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

Optimal. Leaf size=88 \[ \frac{\cos ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{3 f (a-b)}-\frac{(3 a-b) \cos (e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{3 f (a-b)^2} \]

[Out]

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

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Rubi [A]  time = 0.101122, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {3664, 453, 264} \[ \frac{\cos ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{3 f (a-b)}-\frac{(3 a-b) \cos (e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{3 f (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 264

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

Rubi steps

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

Mathematica [A]  time = 1.43294, size = 74, normalized size = 0.84 \[ \frac{\cos (e+f x) ((a-b) \cos (2 (e+f x))-5 a+b) \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}}{6 \sqrt{2} f (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]*(-5*a + b + (a - b)*Cos[2*(e + f*x)])*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(
6*Sqrt[2]*(a - b)^2*f)

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Maple [A]  time = 0.221, size = 104, normalized size = 1.2 \begin{align*}{\frac{ \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b-3\,a+b \right ) }{3\,f \left ( a-b \right ) ^{2}\cos \left ( fx+e \right ) }{\frac{1}{\sqrt{{\frac{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b}{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}}}} \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)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.02708, size = 143, normalized size = 1.62 \begin{align*} -\frac{\frac{3 \, \sqrt{a - b + \frac{b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a - b} - \frac{{\left (a - b + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} \cos \left (f x + e\right )^{3} - 3 \, \sqrt{a - b + \frac{b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )}{a^{2} - 2 \, a b + b^{2}}}{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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.85646, size = 174, normalized size = 1.98 \begin{align*} \frac{{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} -{\left (3 \, a - b\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 (a^{2} - 2 \, a b + b^{2}\right )} 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)^(1/2),x, algorithm="fricas")

[Out]

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

[Out]

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