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

Optimal. Leaf size=140 \[ \frac{\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{p+1}}{3 f (a-b)}-\frac{(3 a-2 b (p+1)) \cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^p \left (\frac{b \sec ^2(e+f x)}{a-b}+1\right )^{-p} \text{Hypergeometric2F1}\left (-\frac{1}{2},-p,\frac{1}{2},-\frac{b \sec ^2(e+f x)}{a-b}\right )}{3 f (a-b)} \]

[Out]

(Cos[e + f*x]^3*(a - b + b*Sec[e + f*x]^2)^(1 + p))/(3*(a - b)*f) - ((3*a - 2*b*(1 + p))*Cos[e + f*x]*Hypergeo
metric2F1[-1/2, -p, 1/2, -((b*Sec[e + f*x]^2)/(a - b))]*(a - b + b*Sec[e + f*x]^2)^p)/(3*(a - b)*f*(1 + (b*Sec
[e + f*x]^2)/(a - b))^p)

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Rubi [A]  time = 0.111393, antiderivative size = 140, normalized size of antiderivative = 1., 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 = {3664, 453, 365, 364} \[ \frac{\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{p+1}}{3 f (a-b)}-\frac{(3 a-2 b (p+1)) \cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^p \left (\frac{b \sec ^2(e+f x)}{a-b}+1\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b \sec ^2(e+f x)}{a-b}\right )}{3 f (a-b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]^3*(a - b + b*Sec[e + f*x]^2)^(1 + p))/(3*(a - b)*f) - ((3*a - 2*b*(1 + p))*Cos[e + f*x]*Hypergeo
metric2F1[-1/2, -p, 1/2, -((b*Sec[e + f*x]^2)/(a - b))]*(a - b + b*Sec[e + f*x]^2)^p)/(3*(a - b)*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 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 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 \sin ^3(e+f x) \left (a+b \tan ^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,\sec (e+f x)\right )}{f}\\ &=\frac{\cos ^3(e+f x) \left (a-b+b \sec ^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,\sec (e+f x)\right )}{3 (a-b) f}\\ &=\frac{\cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{3 (a-b) f}+\frac{\left ((3 a-2 b (1+p)) \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{\left (1+\frac{b x^2}{a-b}\right )^p}{x^2} \, dx,x,\sec (e+f x)\right )}{3 (a-b) f}\\ &=\frac{\cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{3 (a-b) f}-\frac{(3 a-2 b (1+p)) \cos (e+f x) \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b \sec ^2(e+f x)}{a-b}\right ) \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 (a-b) f}\\ \end{align*}

Mathematica [A]  time = 3.87714, size = 184, normalized size = 1.31 \[ \frac{\sin (e+f x) \tan (e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left ((2 b (p+1)-3 a) \text{Hypergeometric2F1}\left (-\frac{1}{2},-p,\frac{1}{2},-\frac{b \sec ^2(e+f x)}{a-b}\right )+\left (\frac{a+b \tan ^2(e+f x)}{a-b}\right )^p \left (a \cos ^2(e+f x)+b \sin ^2(e+f x)\right )\right )}{f \left (3 a \sec ^2(e+f x) \left (\frac{a+b \sec ^2(e+f x)-b}{a-b}\right )^p-3 (a-b) \left (\frac{a+b \tan ^2(e+f x)}{a-b}\right )^{p+1}\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.583, size = 0, normalized size = 0. \begin{align*} \int \left ( \sin \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(sin(f*x+e)^3*(a+b*tan(f*x+e)^2)^p,x)

[Out]

int(sin(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} \sin \left (f x + e\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(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*sin(f*x + e)^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (\cos \left (f x + e\right )^{2} - 1\right )}{\left (b \tan \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(sin(f*x+e)^3*(a+b*tan(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

integral(-(cos(f*x + e)^2 - 1)*(b*tan(f*x + e)^2 + a)^p*sin(f*x + e), 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(sin(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} \sin \left (f x + e\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(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*sin(f*x + e)^3, x)