\(\int (a+b \sec ^2(e+f x))^p \sin ^3(e+f x) \, dx\) [134]

Optimal result

Integrand size = 23, antiderivative size = 117 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^3(e+f x) \, dx=\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{3 a f}-\frac {(3 a+b-2 b p) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sec ^2(e+f x)}{a}\right ) \left (a+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \sec ^2(e+f x)}{a}\right )^{-p}}{3 a f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {4219, 464, 372, 371} \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^3(e+f x) \, dx=\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{3 a f}-\frac {(3 a-2 b p+b) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \left (\frac {b \sec ^2(e+f x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sec ^2(e+f x)}{a}\right )}{3 a f} \]

[In]

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

[Out]

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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 464

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 4219

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])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (-1+x^2\right ) \left (a+b x^2\right )^p}{x^4} \, dx,x,\sec (e+f x)\right )}{f} \\ & = \frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{3 a f}+\frac {(3 a+b-2 b p) \text {Subst}\left (\int \frac {\left (a+b x^2\right )^p}{x^2} \, dx,x,\sec (e+f x)\right )}{3 a f} \\ & = \frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{3 a f}+\frac {\left ((3 a+b-2 b p) \left (a+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \sec ^2(e+f x)}{a}\right )^{-p}\right ) \text {Subst}\left (\int \frac {\left (1+\frac {b x^2}{a}\right )^p}{x^2} \, dx,x,\sec (e+f x)\right )}{3 a f} \\ & = \frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{3 a f}-\frac {(3 a+b-2 b p) \cos (e+f x) \operatorname {Hypergeometric2F1}\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}}{3 a f} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.40 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.52 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^3(e+f x) \, dx=-\frac {\cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \sin ^2(e+f x) \left (-2 (3 a+b-2 b p) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sec ^2(e+f x)}{a}\right )+(a+2 b+a \cos (2 (e+f x))) \left (\frac {a+b+b \tan ^2(e+f x)}{a}\right )^p\right )}{3 a f \left (-2 \left (1+\frac {b \sec ^2(e+f x)}{a}\right )^p+\left (\frac {a+b+b \tan ^2(e+f x)}{a}\right )^p+\cos (2 (e+f x)) \left (\frac {a+b+b \tan ^2(e+f x)}{a}\right )^p\right )} \]

[In]

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

[Out]

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

Maple [F]

\[\int \left (a +b \sec \left (f x +e \right )^{2}\right )^{p} \sin \left (f x +e \right )^{3}d x\]

[In]

int((a+b*sec(f*x+e)^2)^p*sin(f*x+e)^3,x)

[Out]

int((a+b*sec(f*x+e)^2)^p*sin(f*x+e)^3,x)

Fricas [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^3(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{3} \,d x } \]

[In]

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

[Out]

integral(-(cos(f*x + e)^2 - 1)*(b*sec(f*x + e)^2 + a)^p*sin(f*x + e), x)

Sympy [F(-1)]

Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^3(e+f x) \, dx=\text {Timed out} \]

[In]

integrate((a+b*sec(f*x+e)**2)**p*sin(f*x+e)**3,x)

[Out]

Timed out

Maxima [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^3(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{3} \,d x } \]

[In]

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

[Out]

integrate((b*sec(f*x + e)^2 + a)^p*sin(f*x + e)^3, x)

Giac [F]

\[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^3(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{3} \,d x } \]

[In]

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

[Out]

integrate((b*sec(f*x + e)^2 + a)^p*sin(f*x + e)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^3(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^3\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p \,d x \]

[In]

int(sin(e + f*x)^3*(a + b/cos(e + f*x)^2)^p,x)

[Out]

int(sin(e + f*x)^3*(a + b/cos(e + f*x)^2)^p, x)