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

Optimal. Leaf size=131 \[ -\frac{2 b (3 a+b) \sec (e+f x)}{3 f (a-b)^3 \sqrt{a+b \sec ^2(e+f x)-b}}+\frac{\cos ^3(e+f x)}{3 f (a-b) \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{(3 a+b) \cos (e+f x)}{3 f (a-b)^2 \sqrt{a+b \sec ^2(e+f x)-b}} \]

[Out]

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

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Rubi [A]  time = 0.134308, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3664, 453, 271, 191} \[ -\frac{2 b (3 a+b) \sec (e+f x)}{3 f (a-b)^3 \sqrt{a+b \sec ^2(e+f x)-b}}+\frac{\cos ^3(e+f x)}{3 f (a-b) \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{(3 a+b) \cos (e+f x)}{3 f (a-b)^2 \sqrt{a+b \sec ^2(e+f x)-b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 1.13964, size = 106, normalized size = 0.81 \[ -\frac{\sec (e+f x) \left (8 \left (a^2-b^2\right ) \cos (2 (e+f x))+9 a^2-(a-b)^2 \cos (4 (e+f x))+46 a b+9 b^2\right )}{12 \sqrt{2} f (a-b)^3 \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((9*a^2 + 46*a*b + 9*b^2 + 8*(a^2 - b^2)*Cos[2*(e + f*x)] - (a - b)^2*Cos[4*(e + f*x)])*Sec[e + f*x])/(12*Sqr
t[2]*(a - b)^3*f*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])

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Maple [B]  time = 1.089, size = 14991, normalized size = 114.4 \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)^3/(a+b*tan(f*x+e)^2)^(3/2),x)

[Out]

result too large to display

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Maxima [A]  time = 1.09813, size = 292, normalized size = 2.23 \begin{align*} -\frac{\frac{3 \, \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{{\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 )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac{3 \, 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{3 \, 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 )}}{3 \, f} \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)^(3/2),x, algorithm="maxima")

[Out]

-1/3*(3*sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e)/(a^2 - 2*a*b + b^2) - ((a - b + b/cos(f*x + e)^2)^(3/2)*co
s(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) + 3*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)) + 3*b/((a^2 - 2*a*b + b^2)*sqrt(a - b
+ b/cos(f*x + e)^2)*cos(f*x + e)))/f

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Fricas [A]  time = 2.74309, size = 358, normalized size = 2.73 \begin{align*} \frac{{\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} -{\left (3 \, a^{2} - 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{3} - 2 \,{\left (3 \, a b + b^{2}\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}}}}{3 \,{\left ({\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}\right )} f\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)^(3/2),x, algorithm="fricas")

[Out]

1/3*((a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - (3*a^2 - 2*a*b - b^2)*cos(f*x + e)^3 - 2*(3*a*b + b^2)*cos(f*x + e))
*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*f*cos(f*x + e)
^2 + (a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4)*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)**3/(a+b*tan(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 )^{3}}{{\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)^3/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

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