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

Optimal. Leaf size=199 \[ -\frac{2 b \left (15 a^2+10 a b-b^2\right ) \sec (e+f x)}{15 f (a-b)^4 \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{\left (15 a^2+10 a b-b^2\right ) \cos (e+f x)}{15 f (a-b)^3 \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{\cos ^5(e+f x)}{5 f (a-b) \sqrt{a+b \sec ^2(e+f x)-b}}+\frac{2 (5 a-2 b) \cos ^3(e+f x)}{15 f (a-b)^2 \sqrt{a+b \sec ^2(e+f x)-b}} \]

[Out]

-((15*a^2 + 10*a*b - b^2)*Cos[e + f*x])/(15*(a - b)^3*f*Sqrt[a - b + b*Sec[e + f*x]^2]) + (2*(5*a - 2*b)*Cos[e
+ f*x]^3)/(15*(a - b)^2*f*Sqrt[a - b + b*Sec[e + f*x]^2]) - Cos[e + f*x]^5/(5*(a - b)*f*Sqrt[a - b + b*Sec[e
+ f*x]^2]) - (2*b*(15*a^2 + 10*a*b - b^2)*Sec[e + f*x])/(15*(a - b)^4*f*Sqrt[a - b + b*Sec[e + f*x]^2])

________________________________________________________________________________________

Rubi [A]  time = 0.187187, antiderivative size = 199, 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 = {3664, 462, 453, 271, 191} \[ -\frac{2 b \left (15 a^2+10 a b-b^2\right ) \sec (e+f x)}{15 f (a-b)^4 \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{\left (15 a^2+10 a b-b^2\right ) \cos (e+f x)}{15 f (a-b)^3 \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{\cos ^5(e+f x)}{5 f (a-b) \sqrt{a+b \sec ^2(e+f x)-b}}+\frac{2 (5 a-2 b) \cos ^3(e+f x)}{15 f (a-b)^2 \sqrt{a+b \sec ^2(e+f x)-b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((15*a^2 + 10*a*b - b^2)*Cos[e + f*x])/(15*(a - b)^3*f*Sqrt[a - b + b*Sec[e + f*x]^2]) + (2*(5*a - 2*b)*Cos[e
+ f*x]^3)/(15*(a - b)^2*f*Sqrt[a - b + b*Sec[e + f*x]^2]) - Cos[e + f*x]^5/(5*(a - b)*f*Sqrt[a - b + b*Sec[e
+ f*x]^2]) - (2*b*(15*a^2 + 10*a*b - b^2)*Sec[e + f*x])/(15*(a - b)^4*f*Sqrt[a - b + b*Sec[e + f*x]^2])

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 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 \tan ^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+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cos ^5(e+f x)}{5 (a-b) f \sqrt{a-b+b \sec ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{-2 (5 a-2 b)+5 (a-b) x^2}{x^4 \left (a-b+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{5 (a-b) f}\\ &=\frac{2 (5 a-2 b) \cos ^3(e+f x)}{15 (a-b)^2 f \sqrt{a-b+b \sec ^2(e+f x)}}-\frac{\cos ^5(e+f x)}{5 (a-b) f \sqrt{a-b+b \sec ^2(e+f x)}}+\frac{\left (15 a^2+10 a b-b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a-b+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{15 (a-b)^2 f}\\ &=-\frac{\left (15 a^2+10 a b-b^2\right ) \cos (e+f x)}{15 (a-b)^3 f \sqrt{a-b+b \sec ^2(e+f x)}}+\frac{2 (5 a-2 b) \cos ^3(e+f x)}{15 (a-b)^2 f \sqrt{a-b+b \sec ^2(e+f x)}}-\frac{\cos ^5(e+f x)}{5 (a-b) f \sqrt{a-b+b \sec ^2(e+f x)}}-\frac{\left (2 b \left (15 a^2+10 a b-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a-b+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{15 (a-b)^3 f}\\ &=-\frac{\left (15 a^2+10 a b-b^2\right ) \cos (e+f x)}{15 (a-b)^3 f \sqrt{a-b+b \sec ^2(e+f x)}}+\frac{2 (5 a-2 b) \cos ^3(e+f x)}{15 (a-b)^2 f \sqrt{a-b+b \sec ^2(e+f x)}}-\frac{\cos ^5(e+f x)}{5 (a-b) f \sqrt{a-b+b \sec ^2(e+f x)}}-\frac{2 b \left (15 a^2+10 a b-b^2\right ) \sec (e+f x)}{15 (a-b)^4 f \sqrt{a-b+b \sec ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.84405, size = 186, normalized size = 0.93 \[ -\frac{\sec (e+f x) \left (\left (169 a^2 b+125 a^3-329 a b^2+35 b^3\right ) \cos (2 (e+f x))-9 a^2 b \cos (6 (e+f x))+1078 a^2 b+3 a^3 \cos (6 (e+f x))+150 a^3+9 a b^2 \cos (6 (e+f x))+338 a b^2-2 (a-b)^2 (11 a+b) \cos (4 (e+f x))-3 b^3 \cos (6 (e+f x))-30 b^3\right )}{240 \sqrt{2} f (a-b)^4 \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((150*a^3 + 1078*a^2*b + 338*a*b^2 - 30*b^3 + (125*a^3 + 169*a^2*b - 329*a*b^2 + 35*b^3)*Cos[2*(e + f*x)] - 2
*(a - b)^2*(11*a + b)*Cos[4*(e + f*x)] + 3*a^3*Cos[6*(e + f*x)] - 9*a^2*b*Cos[6*(e + f*x)] + 9*a*b^2*Cos[6*(e
+ f*x)] - 3*b^3*Cos[6*(e + f*x)])*Sec[e + f*x])/(240*Sqrt[2]*(a - b)^4*f*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)
])*Sec[e + f*x]^2])

________________________________________________________________________________________

Maple [B]  time = 3.807, size = 67748, normalized size = 340.4 \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*tan(f*x+e)^2)^(3/2),x)

[Out]

result too large to display

________________________________________________________________________________________

Maxima [B]  time = 1.14598, size = 525, normalized size = 2.64 \begin{align*} -\frac{\frac{15 \, b^{3}}{{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \sqrt{a - b + \frac{b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )} + \frac{15 \, \sqrt{a - b + \frac{b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{2} - 2 \, a b + b^{2}} + \frac{3 \,{\left ({\left (a - b + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{5}{2}} \cos \left (f x + e\right )^{5} - 5 \,{\left (a - b + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} b \cos \left (f x + e\right )^{3} + 15 \, \sqrt{a - b + \frac{b}{\cos \left (f x + e\right )^{2}}} b^{2} \cos \left (f x + e\right )\right )}}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} - \frac{10 \,{\left ({\left (a - b + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} \cos \left (f x + e\right )^{3} - 6 \, \sqrt{a - b + \frac{b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac{30 \, b^{2}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt{a - b + \frac{b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )} + \frac{15 \, b}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt{a - b + \frac{b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}}{15 \, f} \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)^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 3.59056, size = 528, normalized size = 2.65 \begin{align*} -\frac{{\left (3 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{7} - 2 \,{\left (5 \, a^{3} - 12 \, a^{2} b + 9 \, a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )^{5} +{\left (15 \, a^{3} - 5 \, a^{2} b - 11 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} + 2 \,{\left (15 \, a^{2} b + 10 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \,{\left ({\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{4} b - 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 4 \, a b^{4} + b^{5}\right )} f\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)^(3/2),x, algorithm="fricas")

[Out]

-1/15*(3*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^7 - 2*(5*a^3 - 12*a^2*b + 9*a*b^2 - 2*b^3)*cos(f*x + e)^
5 + (15*a^3 - 5*a^2*b - 11*a*b^2 + b^3)*cos(f*x + e)^3 + 2*(15*a^2*b + 10*a*b^2 - b^3)*cos(f*x + e))*sqrt(((a
- b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*f*cos(f*x
+ e)^2 + (a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)*f)

________________________________________________________________________________________

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)**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{5}}{{\left (b \tan \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*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

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