3.158 \(\int \csc ^3(e+f x) (a+b \tan ^2(e+f x))^p \, dx\)

Optimal. Leaf size=92 \[ \frac{\sec ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^p \left (\frac{b \sec ^2(e+f x)}{a-b}+1\right )^{-p} F_1\left (\frac{3}{2};2,-p;\frac{5}{2};\sec ^2(e+f x),-\frac{b \sec ^2(e+f x)}{a-b}\right )}{3 f} \]

[Out]

(AppellF1[3/2, 2, -p, 5/2, Sec[e + f*x]^2, -((b*Sec[e + f*x]^2)/(a - b))]*Sec[e + f*x]^3*(a - b + b*Sec[e + f*
x]^2)^p)/(3*f*(1 + (b*Sec[e + f*x]^2)/(a - b))^p)

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Rubi [A]  time = 0.123787, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3664, 511, 510} \[ \frac{\sec ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^p \left (\frac{b \sec ^2(e+f x)}{a-b}+1\right )^{-p} F_1\left (\frac{3}{2};2,-p;\frac{5}{2};\sec ^2(e+f x),-\frac{b \sec ^2(e+f x)}{a-b}\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^3*(a + b*Tan[e + f*x]^2)^p,x]

[Out]

(AppellF1[3/2, 2, -p, 5/2, Sec[e + f*x]^2, -((b*Sec[e + f*x]^2)/(a - b))]*Sec[e + f*x]^3*(a - b + b*Sec[e + f*
x]^2)^p)/(3*f*(1 + (b*Sec[e + f*x]^2)/(a - b))^p)

Rule 3664

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

Rule 511

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

Rule 510

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

Rubi steps

\begin{align*} \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (a-b+b x^2\right )^p}{\left (-1+x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac{\left (\left (a-b+b \sec ^2(e+f x)\right )^p \left (1+\frac{b \sec ^2(e+f x)}{a-b}\right )^{-p}\right ) \operatorname{Subst}\left (\int \frac{x^2 \left (1+\frac{b x^2}{a-b}\right )^p}{\left (-1+x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac{F_1\left (\frac{3}{2};2,-p;\frac{5}{2};\sec ^2(e+f x),-\frac{b \sec ^2(e+f x)}{a-b}\right ) \sec ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^p \left (1+\frac{b \sec ^2(e+f x)}{a-b}\right )^{-p}}{3 f}\\ \end{align*}

Mathematica [B]  time = 2.17901, size = 252, normalized size = 2.74 \[ \frac{b (2 p-3) \cot (e+f x) \csc (e+f x) \left (a+b \tan ^2(e+f x)\right )^p F_1\left (\frac{1}{2}-p;-\frac{1}{2},-p;\frac{3}{2}-p;-\cot ^2(e+f x),-\frac{a \cot ^2(e+f x)}{b}\right )}{f (2 p-1) \left (b (2 p-3) F_1\left (\frac{1}{2}-p;-\frac{1}{2},-p;\frac{3}{2}-p;-\cot ^2(e+f x),-\frac{a \cot ^2(e+f x)}{b}\right )-\cot ^2(e+f x) \left (2 a p F_1\left (\frac{3}{2}-p;-\frac{1}{2},1-p;\frac{5}{2}-p;-\cot ^2(e+f x),-\frac{a \cot ^2(e+f x)}{b}\right )+b F_1\left (\frac{3}{2}-p;\frac{1}{2},-p;\frac{5}{2}-p;-\cot ^2(e+f x),-\frac{a \cot ^2(e+f x)}{b}\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Csc[e + f*x]^3*(a + b*Tan[e + f*x]^2)^p,x]

[Out]

(b*(-3 + 2*p)*AppellF1[1/2 - p, -1/2, -p, 3/2 - p, -Cot[e + f*x]^2, -((a*Cot[e + f*x]^2)/b)]*Cot[e + f*x]*Csc[
e + f*x]*(a + b*Tan[e + f*x]^2)^p)/(f*(-1 + 2*p)*(b*(-3 + 2*p)*AppellF1[1/2 - p, -1/2, -p, 3/2 - p, -Cot[e + f
*x]^2, -((a*Cot[e + f*x]^2)/b)] - (2*a*p*AppellF1[3/2 - p, -1/2, 1 - p, 5/2 - p, -Cot[e + f*x]^2, -((a*Cot[e +
f*x]^2)/b)] + b*AppellF1[3/2 - p, 1/2, -p, 5/2 - p, -Cot[e + f*x]^2, -((a*Cot[e + f*x]^2)/b)])*Cot[e + f*x]^2
))

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Maple [F]  time = 0.228, size = 0, normalized size = 0. \begin{align*} \int \left ( \csc \left ( fx+e \right ) \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^p,x)

[Out]

int(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^p,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^p,x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e)^2 + a)^p*csc(f*x + e)^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

integral((b*tan(f*x + e)^2 + a)^p*csc(f*x + e)^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(csc(f*x+e)**3*(a+b*tan(f*x+e)**2)**p,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^p,x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e)^2 + a)^p*csc(f*x + e)^3, x)