3.62 \(\int \frac{\sin ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx\)

Optimal. Leaf size=129 \[ \frac{x \left (3 a^2+6 a b-b^2\right )}{8 (a-b)^3}-\frac{a^{3/2} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{f (a-b)^3}+\frac{\sin (e+f x) \cos ^3(e+f x)}{4 f (a-b)}-\frac{(5 a-b) \sin (e+f x) \cos (e+f x)}{8 f (a-b)^2} \]

[Out]

((3*a^2 + 6*a*b - b^2)*x)/(8*(a - b)^3) - (a^(3/2)*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/((a - b)^3*
f) - ((5*a - b)*Cos[e + f*x]*Sin[e + f*x])/(8*(a - b)^2*f) + (Cos[e + f*x]^3*Sin[e + f*x])/(4*(a - b)*f)

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Rubi [A]  time = 0.151545, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3663, 470, 527, 522, 203, 205} \[ \frac{x \left (3 a^2+6 a b-b^2\right )}{8 (a-b)^3}-\frac{a^{3/2} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{f (a-b)^3}+\frac{\sin (e+f x) \cos ^3(e+f x)}{4 f (a-b)}-\frac{(5 a-b) \sin (e+f x) \cos (e+f x)}{8 f (a-b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^4/(a + b*Tan[e + f*x]^2),x]

[Out]

((3*a^2 + 6*a*b - b^2)*x)/(8*(a - b)^3) - (a^(3/2)*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/((a - b)^3*
f) - ((5*a - b)*Cos[e + f*x]*Sin[e + f*x])/(8*(a - b)^2*f) + (Cos[e + f*x]^3*Sin[e + f*x])/(4*(a - b)*f)

Rule 3663

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff^(m + 1))/f, Subst[Int[(x^m*(a + b*(ff*x)^n)^p)/(c^2 + ff^2*x^2
)^(m/2 + 1), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2]

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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 ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right )^3 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f}-\frac{\operatorname{Subst}\left (\int \frac{a+(-4 a+b) x^2}{\left (1+x^2\right )^2 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{4 (a-b) f}\\ &=-\frac{(5 a-b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f}+\frac{\operatorname{Subst}\left (\int \frac{a (3 a+b)-(5 a-b) b x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 (a-b)^2 f}\\ &=-\frac{(5 a-b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f}-\frac{\left (a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{(a-b)^3 f}+\frac{\left (3 a^2+6 a b-b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 (a-b)^3 f}\\ &=\frac{\left (3 a^2+6 a b-b^2\right ) x}{8 (a-b)^3}-\frac{a^{3/2} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{(a-b)^3 f}-\frac{(5 a-b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f}\\ \end{align*}

Mathematica [A]  time = 0.261784, size = 99, normalized size = 0.77 \[ \frac{4 \left (3 a^2+6 a b-b^2\right ) (e+f x)-32 a^{3/2} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )+(a-b)^2 \sin (4 (e+f x))-8 a (a-b) \sin (2 (e+f x))}{32 f (a-b)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^4/(a + b*Tan[e + f*x]^2),x]

[Out]

(4*(3*a^2 + 6*a*b - b^2)*(e + f*x) - 32*a^(3/2)*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]] - 8*a*(a - b)*S
in[2*(e + f*x)] + (a - b)^2*Sin[4*(e + f*x)])/(32*(a - b)^3*f)

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Maple [B]  time = 0.063, size = 304, normalized size = 2.4 \begin{align*} -{\frac{{a}^{2}b}{f \left ( a-b \right ) ^{3}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{5\, \left ( \tan \left ( fx+e \right ) \right ) ^{3}{a}^{2}}{8\,f \left ( a-b \right ) ^{3} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{3\, \left ( \tan \left ( fx+e \right ) \right ) ^{3}ab}{4\,f \left ( a-b \right ) ^{3} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{3}{b}^{2}}{8\,f \left ( a-b \right ) ^{3} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{3\,\tan \left ( fx+e \right ){a}^{2}}{8\,f \left ( a-b \right ) ^{3} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{\tan \left ( fx+e \right ) ab}{4\,f \left ( a-b \right ) ^{3} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{\tan \left ( fx+e \right ){b}^{2}}{8\,f \left ( a-b \right ) ^{3} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{3\,\arctan \left ( \tan \left ( fx+e \right ) \right ){a}^{2}}{8\,f \left ( a-b \right ) ^{3}}}+{\frac{3\,\arctan \left ( \tan \left ( fx+e \right ) \right ) ab}{4\,f \left ( a-b \right ) ^{3}}}-{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ){b}^{2}}{8\,f \left ( a-b \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^4/(a+b*tan(f*x+e)^2),x)

[Out]

-1/f*a^2*b/(a-b)^3/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))-5/8/f/(a-b)^3/(1+tan(f*x+e)^2)^2*tan(f*x+e)^3*
a^2+3/4/f/(a-b)^3/(1+tan(f*x+e)^2)^2*tan(f*x+e)^3*a*b-1/8/f/(a-b)^3/(1+tan(f*x+e)^2)^2*tan(f*x+e)^3*b^2-3/8/f/
(a-b)^3/(1+tan(f*x+e)^2)^2*tan(f*x+e)*a^2+1/4/f/(a-b)^3/(1+tan(f*x+e)^2)^2*tan(f*x+e)*a*b+1/8/f/(a-b)^3/(1+tan
(f*x+e)^2)^2*tan(f*x+e)*b^2+3/8/f/(a-b)^3*arctan(tan(f*x+e))*a^2+3/4/f/(a-b)^3*arctan(tan(f*x+e))*a*b-1/8/f/(a
-b)^3*arctan(tan(f*x+e))*b^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)^4/(a+b*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.07707, size = 898, normalized size = 6.96 \begin{align*} \left [\frac{{\left (3 \, a^{2} + 6 \, a b - b^{2}\right )} f x - 2 \, \sqrt{-a b} a \log \left (\frac{{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \,{\left ({\left (a + b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt{-a b} \sin \left (f x + e\right ) + b^{2}}{{\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 ) +{\left (2 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} -{\left (5 \, a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f}, \frac{{\left (3 \, a^{2} + 6 \, a b - b^{2}\right )} f x + 4 \, \sqrt{a b} a \arctan \left (\frac{{\left ({\left (a + b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt{a b}}{2 \, a b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) +{\left (2 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} -{\left (5 \, a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*tan(f*x+e)^2),x, algorithm="fricas")

[Out]

[1/8*((3*a^2 + 6*a*b - b^2)*f*x - 2*sqrt(-a*b)*a*log(((a^2 + 6*a*b + b^2)*cos(f*x + e)^4 - 2*(3*a*b + b^2)*cos
(f*x + e)^2 - 4*((a + b)*cos(f*x + e)^3 - b*cos(f*x + e))*sqrt(-a*b)*sin(f*x + e) + b^2)/((a^2 - 2*a*b + b^2)*
cos(f*x + e)^4 + 2*(a*b - b^2)*cos(f*x + e)^2 + b^2)) + (2*(a^2 - 2*a*b + b^2)*cos(f*x + e)^3 - (5*a^2 - 6*a*b
+ b^2)*cos(f*x + e))*sin(f*x + e))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*f), 1/8*((3*a^2 + 6*a*b - b^2)*f*x + 4*sq
rt(a*b)*a*arctan(1/2*((a + b)*cos(f*x + e)^2 - b)*sqrt(a*b)/(a*b*cos(f*x + e)*sin(f*x + e))) + (2*(a^2 - 2*a*b
+ b^2)*cos(f*x + e)^3 - (5*a^2 - 6*a*b + b^2)*cos(f*x + e))*sin(f*x + e))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*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)**4/(a+b*tan(f*x+e)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.44372, size = 257, normalized size = 1.99 \begin{align*} -\frac{\frac{8 \,{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b}}\right )\right )} a^{2} b}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt{a b}} - \frac{{\left (3 \, a^{2} + 6 \, a b - b^{2}\right )}{\left (f x + e\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac{5 \, a \tan \left (f x + e\right )^{3} - b \tan \left (f x + e\right )^{3} + 3 \, a \tan \left (f x + e\right ) + b \tan \left (f x + e\right )}{{\left (a^{2} - 2 \, a b + b^{2}\right )}{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{2}}}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*tan(f*x+e)^2),x, algorithm="giac")

[Out]

-1/8*(8*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))*a^2*b/((a^3 - 3*a^2*b + 3*a*b
^2 - b^3)*sqrt(a*b)) - (3*a^2 + 6*a*b - b^2)*(f*x + e)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (5*a*tan(f*x + e)^3 -
b*tan(f*x + e)^3 + 3*a*tan(f*x + e) + b*tan(f*x + e))/((a^2 - 2*a*b + b^2)*(tan(f*x + e)^2 + 1)^2))/f