3.106 \(\int \frac{\sin ^5(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=171 \[ -\frac{2 b \left (15 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{15 a^4 f \sqrt{a+b \sec ^2(e+f x)}}-\frac{\left (15 a^2+40 a b+24 b^2\right ) \cos (e+f x)}{15 a^3 f \sqrt{a+b \sec ^2(e+f x)}}+\frac{2 (5 a+3 b) \cos ^3(e+f x)}{15 a^2 f \sqrt{a+b \sec ^2(e+f x)}}-\frac{\cos ^5(e+f x)}{5 a f \sqrt{a+b \sec ^2(e+f x)}} \]

[Out]

-((15*a^2 + 40*a*b + 24*b^2)*Cos[e + f*x])/(15*a^3*f*Sqrt[a + b*Sec[e + f*x]^2]) + (2*(5*a + 3*b)*Cos[e + f*x]
^3)/(15*a^2*f*Sqrt[a + b*Sec[e + f*x]^2]) - Cos[e + f*x]^5/(5*a*f*Sqrt[a + b*Sec[e + f*x]^2]) - (2*b*(15*a^2 +
40*a*b + 24*b^2)*Sec[e + f*x])/(15*a^4*f*Sqrt[a + b*Sec[e + f*x]^2])

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Rubi [A]  time = 0.187001, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4134, 462, 453, 271, 191} \[ -\frac{2 b \left (15 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{15 a^4 f \sqrt{a+b \sec ^2(e+f x)}}+\frac{2 (5 a+3 b) \cos ^3(e+f x)}{15 a^2 f \sqrt{a+b \sec ^2(e+f x)}}-\frac{\left (\frac{8 b (5 a+3 b)}{a^2}+15\right ) \cos (e+f x)}{15 a f \sqrt{a+b \sec ^2(e+f x)}}-\frac{\cos ^5(e+f x)}{5 a f \sqrt{a+b \sec ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^5/(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

-((15 + (8*b*(5*a + 3*b))/a^2)*Cos[e + f*x])/(15*a*f*Sqrt[a + b*Sec[e + f*x]^2]) + (2*(5*a + 3*b)*Cos[e + f*x]
^3)/(15*a^2*f*Sqrt[a + b*Sec[e + f*x]^2]) - Cos[e + f*x]^5/(5*a*f*Sqrt[a + b*Sec[e + f*x]^2]) - (2*b*(15*a^2 +
40*a*b + 24*b^2)*Sec[e + f*x])/(15*a^4*f*Sqrt[a + b*Sec[e + f*x]^2])

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 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 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int \frac{\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^2}{x^6 \left (a+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cos ^5(e+f x)}{5 a f \sqrt{a+b \sec ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{-2 (5 a+3 b)+5 a x^2}{x^4 \left (a+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{5 a f}\\ &=\frac{2 (5 a+3 b) \cos ^3(e+f x)}{15 a^2 f \sqrt{a+b \sec ^2(e+f x)}}-\frac{\cos ^5(e+f x)}{5 a f \sqrt{a+b \sec ^2(e+f x)}}-\frac{\left (-15 a^2-8 b (5 a+3 b)\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{15 a^2 f}\\ &=-\frac{\left (15 a^2+8 b (5 a+3 b)\right ) \cos (e+f x)}{15 a^3 f \sqrt{a+b \sec ^2(e+f x)}}+\frac{2 (5 a+3 b) \cos ^3(e+f x)}{15 a^2 f \sqrt{a+b \sec ^2(e+f x)}}-\frac{\cos ^5(e+f x)}{5 a f \sqrt{a+b \sec ^2(e+f x)}}+\frac{\left (2 b \left (-15 a^2-8 b (5 a+3 b)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{15 a^3 f}\\ &=-\frac{\left (15 a^2+8 b (5 a+3 b)\right ) \cos (e+f x)}{15 a^3 f \sqrt{a+b \sec ^2(e+f x)}}+\frac{2 (5 a+3 b) \cos ^3(e+f x)}{15 a^2 f \sqrt{a+b \sec ^2(e+f x)}}-\frac{\cos ^5(e+f x)}{5 a f \sqrt{a+b \sec ^2(e+f x)}}-\frac{2 b \left (15 a^2+8 b (5 a+3 b)\right ) \sec (e+f x)}{15 a^4 f \sqrt{a+b \sec ^2(e+f x)}}\\ \end{align*}

Mathematica [B]  time = 7.61137, size = 432, normalized size = 2.53 \[ -\frac{\sec ^3(e+f x) \left (-2 a^2 \cos (4 (e+f x))+27 a^2+16 a (a+2 b) \cos (2 (e+f x))+128 a b+128 b^2\right ) (a \cos (2 e+2 f x)+a+2 b)^{3/2}}{192 a^3 f \sqrt{a \cos (2 (e+f x))+a+2 b} \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\sec ^3(e+f x) \left (a \left (25 a^2+128 a b+128 b^2\right ) \cos (2 (e+f x))-4 a^2 (a+2 b) \cos (4 (e+f x))+336 a^2 b+a^3 \cos (6 (e+f x))+40 a^3+768 a b^2+512 b^3\right ) (a \cos (2 e+2 f x)+a+2 b)^{3/2}}{320 a^4 f \sqrt{a \cos (2 (e+f x))+a+2 b} \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac{\sec ^3(e+f x) (a \cos (2 (e+f x))+2 a+4 b) (a \cos (2 e+2 f x)+a+2 b)^{3/2}}{32 a^2 f \sqrt{a \cos (2 (e+f x))+a+2 b} \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac{3 \sec ^3(e+f x) (a \cos (2 e+2 f x)+a+2 b)^{3/2}}{64 a f \sqrt{a \cos (2 (e+f x))+a+2 b} \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^5/(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

(3*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2)*Sec[e + f*x]^3)/(64*a*f*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(a + b*Sec[
e + f*x]^2)^(3/2)) + ((2*a + 4*b + a*Cos[2*(e + f*x)])*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2)*Sec[e + f*x]^3)/(3
2*a^2*f*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(a + b*Sec[e + f*x]^2)^(3/2)) - ((27*a^2 + 128*a*b + 128*b^2 + 16*a
*(a + 2*b)*Cos[2*(e + f*x)] - 2*a^2*Cos[4*(e + f*x)])*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2)*Sec[e + f*x]^3)/(19
2*a^3*f*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(a + b*Sec[e + f*x]^2)^(3/2)) - ((40*a^3 + 336*a^2*b + 768*a*b^2 +
512*b^3 + a*(25*a^2 + 128*a*b + 128*b^2)*Cos[2*(e + f*x)] - 4*a^2*(a + 2*b)*Cos[4*(e + f*x)] + a^3*Cos[6*(e +
f*x)])*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2)*Sec[e + f*x]^3)/(320*a^4*f*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*(a +
b*Sec[e + f*x]^2)^(3/2))

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Maple [B]  time = 2.064, size = 35190, normalized size = 205.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^(3/2),x)

[Out]

result too large to display

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Maxima [A]  time = 1.04516, size = 338, normalized size = 1.98 \begin{align*} -\frac{\frac{15 \, \sqrt{a + \frac{b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{2}} - \frac{10 \,{\left ({\left (a + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} \cos \left (f x + e\right )^{3} - 6 \, \sqrt{a + \frac{b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )\right )}}{a^{3}} + \frac{15 \, b}{\sqrt{a + \frac{b}{\cos \left (f x + e\right )^{2}}} a^{2} \cos \left (f x + e\right )} + \frac{30 \, b^{2}}{\sqrt{a + \frac{b}{\cos \left (f x + e\right )^{2}}} a^{3} \cos \left (f x + e\right )} + \frac{15 \, b^{3}}{\sqrt{a + \frac{b}{\cos \left (f x + e\right )^{2}}} a^{4} \cos \left (f x + e\right )} + \frac{3 \,{\left ({\left (a + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{5}{2}} \cos \left (f x + e\right )^{5} - 5 \,{\left (a + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} b \cos \left (f x + e\right )^{3} + 15 \, \sqrt{a + \frac{b}{\cos \left (f x + e\right )^{2}}} b^{2} \cos \left (f x + e\right )\right )}}{a^{4}}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

-1/15*(15*sqrt(a + b/cos(f*x + e)^2)*cos(f*x + e)/a^2 - 10*((a + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3 - 6*sq
rt(a + b/cos(f*x + e)^2)*b*cos(f*x + e))/a^3 + 15*b/(sqrt(a + b/cos(f*x + e)^2)*a^2*cos(f*x + e)) + 30*b^2/(sq
rt(a + b/cos(f*x + e)^2)*a^3*cos(f*x + e)) + 15*b^3/(sqrt(a + b/cos(f*x + e)^2)*a^4*cos(f*x + e)) + 3*((a + b/
cos(f*x + e)^2)^(5/2)*cos(f*x + e)^5 - 5*(a + b/cos(f*x + e)^2)^(3/2)*b*cos(f*x + e)^3 + 15*sqrt(a + b/cos(f*x
+ e)^2)*b^2*cos(f*x + e))/a^4)/f

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Fricas [A]  time = 0.863724, size = 324, normalized size = 1.89 \begin{align*} -\frac{{\left (3 \, a^{3} \cos \left (f x + e\right )^{7} - 2 \,{\left (5 \, a^{3} + 3 \, a^{2} b\right )} \cos \left (f x + e\right )^{5} +{\left (15 \, a^{3} + 40 \, a^{2} b + 24 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} + 2 \,{\left (15 \, a^{2} b + 40 \, a b^{2} + 24 \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \,{\left (a^{5} f \cos \left (f x + e\right )^{2} + a^{4} b f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

-1/15*(3*a^3*cos(f*x + e)^7 - 2*(5*a^3 + 3*a^2*b)*cos(f*x + e)^5 + (15*a^3 + 40*a^2*b + 24*a*b^2)*cos(f*x + e)
^3 + 2*(15*a^2*b + 40*a*b^2 + 24*b^3)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(a^5*f*cos(f*x
+ e)^2 + a^4*b*f)

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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*sec(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{5}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate(sin(f*x + e)^5/(b*sec(f*x + e)^2 + a)^(3/2), x)