In]

Int[F^(c*(a + b*x))*Csc[d + e*x]^3,x]

[Out]

-(F^(c*(a + b*x))*Cot[d + e*x]*Csc[d + e*x])/(2*e) - (b*c*F^(c*(a + b*x))*Csc[d + e*x]*Log[F])/(2*e^2) - (E^(I
*(d + e*x))*F^(c*(a + b*x))*Hypergeometric2F1[1, (e - I*b*c*Log[F])/(2*e), (3 - (I*b*c*Log[F])/e)/2, E^((2*I)*
(d + e*x))]*(e + I*b*c*Log[F]))/e^2

Rule 4449

Int[Csc[(d_.) + (e_.)*(x_)]^(n_)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> -Simp[(b*c*Log[F]*F^(c*(a + b
*x))*Csc[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))*Csc[d + e*x]^(n - 2), x], x] - Simp[(F^(c*(a + b*x))*Csc[d + e*x]^(n - 1)*Cos[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 4453

Int[Csc[(d_.) + (e_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Simp[(-2*I)^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*L
og[F])/(2*e), E^(2*I*(d + e*x))], x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]

Rubi steps

\begin{align*} \int F^{c (a+b x)} \csc ^3(d+e x) \, dx &=-\frac{F^{c (a+b x)} \cot (d+e x) \csc (d+e x)}{2 e}-\frac{b c F^{c (a+b x)} \csc (d+e x) \log (F)}{2 e^2}+\frac{1}{2} \left (1+\frac{b^2 c^2 \log ^2(F)}{e^2}\right ) \int F^{c (a+b x)} \csc (d+e x) \, dx\\ &=-\frac{F^{c (a+b x)} \cot (d+e x) \csc (d+e x)}{2 e}-\frac{b c F^{c (a+b x)} \csc (d+e x) \log (F)}{2 e^2}-\frac{e^{i (d+e x)} F^{c (a+b x)} \, _2F_1\left (1,\frac{e-i b c \log (F)}{2 e};\frac{1}{2} \left (3-\frac{i b c \log (F)}{e}\right );e^{2 i (d+e x)}\right ) (e+i b c \log (F))}{e^2}\\ \end{align*}

Mathematica [B]  time = 7.89086, size = 334, normalized size = 2.44 \[ \frac{F^{c (a+b x)} \left (-\frac{4 i \left (b^2 c^2 \log ^2(F)+e^2\right ) \left (1+(i \sin (d)+\cos (d)-1) \text{Hypergeometric2F1}\left (1,-\frac{i b c \log (F)}{e},1-\frac{i b c \log (F)}{e},\cos (d+e x)+i \sin (d+e x)\right )\right )}{b c \log (F) (i \sin (d)+\cos (d)-1)}-\frac{4 i \left (b^2 c^2 \log ^2(F)+e^2\right ) \left (1-(i \sin (d)+\cos (d)+1) \text{Hypergeometric2F1}\left (1,-\frac{i b c \log (F)}{e},1-\frac{i b c \log (F)}{e},-\cos (d+e x)-i \sin (d+e x)\right )\right )}{b c \log (F) (i \sin (d)+\cos (d)+1)}+\csc (d) \left (\frac{4 e^2}{b c \log (F)}+4 b c \log (F)\right )-2 b c \sec \left (\frac{d}{2}\right ) \log (F) \sin \left (\frac{e x}{2}\right ) \sec \left (\frac{1}{2} (d+e x)\right )+2 b c \csc \left (\frac{d}{2}\right ) \log (F) \sin \left (\frac{e x}{2}\right ) \csc \left (\frac{1}{2} (d+e x)\right )-4 b c \csc (d) \log (F)-e \csc ^2\left (\frac{1}{2} (d+e x)\right )+e \sec ^2\left (\frac{1}{2} (d+e x)\right )\right )}{8 e^2} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(c*(a + b*x))*Csc[d + e*x]^3,x]

[Out]

(F^(c*(a + b*x))*(-(e*Csc[(d + e*x)/2]^2) - 4*b*c*Csc[d]*Log[F] + Csc[d]*((4*e^2)/(b*c*Log[F]) + 4*b*c*Log[F])
+ e*Sec[(d + e*x)/2]^2 - ((4*I)*(e^2 + b^2*c^2*Log[F]^2)*(1 + Hypergeometric2F1[1, ((-I)*b*c*Log[F])/e, 1 - (
I*b*c*Log[F])/e, Cos[d + e*x] + I*Sin[d + e*x]]*(-1 + Cos[d] + I*Sin[d])))/(b*c*Log[F]*(-1 + Cos[d] + I*Sin[d]
)) - ((4*I)*(e^2 + b^2*c^2*Log[F]^2)*(1 - Hypergeometric2F1[1, ((-I)*b*c*Log[F])/e, 1 - (I*b*c*Log[F])/e, -Cos
[d + e*x] - I*Sin[d + e*x]]*(1 + Cos[d] + I*Sin[d])))/(b*c*Log[F]*(1 + Cos[d] + I*Sin[d])) + 2*b*c*Csc[d/2]*Cs
c[(d + e*x)/2]*Log[F]*Sin[(e*x)/2] - 2*b*c*Log[F]*Sec[d/2]*Sec[(d + e*x)/2]*Sin[(e*x)/2]))/(8*e^2)

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Maple [F]  time = 0.117, size = 0, normalized size = 0. \begin{align*} \int{F}^{c \left ( bx+a \right ) } \left ( \csc \left ( ex+d \right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(b*x+a))*csc(e*x+d)^3,x)

[Out]

int(F^(c*(b*x+a))*csc(e*x+d)^3,x)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*csc(e*x+d)^3,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (F^{b c x + a c} \csc \left (e x + d\right )^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*csc(e*x+d)^3,x, algorithm="fricas")

[Out]

integral(F^(b*c*x + a*c)*csc(e*x + d)^3, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(c*(b*x+a))*csc(e*x+d)**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int F^{{\left (b x + a\right )} c} \csc \left (e x + d\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*csc(e*x+d)^3,x, algorithm="giac")

[Out]

integrate(F^((b*x + a)*c)*csc(e*x + d)^3, x)