Integrand size = 25, antiderivative size = 177 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {(a-5 b) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{2 a^{7/2} f}-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {5 b \sec (e+f x)}{6 a^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {(13 a-15 b) b \sec (e+f x)}{6 a^3 (a-b) f \sqrt {a-b+b \sec ^2(e+f x)}} \]
-1/2*(a-5*b)*arctanh(sec(f*x+e)*a^(1/2)/(a-b+b*sec(f*x+e)^2)^(1/2))/a^(7/2 )/f-1/2*cot(f*x+e)*csc(f*x+e)/a/f/(a-b+b*sec(f*x+e)^2)^(3/2)-5/6*b*sec(f*x +e)/a^2/f/(a-b+b*sec(f*x+e)^2)^(3/2)-1/6*(13*a-15*b)*b*sec(f*x+e)/a^3/(a-b )/f/(a-b+b*sec(f*x+e)^2)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(380\) vs. \(2(177)=354\).
Time = 5.78 (sec) , antiderivative size = 380, normalized size of antiderivative = 2.15 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\frac {\frac {\sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \left (8 a b^2 \cos (e+f x)-24 (a-b) b \cos (e+f x) (a+b+(a-b) \cos (2 (e+f x)))-3 (a-b) (a+b+(a-b) \cos (2 (e+f x)))^2 \cot (e+f x) \csc (e+f x)\right )}{3 a^3 (a-b) (a+b+(a-b) \cos (2 (e+f x)))^2}+\frac {(a-5 b) \cos (e+f x) \left (2 \text {arctanh}\left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\frac {\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{\sqrt {a}}\right )+\log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )\right ) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{2 a^{7/2} \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )}}}{2 f} \]
((Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*(8*a*b^2 *Cos[e + f*x] - 24*(a - b)*b*Cos[e + f*x]*(a + b + (a - b)*Cos[2*(e + f*x) ]) - 3*(a - b)*(a + b + (a - b)*Cos[2*(e + f*x)])^2*Cot[e + f*x]*Csc[e + f *x]))/(3*a^3*(a - b)*(a + b + (a - b)*Cos[2*(e + f*x)])^2) + ((a - 5*b)*Co s[e + f*x]*(2*ArcTanh[Tan[(e + f*x)/2]^2 - Sqrt[4*b*Tan[(e + f*x)/2]^2 + a *(-1 + Tan[(e + f*x)/2]^2)^2]/Sqrt[a]] + Log[a - 2*b - a*Tan[(e + f*x)/2]^ 2 + Sqrt[a]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]]) *Sec[(e + f*x)/2]^2*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2 ])/(2*a^(7/2)*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[(e + f*x)/2]^4]) )/(2*f)
Time = 0.37 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4147, 373, 402, 25, 27, 402, 27, 291, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x)^3 \left (a+b \tan (e+f x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int \frac {\sec ^2(e+f x)}{\left (1-\sec ^2(e+f x)\right )^2 \left (b \sec ^2(e+f x)+a-b\right )^{5/2}}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 373 |
| \(\displaystyle \frac {\frac {\sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\int \frac {-4 b \sec ^2(e+f x)+a-b}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{5/2}}d\sec (e+f x)}{2 a}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {5 b \sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\int -\frac {(a-b) \left (-10 b \sec ^2(e+f x)+3 a-5 b\right )}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{3/2}}d\sec (e+f x)}{3 a (a-b)}}{2 a}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {\int \frac {(a-b) \left (-10 b \sec ^2(e+f x)+3 a-5 b\right )}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{3/2}}d\sec (e+f x)}{3 a (a-b)}+\frac {5 b \sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{2 a}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {\int \frac {-10 b \sec ^2(e+f x)+3 a-5 b}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^{3/2}}d\sec (e+f x)}{3 a}+\frac {5 b \sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{2 a}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {\frac {b (13 a-15 b) \sec (e+f x)}{a (a-b) \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {\int -\frac {3 (a-5 b) (a-b)}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{a (a-b)}}{3 a}+\frac {5 b \sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{2 a}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {\frac {3 (a-5 b) \int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)}{a}+\frac {b (13 a-15 b) \sec (e+f x)}{a (a-b) \sqrt {a+b \sec ^2(e+f x)-b}}}{3 a}+\frac {5 b \sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{2 a}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {\frac {3 (a-5 b) \int \frac {1}{1-\frac {a \sec ^2(e+f x)}{b \sec ^2(e+f x)+a-b}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a-b}}}{a}+\frac {b (13 a-15 b) \sec (e+f x)}{a (a-b) \sqrt {a+b \sec ^2(e+f x)-b}}}{3 a}+\frac {5 b \sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{2 a}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\sec (e+f x)}{2 a \left (1-\sec ^2(e+f x)\right ) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {\frac {3 (a-5 b) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{a^{3/2}}+\frac {b (13 a-15 b) \sec (e+f x)}{a (a-b) \sqrt {a+b \sec ^2(e+f x)-b}}}{3 a}+\frac {5 b \sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}}{2 a}}{f}\) |
(Sec[e + f*x]/(2*a*(1 - Sec[e + f*x]^2)*(a - b + b*Sec[e + f*x]^2)^(3/2)) - ((5*b*Sec[e + f*x])/(3*a*(a - b + b*Sec[e + f*x]^2)^(3/2)) + ((3*(a - 5* b)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/a^(3/2) + ((13*a - 15*b)*b*Sec[e + f*x])/(a*(a - b)*Sqrt[a - b + b*Sec[e + f*x]^2 ]))/(3*a))/(2*a))/f
3.2.44.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1)) Int[(e *x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Simp[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]
Leaf count of result is larger than twice the leaf count of optimal. \(26226\) vs. \(2(157)=314\).
Time = 6.77 (sec) , antiderivative size = 26227, normalized size of antiderivative = 148.18
Leaf count of result is larger than twice the leaf count of optimal. 431 vs. \(2 (157) = 314\).
Time = 0.47 (sec) , antiderivative size = 889, normalized size of antiderivative = 5.02 \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\left [-\frac {3 \, {\left ({\left (a^{4} - 8 \, a^{3} b + 18 \, a^{2} b^{2} - 16 \, a b^{3} + 5 \, b^{4}\right )} \cos \left (f x + e\right )^{6} - {\left (a^{4} - 10 \, a^{3} b + 32 \, a^{2} b^{2} - 38 \, a b^{3} + 15 \, b^{4}\right )} \cos \left (f x + e\right )^{4} - a^{2} b^{2} + 6 \, a b^{3} - 5 \, b^{4} - {\left (2 \, a^{3} b - 15 \, a^{2} b^{2} + 28 \, a b^{3} - 15 \, b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \log \left (-\frac {2 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) - 2 \, {\left (3 \, {\left (a^{4} - 7 \, a^{3} b + 11 \, a^{2} b^{2} - 5 \, a b^{3}\right )} \cos \left (f x + e\right )^{5} + 2 \, {\left (9 \, a^{3} b - 23 \, a^{2} b^{2} + 15 \, a b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (13 \, a^{2} b^{2} - 15 \, a b^{3}\right )} \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}}}}{12 \, {\left ({\left (a^{7} - 3 \, a^{6} b + 3 \, a^{5} b^{2} - a^{4} b^{3}\right )} f \cos \left (f x + e\right )^{6} - {\left (a^{7} - 5 \, a^{6} b + 7 \, a^{5} b^{2} - 3 \, a^{4} b^{3}\right )} f \cos \left (f x + e\right )^{4} - {\left (2 \, a^{6} b - 5 \, a^{5} b^{2} + 3 \, a^{4} b^{3}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{5} b^{2} - a^{4} b^{3}\right )} f\right )}}, \frac {3 \, {\left ({\left (a^{4} - 8 \, a^{3} b + 18 \, a^{2} b^{2} - 16 \, a b^{3} + 5 \, b^{4}\right )} \cos \left (f x + e\right )^{6} - {\left (a^{4} - 10 \, a^{3} b + 32 \, a^{2} b^{2} - 38 \, a b^{3} + 15 \, b^{4}\right )} \cos \left (f x + e\right )^{4} - a^{2} b^{2} + 6 \, a b^{3} - 5 \, b^{4} - {\left (2 \, a^{3} b - 15 \, a^{2} b^{2} + 28 \, a b^{3} - 15 \, b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a}\right ) + {\left (3 \, {\left (a^{4} - 7 \, a^{3} b + 11 \, a^{2} b^{2} - 5 \, a b^{3}\right )} \cos \left (f x + e\right )^{5} + 2 \, {\left (9 \, a^{3} b - 23 \, a^{2} b^{2} + 15 \, a b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (13 \, a^{2} b^{2} - 15 \, a b^{3}\right )} \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}}}}{6 \, {\left ({\left (a^{7} - 3 \, a^{6} b + 3 \, a^{5} b^{2} - a^{4} b^{3}\right )} f \cos \left (f x + e\right )^{6} - {\left (a^{7} - 5 \, a^{6} b + 7 \, a^{5} b^{2} - 3 \, a^{4} b^{3}\right )} f \cos \left (f x + e\right )^{4} - {\left (2 \, a^{6} b - 5 \, a^{5} b^{2} + 3 \, a^{4} b^{3}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{5} b^{2} - a^{4} b^{3}\right )} f\right )}}\right ] \]
[-1/12*(3*((a^4 - 8*a^3*b + 18*a^2*b^2 - 16*a*b^3 + 5*b^4)*cos(f*x + e)^6 - (a^4 - 10*a^3*b + 32*a^2*b^2 - 38*a*b^3 + 15*b^4)*cos(f*x + e)^4 - a^2*b ^2 + 6*a*b^3 - 5*b^4 - (2*a^3*b - 15*a^2*b^2 + 28*a*b^3 - 15*b^4)*cos(f*x + e)^2)*sqrt(a)*log(-2*((a - b)*cos(f*x + e)^2 + 2*sqrt(a)*sqrt(((a - b)*c os(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + a + b)/(cos(f*x + e)^2 - 1)) - 2*(3*(a^4 - 7*a^3*b + 11*a^2*b^2 - 5*a*b^3)*cos(f*x + e)^5 + 2*(9*a ^3*b - 23*a^2*b^2 + 15*a*b^3)*cos(f*x + e)^3 + (13*a^2*b^2 - 15*a*b^3)*cos (f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^7 - 3*a^ 6*b + 3*a^5*b^2 - a^4*b^3)*f*cos(f*x + e)^6 - (a^7 - 5*a^6*b + 7*a^5*b^2 - 3*a^4*b^3)*f*cos(f*x + e)^4 - (2*a^6*b - 5*a^5*b^2 + 3*a^4*b^3)*f*cos(f*x + e)^2 - (a^5*b^2 - a^4*b^3)*f), 1/6*(3*((a^4 - 8*a^3*b + 18*a^2*b^2 - 16 *a*b^3 + 5*b^4)*cos(f*x + e)^6 - (a^4 - 10*a^3*b + 32*a^2*b^2 - 38*a*b^3 + 15*b^4)*cos(f*x + e)^4 - a^2*b^2 + 6*a*b^3 - 5*b^4 - (2*a^3*b - 15*a^2*b^ 2 + 28*a*b^3 - 15*b^4)*cos(f*x + e)^2)*sqrt(-a)*arctan(sqrt(-a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/a) + (3*(a^4 - 7*a^3* b + 11*a^2*b^2 - 5*a*b^3)*cos(f*x + e)^5 + 2*(9*a^3*b - 23*a^2*b^2 + 15*a* b^3)*cos(f*x + e)^3 + (13*a^2*b^2 - 15*a*b^3)*cos(f*x + e))*sqrt(((a - b)* cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^7 - 3*a^6*b + 3*a^5*b^2 - a^4*b^3 )*f*cos(f*x + e)^6 - (a^7 - 5*a^6*b + 7*a^5*b^2 - 3*a^4*b^3)*f*cos(f*x + e )^4 - (2*a^6*b - 5*a^5*b^2 + 3*a^4*b^3)*f*cos(f*x + e)^2 - (a^5*b^2 - a...
\[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\csc ^{3}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\csc \left (f x + e\right )^{3}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {1}{{\sin \left (e+f\,x\right )}^3\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \]