3.2.75 \(\int \csc (e+f x) (a+b \sin ^2(e+f x))^p \, dx\) [175]

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

[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)

________________________________________________________________________________________

Rubi [A]
time = 0.05, 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 = {3265, 441, 440} \begin {gather*} -\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} \end {gather*}

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 440

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 441

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 3265

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 {\text {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 ) \text {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*}

________________________________________________________________________________________

Mathematica [F]
time = 15.14, size = 0, normalized size = 0.00 \begin {gather*} \int \csc (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx \end {gather*}

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]

________________________________________________________________________________________

Maple [F]
time = 0.32, 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)

________________________________________________________________________________________

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(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)

________________________________________________________________________________________

Fricas [F]
time = 0.41, size = 25, normalized size = 0.30 \begin {gather*} {\rm integral}\left ({\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{p} \csc \left (f x + e\right ), x\right ) \end {gather*}

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)

________________________________________________________________________________________

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

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]

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

________________________________________________________________________________________

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(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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________