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

Optimal. Leaf size=68 \[ -\frac{\cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \left (\frac{b \sec ^2(e+f x)}{a}+1\right )^{-p} \text{Hypergeometric2F1}\left (-\frac{1}{2},-p,\frac{1}{2},-\frac{b \sec ^2(e+f x)}{a}\right )}{f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0486125, antiderivative size = 68, normalized size of antiderivative = 1., 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 = {4134, 365, 364} \[ -\frac{\cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \left (\frac{b \sec ^2(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b \sec ^2(e+f x)}{a}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4134

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Cos[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^
n)^p)/x^(m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (Gt
Q[m, 0] || EqQ[n, 2] || EqQ[n, 4])

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 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])

Rubi steps

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

Mathematica [A]  time = 1.66542, size = 68, normalized size = 1. \[ -\frac{\cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \left (\frac{b \sec ^2(e+f x)}{a}+1\right )^{-p} \text{Hypergeometric2F1}\left (-\frac{1}{2},-p,\frac{1}{2},-\frac{b \sec ^2(e+f x)}{a}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.497, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2} \right ) ^{p}\sin \left ( fx+e \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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