3.154 \(\int \sin ^5(e+f x) (a+b \tan ^2(e+f x))^p \, dx\)

Optimal. Leaf size=208 \[ -\frac{\left (15 a^2-20 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^p \left (\frac{b \sec ^2(e+f x)}{a-b}+1\right )^{-p} \text{Hypergeometric2F1}\left (-\frac{1}{2},-p,\frac{1}{2},-\frac{b \sec ^2(e+f x)}{a-b}\right )}{15 f (a-b)^2}-\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{p+1}}{5 f (a-b)}+\frac{(10 a-2 b p-7 b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{p+1}}{15 f (a-b)^2} \]

[Out]

((10*a - 7*b - 2*b*p)*Cos[e + f*x]^3*(a - b + b*Sec[e + f*x]^2)^(1 + p))/(15*(a - b)^2*f) - (Cos[e + f*x]^5*(a
 - b + b*Sec[e + f*x]^2)^(1 + p))/(5*(a - b)*f) - ((15*a^2 - 20*a*b*(1 + p) + 4*b^2*(2 + 3*p + p^2))*Cos[e + f
*x]*Hypergeometric2F1[-1/2, -p, 1/2, -((b*Sec[e + f*x]^2)/(a - b))]*(a - b + b*Sec[e + f*x]^2)^p)/(15*(a - b)^
2*f*(1 + (b*Sec[e + f*x]^2)/(a - b))^p)

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Rubi [A]  time = 0.222801, antiderivative size = 208, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3664, 462, 453, 365, 364} \[ -\frac{\left (15 a^2-20 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^p \left (\frac{b \sec ^2(e+f x)}{a-b}+1\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b \sec ^2(e+f x)}{a-b}\right )}{15 f (a-b)^2}-\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{p+1}}{5 f (a-b)}+\frac{(10 a-2 b p-7 b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{p+1}}{15 f (a-b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^5*(a + b*Tan[e + f*x]^2)^p,x]

[Out]

((10*a - 7*b - 2*b*p)*Cos[e + f*x]^3*(a - b + b*Sec[e + f*x]^2)^(1 + p))/(15*(a - b)^2*f) - (Cos[e + f*x]^5*(a
 - b + b*Sec[e + f*x]^2)^(1 + p))/(5*(a - b)*f) - ((15*a^2 - 20*a*b*(1 + p) + 4*b^2*(2 + 3*p + p^2))*Cos[e + f
*x]*Hypergeometric2F1[-1/2, -p, 1/2, -((b*Sec[e + f*x]^2)/(a - b))]*(a - b + b*Sec[e + f*x]^2)^p)/(15*(a - b)^
2*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 462

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(c^2*(e*x)^(
m + 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 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 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 ^5(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 )^2 \left (a-b+b x^2\right )^p}{x^6} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{5 (a-b) f}+\frac{\operatorname{Subst}\left (\int \frac{\left (-10 a+b (7+2 p)+5 (a-b) x^2\right ) \left (a-b+b x^2\right )^p}{x^4} \, dx,x,\sec (e+f x)\right )}{5 (a-b) f}\\ &=\frac{(10 a-7 b-2 b p) \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{15 (a-b)^2 f}-\frac{\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{5 (a-b) f}+\frac{\left (15 a^2-20 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \operatorname{Subst}\left (\int \frac{\left (a-b+b x^2\right )^p}{x^2} \, dx,x,\sec (e+f x)\right )}{15 (a-b)^2 f}\\ &=\frac{(10 a-7 b-2 b p) \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{15 (a-b)^2 f}-\frac{\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{5 (a-b) f}+\frac{\left (\left (15 a^2-20 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\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}\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 )}{15 (a-b)^2 f}\\ &=\frac{(10 a-7 b-2 b p) \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{15 (a-b)^2 f}-\frac{\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{5 (a-b) f}-\frac{\left (15 a^2-20 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \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}}{15 (a-b)^2 f}\\ \end{align*}

Mathematica [A]  time = 7.42739, size = 283, normalized size = 1.36 \[ -\frac{2^{p+3} \sin ^4(e+f x) \cos (e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (\left (15 a^2-20 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \text{Hypergeometric2F1}\left (-\frac{1}{2},-p,\frac{1}{2},-\frac{b \sec ^2(e+f x)}{a-b}\right )+\frac{1}{4} ((a-b) \cos (2 (e+f x))+a+b) (3 (a-b) \cos (2 (e+f x))-17 a+b (4 p+11)) \left (\frac{a+b \tan ^2(e+f x)}{a-b}\right )^p\right )}{15 f (a-b)^2 \left (-2^{p+2} \cos (2 (e+f x)) \left (\frac{a+b \tan ^2(e+f x)}{a-b}\right )^p+2^p \cos (4 (e+f x)) \left (\frac{a+b \tan ^2(e+f x)}{a-b}\right )^p+3 \left (\frac{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{a-b}\right )^p\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sin[e + f*x]^5*(a + b*Tan[e + f*x]^2)^p,x]

[Out]

-(2^(3 + p)*Cos[e + f*x]*Sin[e + f*x]^4*(a + b*Tan[e + f*x]^2)^p*((15*a^2 - 20*a*b*(1 + p) + 4*b^2*(2 + 3*p +
p^2))*Hypergeometric2F1[-1/2, -p, 1/2, -((b*Sec[e + f*x]^2)/(a - b))] + ((a + b + (a - b)*Cos[2*(e + f*x)])*(-
17*a + b*(11 + 4*p) + 3*(a - b)*Cos[2*(e + f*x)])*((a + b*Tan[e + f*x]^2)/(a - b))^p)/4))/(15*(a - b)^2*f*(3*(
((a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2)/(a - b))^p - 2^(2 + p)*Cos[2*(e + f*x)]*((a + b*Tan[e + f*
x]^2)/(a - b))^p + 2^p*Cos[4*(e + f*x)]*((a + b*Tan[e + f*x]^2)/(a - b))^p))

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Maple [F]  time = 0.885, size = 0, normalized size = 0. \begin{align*} \int \left ( \sin \left ( fx+e \right ) \right ) ^{5} \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)^5*(a+b*tan(f*x+e)^2)^p,x)

[Out]

int(sin(f*x+e)^5*(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 )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5*(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)^5, 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 )^{4} - 2 \, \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)^5*(a+b*tan(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

integral((cos(f*x + e)^4 - 2*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)**5*(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 )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5*(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)^5, x)