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

Optimal. Leaf size=138 \[ -\frac{15 \cos (e+f x)}{8 f (a-b)^3}+\frac{5 \cos (e+f x)}{8 f (a-b)^2 \left (a+b \sec ^2(e+f x)-b\right )}+\frac{\cos (e+f x)}{4 f (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2}-\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{8 f (a-b)^{7/2}} \]

[Out]

(-15*Sqrt[b]*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(8*(a - b)^(7/2)*f) - (15*Cos[e + f*x])/(8*(a - b)^3*
f) + Cos[e + f*x]/(4*(a - b)*f*(a - b + b*Sec[e + f*x]^2)^2) + (5*Cos[e + f*x])/(8*(a - b)^2*f*(a - b + b*Sec[
e + f*x]^2))

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Rubi [A]  time = 0.0907196, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3664, 290, 325, 205} \[ -\frac{15 \cos (e+f x)}{8 f (a-b)^3}+\frac{5 \cos (e+f x)}{8 f (a-b)^2 \left (a+b \sec ^2(e+f x)-b\right )}+\frac{\cos (e+f x)}{4 f (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2}-\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{8 f (a-b)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*Sqrt[b]*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(8*(a - b)^(7/2)*f) - (15*Cos[e + f*x])/(8*(a - b)^3*
f) + Cos[e + f*x]/(4*(a - b)*f*(a - b + b*Sec[e + f*x]^2)^2) + (5*Cos[e + f*x])/(8*(a - b)^2*f*(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 290

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

Rule 325

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] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\sin (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (a-b+b x^2\right )^3} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac{\cos (e+f x)}{4 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a-b+b x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{4 (a-b) f}\\ &=\frac{\cos (e+f x)}{4 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}+\frac{5 \cos (e+f x)}{8 (a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )}+\frac{15 \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{8 (a-b)^2 f}\\ &=-\frac{15 \cos (e+f x)}{8 (a-b)^3 f}+\frac{\cos (e+f x)}{4 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}+\frac{5 \cos (e+f x)}{8 (a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{(15 b) \operatorname{Subst}\left (\int \frac{1}{a-b+b x^2} \, dx,x,\sec (e+f x)\right )}{8 (a-b)^3 f}\\ &=-\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{8 (a-b)^{7/2} f}-\frac{15 \cos (e+f x)}{8 (a-b)^3 f}+\frac{\cos (e+f x)}{4 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}+\frac{5 \cos (e+f x)}{8 (a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.61036, size = 170, normalized size = 1.23 \[ \frac{\frac{2 \cos (e+f x) \left (\frac{4 b^2}{((a-b) \cos (2 (e+f x))+a+b)^2}-\frac{9 b}{(a-b) \cos (2 (e+f x))+a+b}-4\right )}{(a-b)^3}+\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a-b}-\sqrt{a} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{b}}\right )}{(a-b)^{7/2}}+\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a-b}+\sqrt{a} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{b}}\right )}{(a-b)^{7/2}}}{8 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((15*Sqrt[b]*ArcTan[(Sqrt[a - b] - Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(7/2) + (15*Sqrt[b]*ArcTan[(Sqr
t[a - b] + Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(7/2) + (2*Cos[e + f*x]*(-4 + (4*b^2)/(a + b + (a - b)*
Cos[2*(e + f*x)])^2 - (9*b)/(a + b + (a - b)*Cos[2*(e + f*x)])))/(a - b)^3)/(8*f)

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Maple [A]  time = 0.073, size = 221, normalized size = 1.6 \begin{align*} -{\frac{\cos \left ( fx+e \right ) }{f \left ({a}^{3}-3\,{a}^{2}b+3\,a{b}^{2}-{b}^{3} \right ) }}-{\frac{9\,ab \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{8\,f \left ( a-b \right ) ^{3} \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) ^{2}}}+{\frac{9\,{b}^{2} \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{8\,f \left ( a-b \right ) ^{3} \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) ^{2}}}-{\frac{7\,{b}^{2}\cos \left ( fx+e \right ) }{8\,f \left ( a-b \right ) ^{3} \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) ^{2}}}+{\frac{15\,b}{8\,f \left ( a-b \right ) ^{3}}\arctan \left ({ \left ( a-b \right ) \cos \left ( fx+e \right ){\frac{1}{\sqrt{b \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{b \left ( a-b \right ) }}}} \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)^3,x)

[Out]

-1/f/(a^3-3*a^2*b+3*a*b^2-b^3)*cos(f*x+e)-9/8/f*b/(a-b)^3/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2*a*cos(f*x+e)^3+9
/8/f*b^2/(a-b)^3/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2*cos(f*x+e)^3-7/8/f*b^2/(a-b)^3/(a*cos(f*x+e)^2-cos(f*x+e)
^2*b+b)^2*cos(f*x+e)+15/8/f*b/(a-b)^3/(b*(a-b))^(1/2)*arctan((a-b)*cos(f*x+e)/(b*(a-b))^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.72399, size = 1249, normalized size = 9.05 \begin{align*} \left [-\frac{16 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} + 50 \,{\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{3} + 30 \, b^{2} \cos \left (f x + e\right ) + 15 \,{\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \,{\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt{-\frac{b}{a - b}} \log \left (-\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \,{\left (a - b\right )} \sqrt{-\frac{b}{a - b}} \cos \left (f x + e\right ) - b}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right )}{16 \,{\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 )}}, -\frac{8 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} + 25 \,{\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{3} + 15 \, b^{2} \cos \left (f x + e\right ) + 15 \,{\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \,{\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt{\frac{b}{a - b}} \arctan \left (-\frac{{\left (a - b\right )} \sqrt{\frac{b}{a - b}} \cos \left (f x + e\right )}{b}\right )}{8 \,{\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 )}}\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)^3,x, algorithm="fricas")

[Out]

[-1/16*(16*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 + 50*(a*b - b^2)*cos(f*x + e)^3 + 30*b^2*cos(f*x + e) + 15*((a^2
- 2*a*b + b^2)*cos(f*x + e)^4 + 2*(a*b - b^2)*cos(f*x + e)^2 + b^2)*sqrt(-b/(a - b))*log(-((a - b)*cos(f*x +
e)^2 - 2*(a - b)*sqrt(-b/(a - b))*cos(f*x + e) - b)/((a - b)*cos(f*x + e)^2 + b)))/((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), -1/8*(8*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 + 25*(a*b - b^2
)*cos(f*x + e)^3 + 15*b^2*cos(f*x + e) + 15*((a^2 - 2*a*b + b^2)*cos(f*x + e)^4 + 2*(a*b - b^2)*cos(f*x + e)^2
+ b^2)*sqrt(b/(a - b))*arctan(-(a - b)*sqrt(b/(a - b))*cos(f*x + e)/b))/((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)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.63463, size = 301, normalized size = 2.18 \begin{align*} -\frac{f^{5} \cos \left (f x + e\right )}{a^{3} f^{6} - 3 \, a^{2} b f^{6} + 3 \, a b^{2} f^{6} - b^{3} f^{6}} + \frac{15 \, b \arctan \left (\frac{a \cos \left (f x + e\right ) - b \cos \left (f x + e\right )}{\sqrt{a b - b^{2}}}\right )}{8 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt{a b - b^{2}} f} - \frac{\frac{9 \, a b \cos \left (f x + e\right )^{3}}{f} - \frac{9 \, b^{2} \cos \left (f x + e\right )^{3}}{f} + \frac{7 \, b^{2} \cos \left (f x + e\right )}{f}}{8 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}{\left (a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b\right )}^{2}} \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)^3,x, algorithm="giac")

[Out]

-f^5*cos(f*x + e)/(a^3*f^6 - 3*a^2*b*f^6 + 3*a*b^2*f^6 - b^3*f^6) + 15/8*b*arctan((a*cos(f*x + e) - b*cos(f*x
+ e))/sqrt(a*b - b^2))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*sqrt(a*b - b^2)*f) - 1/8*(9*a*b*cos(f*x + e)^3/f - 9*b
^2*cos(f*x + e)^3/f + 7*b^2*cos(f*x + e)/f)/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(a*cos(f*x + e)^2 - b*cos(f*x + e
)^2 + b)^2)