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]

-15/8*cos(f*x+e)/(a-b)^3/f+1/4*cos(f*x+e)/(a-b)/f/(a-b+b*sec(f*x+e)^2)^2+5/8*cos(f*x+e)/(a-b)^2/f/(a-b+b*sec(f
*x+e)^2)-15/8*arctan(sec(f*x+e)*b^(1/2)/(a-b)^(1/2))*b^(1/2)/(a-b)^(7/2)/f

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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 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]

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 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]

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 = 1.77, 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)

________________________________________________________________________________________

fricas [B]  time = 0.61, size = 556, normalized size = 4.03 \[ \left [-\frac {16 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} + 50 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{3} + 30 \, b^{2} \cos \left (f x + e\right ) + 15 \, {\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-\frac {b}{a - b}} \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (a - b\right )} \sqrt {-\frac {b}{a - b}} \cos \left (f x + e\right ) - b}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right )}{16 \, {\left ({\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b - 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 4 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{2} - 3 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5}\right )} f\right )}}, -\frac {8 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} + 25 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{3} + 15 \, b^{2} \cos \left (f x + e\right ) + 15 \, {\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {\frac {b}{a - b}} \arctan \left (-\frac {{\left (a - b\right )} \sqrt {\frac {b}{a - b}} \cos \left (f x + e\right )}{b}\right )}{8 \, {\left ({\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b - 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 4 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{2} - 3 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5}\right )} f\right )}}\right ] \]

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)]

________________________________________________________________________________________

giac [A]  time = 4.68, size = 223, normalized size = 1.62 \[ -\frac {f^{5} \cos \left (f x + e\right )}{a^{3} f^{6} - 3 \, a^{2} b f^{6} + 3 \, a b^{2} f^{6} - b^{3} f^{6}} + \frac {15 \, b \arctan \left (\frac {a \cos \left (f x + e\right ) - b \cos \left (f x + e\right )}{\sqrt {a b - b^{2}}}\right )}{8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a b - b^{2}} f} - \frac {\frac {9 \, a b \cos \left (f x + e\right )^{3}}{f} - \frac {9 \, b^{2} \cos \left (f x + e\right )^{3}}{f} + \frac {7 \, b^{2} \cos \left (f x + e\right )}{f}}{8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} {\left (a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b\right )}^{2}} \]

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)

________________________________________________________________________________________

maple [A]  time = 0.47, size = 221, normalized size = 1.60 \[ -\frac {\cos \left (f x +e \right )}{f \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}-\frac {9 b a \left (\cos ^{3}\left (f x +e \right )\right )}{8 f \left (a -b \right )^{3} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2}}+\frac {9 b^{2} \left (\cos ^{3}\left (f x +e \right )\right )}{8 f \left (a -b \right )^{3} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2}}-\frac {7 b^{2} \cos \left (f x +e \right )}{8 f \left (a -b \right )^{3} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2}}+\frac {15 b \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right )}{8 f \left (a -b \right )^{3} \sqrt {\left (a -b \right ) b}} \]

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/((a-b)*b)^(1/2)*arctan((a-b)*cos(f*x+e)/((a-b)*b)^(1/2))

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is b-a positive or negative?

________________________________________________________________________________________

mupad [B]  time = 14.87, size = 780, normalized size = 5.65 \[ -\frac {\frac {8\,a^2+9\,a\,b-2\,b^2}{4\,\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (16\,a^4-41\,a^3\,b+27\,a^2\,b^2-40\,a\,b^3+8\,b^4\right )}{2\,a^2\,\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (24\,a^4-64\,a^3\,b+53\,a^2\,b^2+40\,a\,b^3-8\,b^4\right )}{2\,a^2\,\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8\,\left (8\,a^3-9\,a^2\,b+24\,a\,b^2-8\,b^3\right )}{4\,a\,\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (-16\,a^3+23\,a^2\,b+27\,a\,b^2-4\,b^3\right )}{2\,a\,\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (8\,a\,b-3\,a^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8\,\left (8\,a\,b-3\,a^2\right )+a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (2\,a^2-8\,a\,b+16\,b^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (2\,a^2-8\,a\,b+16\,b^2\right )+a^2\right )}-\frac {15\,\sqrt {b}\,\mathrm {atan}\left (\frac {{\left (a-b\right )}^7\,\left (2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {\sqrt {b}\,\left (225\,a^8\,b-1350\,a^7\,b^2+3375\,a^6\,b^3-4500\,a^5\,b^4+3375\,a^4\,b^5-1350\,a^3\,b^6+225\,a^2\,b^7\right )}{a\,{\left (a-b\right )}^{13/2}}+\frac {225\,\sqrt {b}\,\left (a-2\,b\right )\,\left (128\,a^{12}-1408\,a^{11}\,b+6912\,a^{10}\,b^2-19968\,a^9\,b^3+37632\,a^8\,b^4-48384\,a^7\,b^5+43008\,a^6\,b^6-26112\,a^5\,b^7+10368\,a^4\,b^8-2432\,a^3\,b^9+256\,a^2\,b^{10}\right )}{512\,a\,{\left (a-b\right )}^{21/2}}\right )+\frac {225\,\sqrt {b}\,\left (a-2\,b\right )\,\left (-128\,a^{12}+1152\,a^{11}\,b-4608\,a^{10}\,b^2+10752\,a^9\,b^3-16128\,a^8\,b^4+16128\,a^7\,b^5-10752\,a^6\,b^6+4608\,a^5\,b^7-1152\,a^4\,b^8+128\,a^3\,b^9\right )}{256\,a\,{\left (a-b\right )}^{21/2}}\right )}{225\,a^8\,b-1350\,a^7\,b^2+3375\,a^6\,b^3-4500\,a^5\,b^4+3375\,a^4\,b^5-1350\,a^3\,b^6+225\,a^2\,b^7}\right )}{8\,f\,{\left (a-b\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((9*a*b + 8*a^2 - 2*b^2)/(4*(a - b)*(a^2 - 2*a*b + b^2)) - (tan(e/2 + (f*x)/2)^6*(16*a^4 - 41*a^3*b - 40*a*b
^3 + 8*b^4 + 27*a^2*b^2))/(2*a^2*(a - b)*(a^2 - 2*a*b + b^2)) + (tan(e/2 + (f*x)/2)^4*(40*a*b^3 - 64*a^3*b + 2
4*a^4 - 8*b^4 + 53*a^2*b^2))/(2*a^2*(a - b)*(a^2 - 2*a*b + b^2)) + (tan(e/2 + (f*x)/2)^8*(24*a*b^2 - 9*a^2*b +
8*a^3 - 8*b^3))/(4*a*(a - b)*(a^2 - 2*a*b + b^2)) + (tan(e/2 + (f*x)/2)^2*(27*a*b^2 + 23*a^2*b - 16*a^3 - 4*b
^3))/(2*a*(a - b)*(a^2 - 2*a*b + b^2)))/(f*(tan(e/2 + (f*x)/2)^2*(8*a*b - 3*a^2) + tan(e/2 + (f*x)/2)^8*(8*a*b
- 3*a^2) + a^2*tan(e/2 + (f*x)/2)^10 + tan(e/2 + (f*x)/2)^4*(2*a^2 - 8*a*b + 16*b^2) + tan(e/2 + (f*x)/2)^6*(
2*a^2 - 8*a*b + 16*b^2) + a^2)) - (15*b^(1/2)*atan(((a - b)^7*(2*tan(e/2 + (f*x)/2)^2*((b^(1/2)*(225*a^8*b + 2
25*a^2*b^7 - 1350*a^3*b^6 + 3375*a^4*b^5 - 4500*a^5*b^4 + 3375*a^6*b^3 - 1350*a^7*b^2))/(a*(a - b)^(13/2)) + (
225*b^(1/2)*(a - 2*b)*(128*a^12 - 1408*a^11*b + 256*a^2*b^10 - 2432*a^3*b^9 + 10368*a^4*b^8 - 26112*a^5*b^7 +
43008*a^6*b^6 - 48384*a^7*b^5 + 37632*a^8*b^4 - 19968*a^9*b^3 + 6912*a^10*b^2))/(512*a*(a - b)^(21/2))) + (225
*b^(1/2)*(a - 2*b)*(1152*a^11*b - 128*a^12 + 128*a^3*b^9 - 1152*a^4*b^8 + 4608*a^5*b^7 - 10752*a^6*b^6 + 16128
*a^7*b^5 - 16128*a^8*b^4 + 10752*a^9*b^3 - 4608*a^10*b^2))/(256*a*(a - b)^(21/2))))/(225*a^8*b + 225*a^2*b^7 -
1350*a^3*b^6 + 3375*a^4*b^5 - 4500*a^5*b^4 + 3375*a^6*b^3 - 1350*a^7*b^2)))/(8*f*(a - b)^(7/2))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

________________________________________________________________________________________