3.142 \(\int \csc ^4(e+f x) (a+b \sec ^2(e+f x))^p \, dx\)

Optimal. Leaf size=128 \[ -\frac {(3 a+2 b (p+1)) \cot (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right )}{3 f (a+b)}-\frac {\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{3 f (a+b)} \]

[Out]

-1/3*cot(f*x+e)^3*(a+b+b*tan(f*x+e)^2)^(1+p)/(a+b)/f-1/3*(3*a+2*b*(1+p))*cot(f*x+e)*hypergeom([-1/2, -p],[1/2]
,-b*tan(f*x+e)^2/(a+b))*(a+b+b*tan(f*x+e)^2)^p/(a+b)/f/((1+b*tan(f*x+e)^2/(a+b))^p)

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Rubi [A]  time = 0.11, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {4132, 453, 365, 364} \[ -\frac {(3 a+2 b (p+1)) \cot (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right )}{3 f (a+b)}-\frac {\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{3 f (a+b)} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^4*(a + b*Sec[e + f*x]^2)^p,x]

[Out]

-(Cot[e + f*x]^3*(a + b + b*Tan[e + f*x]^2)^(1 + p))/(3*(a + b)*f) - ((3*a + 2*b*(1 + p))*Cot[e + f*x]*Hyperge
ometric2F1[-1/2, -p, 1/2, -((b*Tan[e + f*x]^2)/(a + b))]*(a + b + b*Tan[e + f*x]^2)^p)/(3*(a + b)*f*(1 + (b*Ta
n[e + f*x]^2)/(a + b))^p)

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 4132

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]

Rubi steps

\begin {align*} \int \csc ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right ) \left (a+b+b x^2\right )^p}{x^4} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{3 (a+b) f}+\frac {(3 a+2 b (1+p)) \operatorname {Subst}\left (\int \frac {\left (a+b+b x^2\right )^p}{x^2} \, dx,x,\tan (e+f x)\right )}{3 (a+b) f}\\ &=-\frac {\cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{3 (a+b) f}+\frac {\left ((3 a+2 b (1+p)) \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^{-p}\right ) \operatorname {Subst}\left (\int \frac {\left (1+\frac {b x^2}{a+b}\right )^p}{x^2} \, dx,x,\tan (e+f x)\right )}{3 (a+b) f}\\ &=-\frac {\cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{3 (a+b) f}-\frac {(3 a+2 b (1+p)) \cot (e+f x) \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right ) \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^{-p}}{3 (a+b) f}\\ \end {align*}

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Mathematica [A]  time = 2.25, size = 132, normalized size = 1.03 \[ -\frac {\cot (e+f x) \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \left (a+b \sec ^2(e+f x)\right )^p \left ((3 a+2 b (p+1)) \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right )+\cot ^2(e+f x) \left (a+b \tan ^2(e+f x)+b\right ) \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^p\right )}{3 f (a+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^4*(a + b*Sec[e + f*x]^2)^p,x]

[Out]

-1/3*(Cot[e + f*x]*(a + b*Sec[e + f*x]^2)^p*((3*a + 2*b*(1 + p))*Hypergeometric2F1[-1/2, -p, 1/2, -((b*Tan[e +
f*x]^2)/(a + b))] + Cot[e + f*x]^2*(a + b + b*Tan[e + f*x]^2)*(1 + (b*Tan[e + f*x]^2)/(a + b))^p))/((a + b)*f
*(1 + (b*Tan[e + f*x]^2)/(a + b))^p)

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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

integral((b*sec(f*x + e)^2 + a)^p*csc(f*x + e)^4, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e)^2 + a)^p*csc(f*x + e)^4, x)

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maple [F]  time = 1.68, size = 0, normalized size = 0.00 \[ \int \left (\csc ^{4}\left (f x +e \right )\right ) \left (a +b \left (\sec ^{2}\left (f x +e \right )\right )\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x)

[Out]

int(csc(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e)^2 + a)^p*csc(f*x + e)^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p}{{\sin \left (e+f\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(e + f*x)^2)^p/sin(e + f*x)^4,x)

[Out]

int((a + b/cos(e + f*x)^2)^p/sin(e + f*x)^4, 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)**4*(a+b*sec(f*x+e)**2)**p,x)

[Out]

Timed out

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