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

Optimal. Leaf size=168 \[ -\frac{8 b (a+b) \sec (e+f x)}{3 f (a-b)^4 \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{4 b (a+b) \sec (e+f x)}{3 f (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}+\frac{\cos ^3(e+f x)}{3 f (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac{(a+b) \cos (e+f x)}{f (a-b)^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.159306, antiderivative size = 168, 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, 453, 271, 192, 191} \[ -\frac{8 b (a+b) \sec (e+f x)}{3 f (a-b)^4 \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{4 b (a+b) \sec (e+f x)}{3 f (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}+\frac{\cos ^3(e+f x)}{3 f (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac{(a+b) \cos (e+f x)}{f (a-b)^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a + b)*Cos[e + f*x])/((a - b)^2*f*(a - b + b*Sec[e + f*x]^2)^(3/2))) + Cos[e + f*x]^3/(3*(a - b)*f*(a - b
+ b*Sec[e + f*x]^2)^(3/2)) - (4*b*(a + b)*Sec[e + f*x])/(3*(a - b)^3*f*(a - b + b*Sec[e + f*x]^2)^(3/2)) - (8*
b*(a + b)*Sec[e + f*x])/(3*(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 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 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -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 )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{-1+x^2}{x^4 \left (a-b+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac{\cos ^3(e+f x)}{3 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}+\frac{(a+b) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a-b+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{(a-b) f}\\ &=-\frac{(a+b) \cos (e+f x)}{(a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}+\frac{\cos ^3(e+f x)}{3 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(4 b (a+b)) \operatorname{Subst}\left (\int \frac{1}{\left (a-b+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{(a-b)^2 f}\\ &=-\frac{(a+b) \cos (e+f x)}{(a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}+\frac{\cos ^3(e+f x)}{3 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{4 b (a+b) \sec (e+f x)}{3 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(8 b (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)^3 f}\\ &=-\frac{(a+b) \cos (e+f x)}{(a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}+\frac{\cos ^3(e+f x)}{3 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{4 b (a+b) \sec (e+f x)}{3 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{8 b (a+b) \sec (e+f x)}{3 (a-b)^4 f \sqrt{a-b+b \sec ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.45017, size = 205, normalized size = 1.22 \[ -\frac{\cos (e+f x) \left (3 \left (63 a^2 b+11 a^3-31 a b^2-43 b^3\right ) \cos (2 (e+f x))+3 a^2 b \cos (6 (e+f x))+186 a^2 b+a^3 (-\cos (6 (e+f x)))+26 a^3-3 a b^2 \cos (6 (e+f x))+190 a b^2+6 (a-b)^2 (a+3 b) \cos (4 (e+f x))+b^3 \cos (6 (e+f x))+110 b^3\right ) \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}}{24 \sqrt{2} f (a-b)^4 ((a-b) \cos (2 (e+f x))+a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[e + f*x]*(26*a^3 + 186*a^2*b + 190*a*b^2 + 110*b^3 + 3*(11*a^3 + 63*a^2*b - 31*a*b^2 - 43*b^3)*Cos[2*(e
+ f*x)] + 6*(a - b)^2*(a + 3*b)*Cos[4*(e + f*x)] - a^3*Cos[6*(e + f*x)] + 3*a^2*b*Cos[6*(e + f*x)] - 3*a*b^2*C
os[6*(e + f*x)] + b^3*Cos[6*(e + f*x)])*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(24*Sqrt[2]*(
a - b)^4*f*(a + b + (a - b)*Cos[2*(e + f*x)])^2)

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Maple [A]  time = 0.614, size = 262, normalized size = 1.6 \begin{align*} -{\frac{{a}^{5} \left ( a-b \right ) \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{6}{a}^{3}-3\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}{a}^{2}b+3\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}a{b}^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{6}{b}^{3}-3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{3}+3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{2}b+3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}a{b}^{2}-3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{b}^{3}-12\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{a}^{2}b+12\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{b}^{3}-8\,a{b}^{2}-8\,{b}^{3} \right ) \sqrt{4}}{6\,f \left ( \cos \left ( fx+e \right ) \right ) ^{5}} \left ( \sqrt{-b \left ( a-b \right ) }+a-b \right ) ^{-5} \left ( \sqrt{-b \left ( a-b \right ) }-a+b \right ) ^{-5} \left ({\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}}} \right ) ^{-{\frac{5}{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)^(5/2),x)

[Out]

-1/6/f*a^5/((-b*(a-b))^(1/2)+a-b)^5/((-b*(a-b))^(1/2)-a+b)^5*(a-b)*(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)*(cos(f*x+
e)^6*a^3-3*cos(f*x+e)^6*a^2*b+3*cos(f*x+e)^6*a*b^2-cos(f*x+e)^6*b^3-3*cos(f*x+e)^4*a^3+3*cos(f*x+e)^4*a^2*b+3*
cos(f*x+e)^4*a*b^2-3*cos(f*x+e)^4*b^3-12*cos(f*x+e)^2*a^2*b+12*cos(f*x+e)^2*b^3-8*a*b^2-8*b^3)*4^(1/2)/cos(f*x
+e)^5/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(5/2)

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Maxima [A]  time = 1.06732, size = 416, normalized size = 2.48 \begin{align*} -\frac{\frac{3 \, \sqrt{a - b + \frac{b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac{{\left (a - b + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} \cos \left (f x + e\right )^{3} - 9 \, \sqrt{a - b + \frac{b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} + \frac{9 \,{\left (a - b + \frac{b}{\cos \left (f x + e\right )^{2}}\right )} b^{2} \cos \left (f x + e\right )^{2} - b^{3}}{{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )}{\left (a - b + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} \cos \left (f x + e\right )^{3}} + \frac{6 \,{\left (a - b + \frac{b}{\cos \left (f x + e\right )^{2}}\right )} b \cos \left (f x + e\right )^{2} - b^{2}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}{\left (a - b + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} \cos \left (f x + e\right )^{3}}}{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)^(5/2),x, algorithm="maxima")

[Out]

-1/3*(3*sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - ((a - b + b/cos(f*x + e)
^2)^(3/2)*cos(f*x + e)^3 - 9*sqrt(a - b + b/cos(f*x + e)^2)*b*cos(f*x + e))/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b
^3 + b^4) + (9*(a - b + b/cos(f*x + e)^2)*b^2*cos(f*x + e)^2 - b^3)/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^
4)*(a - b + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3) + (6*(a - b + b/cos(f*x + e)^2)*b*cos(f*x + e)^2 - b^2)/((
a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(a - b + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3))/f

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Fricas [A]  time = 4.4293, size = 598, normalized size = 3.56 \begin{align*} \frac{{\left ({\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{7} - 3 \,{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{5} - 12 \,{\left (a^{2} b - b^{3}\right )} \cos \left (f x + e\right )^{3} - 8 \,{\left (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}}}}{3 \,{\left ({\left (a^{6} - 6 \, a^{5} b + 15 \, a^{4} b^{2} - 20 \, a^{3} b^{3} + 15 \, a^{2} b^{4} - 6 \, a b^{5} + b^{6}\right )} f \cos \left (f x + e\right )^{4} + 2 \,{\left (a^{5} b - 5 \, a^{4} b^{2} + 10 \, a^{3} b^{3} - 10 \, a^{2} b^{4} + 5 \, a b^{5} - b^{6}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{4} b^{2} - 4 \, a^{3} b^{3} + 6 \, a^{2} b^{4} - 4 \, a b^{5} + b^{6}\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)^(5/2),x, algorithm="fricas")

[Out]

1/3*((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^7 - 3*(a^3 - a^2*b - a*b^2 + b^3)*cos(f*x + e)^5 - 12*(a^2*b
- b^3)*cos(f*x + e)^3 - 8*(a*b^2 + b^3)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a^6
- 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4 - 6*a*b^5 + b^6)*f*cos(f*x + e)^4 + 2*(a^5*b - 5*a^4*b^2 + 1
0*a^3*b^3 - 10*a^2*b^4 + 5*a*b^5 - b^6)*f*cos(f*x + e)^2 + (a^4*b^2 - 4*a^3*b^3 + 6*a^2*b^4 - 4*a*b^5 + b^6)*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)**(5/2),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{5}{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)^(5/2),x, algorithm="giac")

[Out]

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