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]

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

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Rubi [A]  time = 0.13, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 1.19, 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]

-1/12*((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])/(S
qrt[2]*(a - b)^3*f*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])

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fricas [A]  time = 0.65, size = 158, normalized size = 1.21 \[ \frac {{\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (3 \, a^{2} - 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{3} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, {\left ({\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}\right )} f\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (f x + e\right )^{3}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

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)

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maple [B]  time = 1.99, size = 14991, normalized size = 114.44 \[ \text {output too large to display} \]

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 = 0.76, size = 216, normalized size = 1.65 \[ -\frac {\frac {3 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {{\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 6 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {3 \, b^{2}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )} + \frac {3 \, b}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}}{3 \, f} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\sin \left (e+f\,x\right )}^3}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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