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

Optimal. Leaf size=118 \[ -\frac{8 b \sec (e+f x)}{3 f (a-b)^3 \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{4 b \sec (e+f x)}{3 f (a-b)^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac{\cos (e+f x)}{f (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}} \]

[Out]

-(Cos[e + f*x]/((a - b)*f*(a - b + b*Sec[e + f*x]^2)^(3/2))) - (4*b*Sec[e + f*x])/(3*(a - b)^2*f*(a - b + b*Se
c[e + f*x]^2)^(3/2)) - (8*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 = 0.0776282, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3664, 271, 192, 191} \[ -\frac{8 b \sec (e+f x)}{3 f (a-b)^3 \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{4 b \sec (e+f x)}{3 f (a-b)^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac{\cos (e+f x)}{f (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[e + f*x]/((a - b)*f*(a - b + b*Sec[e + f*x]^2)^(3/2))) - (4*b*Sec[e + f*x])/(3*(a - b)^2*f*(a - b + b*Se
c[e + f*x]^2)^(3/2)) - (8*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 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 (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (a-b+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cos (e+f x)}{(a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(4 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) f}\\ &=-\frac{\cos (e+f x)}{(a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{4 b \sec (e+f x)}{3 (a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(8 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{\cos (e+f x)}{(a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{4 b \sec (e+f x)}{3 (a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac{8 b \sec (e+f x)}{3 (a-b)^3 f \sqrt{a-b+b \sec ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.21265, size = 124, normalized size = 1.05 \[ -\frac{\cos (e+f x) \left (12 \left (a^2+2 a b-3 b^2\right ) \cos (2 (e+f x))+3 (a-b)^2 \cos (4 (e+f x))+(3 a+5 b)^2\right ) \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}}{6 \sqrt{2} f (a-b)^3 ((a-b) \cos (2 (e+f x))+a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[e + f*x]*((3*a + 5*b)^2 + 12*(a^2 + 2*a*b - 3*b^2)*Cos[2*(e + f*x)] + 3*(a - b)^2*Cos[4*(e + f*x)])*Sqrt
[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(6*Sqrt[2]*(a - b)^3*f*(a + b + (a - b)*Cos[2*(e + f*x)])
^2)

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Maple [A]  time = 0.058, size = 147, normalized size = 1.3 \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 ( 3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{2}-6\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}ab+3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{b}^{2}+12\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}ab-12\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{b}^{2}+8\,{b}^{2} \right ) }{3\,f \left ( \cos \left ( fx+e \right ) \right ) ^{5} \left ( a-b \right ) ^{3}} \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)/(a+b*tan(f*x+e)^2)^(5/2),x)

[Out]

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

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Maxima [A]  time = 1.03534, size = 182, normalized size = 1.54 \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{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)/(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) + (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 = 2.96788, size = 454, normalized size = 3.85 \begin{align*} -\frac{{\left (3 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} + 12 \,{\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{3} + 8 \, b^{2} \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^{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 )^{4} + 2 \,{\left (a^{4} b - 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 4 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{3} b^{2} - 3 \, a^{2} b^{3} + 3 \, 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)/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

-1/3*(3*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 + 12*(a*b - b^2)*cos(f*x + e)^3 + 8*b^2*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)
^4 + 2*(a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)*f*cos(f*x + e)^2 + (a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5
)*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)/(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 )}{{\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)/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

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