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

Optimal. Leaf size=196 \[ \frac{3 x \left (a^2+6 a b+b^2\right )}{8 (a-b)^4}-\frac{3 \sqrt{a} \sqrt{b} (a+b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{2 f (a-b)^4}-\frac{3 b (3 a+b) \tan (e+f x)}{8 f (a-b)^3 \left (a+b \tan ^2(e+f x)\right )}+\frac{\sin (e+f x) \cos ^3(e+f x)}{4 f (a-b) \left (a+b \tan ^2(e+f x)\right )}-\frac{(5 a+b) \sin (e+f x) \cos (e+f x)}{8 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )} \]

[Out]

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

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Rubi [A]  time = 0.249952, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{3 x \left (a^2+6 a b+b^2\right )}{8 (a-b)^4}-\frac{3 \sqrt{a} \sqrt{b} (a+b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{2 f (a-b)^4}-\frac{3 b (3 a+b) \tan (e+f x)}{8 f (a-b)^3 \left (a+b \tan ^2(e+f x)\right )}+\frac{\sin (e+f x) \cos ^3(e+f x)}{4 f (a-b) \left (a+b \tan ^2(e+f x)\right )}-\frac{(5 a+b) \sin (e+f x) \cos (e+f x)}{8 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right )^3 \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{a+(-4 a-b) x^2}{\left (1+x^2\right )^2 \left (a+b x^2\right )^2} \, 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 \left (a+b \tan ^2(e+f x)\right )}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{3 a (a+b)-3 b (5 a+b) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^2} \, 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 \left (a+b \tan ^2(e+f x)\right )}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )}-\frac{3 b (3 a+b) \tan (e+f x)}{8 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{6 a^2 (a+3 b)-6 a b (3 a+b) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{16 a (a-b)^3 f}\\ &=-\frac{(5 a+b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )}-\frac{3 b (3 a+b) \tan (e+f x)}{8 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )}-\frac{(3 a b (a+b)) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{2 (a-b)^4 f}+\frac{\left (3 \left (a^2+6 a b+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 (a-b)^4 f}\\ &=\frac{3 \left (a^2+6 a b+b^2\right ) x}{8 (a-b)^4}-\frac{3 \sqrt{a} \sqrt{b} (a+b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{2 (a-b)^4 f}-\frac{(5 a+b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )}-\frac{3 b (3 a+b) \tan (e+f x)}{8 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.51482, size = 136, normalized size = 0.69 \[ \frac{12 \left (a^2+6 a b+b^2\right ) (e+f x)+(a-b)^2 \sin (4 (e+f x))-8 (a+b) (a-b) \sin (2 (e+f x))-48 \sqrt{a} \sqrt{b} (a+b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )-\frac{16 a b (a-b) \sin (2 (e+f x))}{(a-b) \cos (2 (e+f x))+a+b}}{32 f (a-b)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(a^2 + 6*a*b + b^2)*(e + f*x) - 48*Sqrt[a]*Sqrt[b]*(a + b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]] - 8*(a -
b)*(a + b)*Sin[2*(e + f*x)] - (16*a*(a - b)*b*Sin[2*(e + f*x)])/(a + b + (a - b)*Cos[2*(e + f*x)]) + (a - b)^
2*Sin[4*(e + f*x)])/(32*(a - b)^4*f)

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Maple [B]  time = 0.092, size = 411, normalized size = 2.1 \begin{align*} -{\frac{{a}^{2}b\tan \left ( fx+e \right ) }{2\,f \left ( a-b \right ) ^{4} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{a{b}^{2}\tan \left ( fx+e \right ) }{2\,f \left ( a-b \right ) ^{4} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{3\,{a}^{2}b}{2\,f \left ( a-b \right ) ^{4}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{3\,a{b}^{2}}{2\,f \left ( a-b \right ) ^{4}}\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 ) ^{4} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{3}ab}{4\,f \left ( a-b \right ) ^{4} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{3\, \left ( \tan \left ( fx+e \right ) \right ) ^{3}{b}^{2}}{8\,f \left ( a-b \right ) ^{4} \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 ) ^{4} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{5\,\tan \left ( fx+e \right ){b}^{2}}{8\,f \left ( a-b \right ) ^{4} \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 ) ^{4} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{9\,\arctan \left ( \tan \left ( fx+e \right ) \right ) ab}{4\,f \left ( a-b \right ) ^{4}}}+{\frac{3\,\arctan \left ( \tan \left ( fx+e \right ) \right ){b}^{2}}{8\,f \left ( a-b \right ) ^{4}}}+{\frac{3\,\arctan \left ( \tan \left ( fx+e \right ) \right ){a}^{2}}{8\,f \left ( a-b \right ) ^{4}}} \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)^2,x)

[Out]

-1/2/f*a^2*b/(a-b)^4*tan(f*x+e)/(a+b*tan(f*x+e)^2)+1/2/f*a*b^2/(a-b)^4*tan(f*x+e)/(a+b*tan(f*x+e)^2)-3/2/f*a^2
*b/(a-b)^4/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))-3/2/f*a*b^2/(a-b)^4/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a
*b)^(1/2))-5/8/f/(a-b)^4/(1+tan(f*x+e)^2)^2*tan(f*x+e)^3*a^2+1/4/f/(a-b)^4/(1+tan(f*x+e)^2)^2*tan(f*x+e)^3*a*b
+3/8/f/(a-b)^4/(1+tan(f*x+e)^2)^2*tan(f*x+e)^3*b^2-3/8/f/(a-b)^4/(1+tan(f*x+e)^2)^2*tan(f*x+e)*a^2+5/8/f/(a-b)
^4/(1+tan(f*x+e)^2)^2*tan(f*x+e)*b^2-1/4/f/(a-b)^4/(1+tan(f*x+e)^2)^2*tan(f*x+e)*a*b+9/4/f/(a-b)^4*arctan(tan(
f*x+e))*a*b+3/8/f/(a-b)^4*arctan(tan(f*x+e))*b^2+3/8/f/(a-b)^4*arctan(tan(f*x+e))*a^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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.47919, size = 1600, normalized size = 8.16 \begin{align*} \text{result too large to display} \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)^2,x, algorithm="fricas")

[Out]

[1/8*(3*(a^3 + 5*a^2*b - 5*a*b^2 - b^3)*f*x*cos(f*x + e)^2 + 3*(a^2*b + 6*a*b^2 + b^3)*f*x + 3*((a^2 - b^2)*co
s(f*x + e)^2 + a*b + b^2)*sqrt(-a*b)*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^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^5 - (5*a^3 - 9*a^2
*b + 3*a*b^2 + b^3)*cos(f*x + e)^3 - 3*(3*a^2*b - 2*a*b^2 - b^3)*cos(f*x + e))*sin(f*x + e))/((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)^2 + (a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)*f
), 1/8*(3*(a^3 + 5*a^2*b - 5*a*b^2 - b^3)*f*x*cos(f*x + e)^2 + 3*(a^2*b + 6*a*b^2 + b^3)*f*x + 6*((a^2 - b^2)*
cos(f*x + e)^2 + a*b + b^2)*sqrt(a*b)*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^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^5 - (5*a^3 - 9*a^2*b + 3*a*b^2 + b^3)*cos(f*x + e
)^3 - 3*(3*a^2*b - 2*a*b^2 - b^3)*cos(f*x + e))*sin(f*x + e))/((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)^2 + (a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 4*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)**4/(a+b*tan(f*x+e)**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.3924, size = 359, normalized size = 1.83 \begin{align*} \frac{\frac{3 \,{\left (a^{2} + 6 \, a b + b^{2}\right )}{\left (f x + e\right )}}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} - \frac{4 \, a b \tan \left (f x + e\right )}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}{\left (b \tan \left (f x + e\right )^{2} + a\right )}} - \frac{12 \,{\left (a^{2} b + a b^{2}\right )}{\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 )}}{{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \sqrt{a b}} - \frac{5 \, a \tan \left (f x + e\right )^{3} + 3 \, b \tan \left (f x + e\right )^{3} + 3 \, a \tan \left (f x + e\right ) + 5 \, b \tan \left (f x + e\right )}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\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)^2,x, algorithm="giac")

[Out]

1/8*(3*(a^2 + 6*a*b + b^2)*(f*x + e)/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4) - 4*a*b*tan(f*x + e)/((a^3 -
3*a^2*b + 3*a*b^2 - b^3)*(b*tan(f*x + e)^2 + a)) - 12*(a^2*b + a*b^2)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + a
rctan(b*tan(f*x + e)/sqrt(a*b)))/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*sqrt(a*b)) - (5*a*tan(f*x + e)^3
+ 3*b*tan(f*x + e)^3 + 3*a*tan(f*x + e) + 5*b*tan(f*x + e))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(tan(f*x + e)^2
+ 1)^2))/f