3.175 \(\int \csc (e+f x) (a+b \sin ^2(e+f x))^p \, dx\)

Optimal. Leaf size=83 \[ -\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p} F_1\left (\frac {1}{2};1,-p;\frac {3}{2};\cos ^2(e+f x),\frac {b \cos ^2(e+f x)}{a+b}\right )}{f} \]

[Out]

-AppellF1(1/2,1,-p,3/2,cos(f*x+e)^2,b*cos(f*x+e)^2/(a+b))*cos(f*x+e)*(a+b-b*cos(f*x+e)^2)^p/f/((1-b*cos(f*x+e)
^2/(a+b))^p)

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Rubi [A]  time = 0.08, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3186, 430, 429} \[ -\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p} F_1\left (\frac {1}{2};1,-p;\frac {3}{2};\cos ^2(e+f x),\frac {b \cos ^2(e+f x)}{a+b}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]*(a + b*Sin[e + f*x]^2)^p,x]

[Out]

-((AppellF1[1/2, 1, -p, 3/2, Cos[e + f*x]^2, (b*Cos[e + f*x]^2)/(a + b)]*Cos[e + f*x]*(a + b - b*Cos[e + f*x]^
2)^p)/(f*(1 - (b*Cos[e + f*x]^2)/(a + b))^p))

Rule 429

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 430

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^F
racPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 3186

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos
[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \csc (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\left (a+b-b x^2\right )^p}{1-x^2} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac {\left (\left (a+b-b \cos ^2(e+f x)\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p}\right ) \operatorname {Subst}\left (\int \frac {\left (1-\frac {b x^2}{a+b}\right )^p}{1-x^2} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac {F_1\left (\frac {1}{2};1,-p;\frac {3}{2};\cos ^2(e+f x),\frac {b \cos ^2(e+f x)}{a+b}\right ) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p}}{f}\\ \end {align*}

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Mathematica [F]  time = 4.84, size = 0, normalized size = 0.00 \[ \int \csc (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Csc[e + f*x]*(a + b*Sin[e + f*x]^2)^p,x]

[Out]

Integrate[Csc[e + f*x]*(a + b*Sin[e + f*x]^2)^p, x]

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{p} \csc \left (f x + e\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)*(a+b*sin(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

integral((-b*cos(f*x + e)^2 + a + b)^p*csc(f*x + e), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)*(a+b*sin(f*x+e)^2)^p,x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e)^2 + a)^p*csc(f*x + e), x)

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maple [F]  time = 2.43, size = 0, normalized size = 0.00 \[ \int \csc \left (f x +e \right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)*(a+b*sin(f*x+e)^2)^p,x)

[Out]

int(csc(f*x+e)*(a+b*sin(f*x+e)^2)^p,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)*(a+b*sin(f*x+e)^2)^p,x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e)^2 + a)^p*csc(f*x + e), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^p}{\sin \left (e+f\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(e + f*x)^2)^p/sin(e + f*x),x)

[Out]

int((a + b*sin(e + f*x)^2)^p/sin(e + f*x), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)*(a+b*sin(f*x+e)**2)**p,x)

[Out]

Timed out

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