Integrand size = 23, antiderivative size = 167 \[ \int \frac {\tan ^2(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {11 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{128 \sqrt {2} a^{5/2} f}-\frac {\sec (e+f x)}{6 f (a+a \sin (e+f x))^{5/2}}-\frac {11 \cos (e+f x)}{128 a f (a+a \sin (e+f x))^{3/2}}+\frac {17 \sec (e+f x)}{48 a f (a+a \sin (e+f x))^{3/2}}+\frac {11 \sec (e+f x)}{96 a^2 f \sqrt {a+a \sin (e+f x)}} \] Output:
-11/256*arctanh(1/2*a^(1/2)*cos(f*x+e)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))*2^( 1/2)/a^(5/2)/f-1/6*sec(f*x+e)/f/(a+a*sin(f*x+e))^(5/2)-11/128*cos(f*x+e)/a /f/(a+a*sin(f*x+e))^(3/2)+17/48*sec(f*x+e)/a/f/(a+a*sin(f*x+e))^(3/2)+11/9 6*sec(f*x+e)/a^2/f/(a+a*sin(f*x+e))^(1/2)
Result contains complex when optimal does not.
Time = 1.70 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.70 \[ \int \frac {\tan ^2(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {-32+\frac {64 \sin \left (\frac {1}{2} (e+f x)\right )}{\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )}-104 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+52 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-30 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+15 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+(33+33 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+\frac {48 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5}{\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )}}{384 f (a (1+\sin (e+f x)))^{5/2}} \] Input:
Integrate[Tan[e + f*x]^2/(a + a*Sin[e + f*x])^(5/2),x]
Output:
(-32 + (64*Sin[(e + f*x)/2])/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) - 104*S in[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + 52*(Cos[(e + f*x)/ 2] + Sin[(e + f*x)/2])^2 - 30*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + 15*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 + (33 + 33*I)*(- 1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5 + (48*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) ^5)/(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))/(384*f*(a*(1 + Sin[e + f*x]))^( 5/2))
Time = 0.81 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.05, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {3042, 3191, 27, 3042, 3338, 3042, 3166, 3042, 3129, 3042, 3128, 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.
| \(\displaystyle \int \frac {\tan ^2(e+f x)}{(a \sin (e+f x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (e+f x)^2}{(a \sin (e+f x)+a)^{5/2}}dx\) |
| \(\Big \downarrow \) 3191 |
| \(\displaystyle \frac {\int -\frac {\sec ^2(e+f x) (5 a-12 a \sin (e+f x))}{2 (\sin (e+f x) a+a)^{3/2}}dx}{6 a^2}-\frac {\sec (e+f x)}{6 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\sec ^2(e+f x) (5 a-12 a \sin (e+f x))}{(\sin (e+f x) a+a)^{3/2}}dx}{12 a^2}-\frac {\sec (e+f x)}{6 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {5 a-12 a \sin (e+f x)}{\cos (e+f x)^2 (\sin (e+f x) a+a)^{3/2}}dx}{12 a^2}-\frac {\sec (e+f x)}{6 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3338 |
| \(\displaystyle -\frac {-\frac {11}{8} \int \frac {\sec ^2(e+f x)}{\sqrt {\sin (e+f x) a+a}}dx-\frac {17 a \sec (e+f x)}{4 f (a \sin (e+f x)+a)^{3/2}}}{12 a^2}-\frac {\sec (e+f x)}{6 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {11}{8} \int \frac {1}{\cos (e+f x)^2 \sqrt {\sin (e+f x) a+a}}dx-\frac {17 a \sec (e+f x)}{4 f (a \sin (e+f x)+a)^{3/2}}}{12 a^2}-\frac {\sec (e+f x)}{6 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3166 |
| \(\displaystyle -\frac {-\frac {11}{8} \left (\frac {3}{2} a \int \frac {1}{(\sin (e+f x) a+a)^{3/2}}dx+\frac {\sec (e+f x)}{f \sqrt {a \sin (e+f x)+a}}\right )-\frac {17 a \sec (e+f x)}{4 f (a \sin (e+f x)+a)^{3/2}}}{12 a^2}-\frac {\sec (e+f x)}{6 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {11}{8} \left (\frac {3}{2} a \int \frac {1}{(\sin (e+f x) a+a)^{3/2}}dx+\frac {\sec (e+f x)}{f \sqrt {a \sin (e+f x)+a}}\right )-\frac {17 a \sec (e+f x)}{4 f (a \sin (e+f x)+a)^{3/2}}}{12 a^2}-\frac {\sec (e+f x)}{6 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3129 |
| \(\displaystyle -\frac {-\frac {11}{8} \left (\frac {3}{2} a \left (\frac {\int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx}{4 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}\right )+\frac {\sec (e+f x)}{f \sqrt {a \sin (e+f x)+a}}\right )-\frac {17 a \sec (e+f x)}{4 f (a \sin (e+f x)+a)^{3/2}}}{12 a^2}-\frac {\sec (e+f x)}{6 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {11}{8} \left (\frac {3}{2} a \left (\frac {\int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx}{4 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}\right )+\frac {\sec (e+f x)}{f \sqrt {a \sin (e+f x)+a}}\right )-\frac {17 a \sec (e+f x)}{4 f (a \sin (e+f x)+a)^{3/2}}}{12 a^2}-\frac {\sec (e+f x)}{6 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle -\frac {-\frac {11}{8} \left (\frac {3}{2} a \left (-\frac {\int \frac {1}{2 a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{2 a f}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}\right )+\frac {\sec (e+f x)}{f \sqrt {a \sin (e+f x)+a}}\right )-\frac {17 a \sec (e+f x)}{4 f (a \sin (e+f x)+a)^{3/2}}}{12 a^2}-\frac {\sec (e+f x)}{6 f (a \sin (e+f x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {-\frac {11}{8} \left (\frac {3}{2} a \left (-\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{2 \sqrt {2} a^{3/2} f}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}\right )+\frac {\sec (e+f x)}{f \sqrt {a \sin (e+f x)+a}}\right )-\frac {17 a \sec (e+f x)}{4 f (a \sin (e+f x)+a)^{3/2}}}{12 a^2}-\frac {\sec (e+f x)}{6 f (a \sin (e+f x)+a)^{5/2}}\) |
Input:
Int[Tan[e + f*x]^2/(a + a*Sin[e + f*x])^(5/2),x]
Output:
-1/6*Sec[e + f*x]/(f*(a + a*Sin[e + f*x])^(5/2)) - ((-17*a*Sec[e + f*x])/( 4*f*(a + a*Sin[e + f*x])^(3/2)) - (11*(Sec[e + f*x]/(f*Sqrt[a + a*Sin[e + f*x]]) + (3*a*(-1/2*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin [e + f*x]])]/(Sqrt[2]*a^(3/2)*f) - Cos[e + f*x]/(2*f*(a + a*Sin[e + f*x])^ (3/2))))/2))/8)/(12*a^2)
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_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d*(2*n + 1))), x] + Simp[(n + 1)/(a*(2*n + 1)) Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_. )*(x_)]], x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(a*f*g*(p + 1)*S qrt[a + b*Sin[e + f*x]])), x] + Simp[a*((2*p + 1)/(2*g^2*(p + 1))) Int[(g *Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e , f, g}, x] && EqQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*p]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> Simp[b*((a + b*Sin[e + f*x])^m/(a*f*(2*m - 1)*Cos[e + f*x])), x] - Simp[1/(a^2*(2*m - 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*((a*m - b*(2 *m - 1)*Sin[e + f*x])/Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[m] && LtQ[m, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(2*m + p + 1) )), x] + Simp[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[m + p], 0 ]) && NeQ[2*m + p + 1, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(287\) vs. \(2(140)=280\).
Time = 0.32 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.72
| method | result | size |
| default | \(-\frac {66 \cos \left (f x +e \right )^{2} \sin \left (f x +e \right ) a^{\frac {7}{2}}-33 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {a -\sin \left (f x +e \right ) a}\, a^{3} \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )+154 \cos \left (f x +e \right )^{2} a^{\frac {7}{2}}-99 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {a -\sin \left (f x +e \right ) a}\, a^{3} \cos \left (f x +e \right )^{2}-448 \sin \left (f x +e \right ) a^{\frac {7}{2}}+132 \sqrt {a -\sin \left (f x +e \right ) a}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {2}\, \sin \left (f x +e \right ) a^{3}-320 a^{\frac {7}{2}}+132 \sqrt {a -\sin \left (f x +e \right ) a}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {2}\, a^{3}}{768 a^{\frac {11}{2}} \left (1+\sin \left (f x +e \right )\right )^{2} \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\) | \(288\) |
Input:
int(tan(f*x+e)^2/(a+sin(f*x+e)*a)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/768/a^(11/2)*(66*cos(f*x+e)^2*sin(f*x+e)*a^(7/2)-33*2^(1/2)*arctanh(1/2 *(a-sin(f*x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*(a-sin(f*x+e)*a)^(1/2)*a^3*cos(f* x+e)^2*sin(f*x+e)+154*cos(f*x+e)^2*a^(7/2)-99*2^(1/2)*arctanh(1/2*(a-sin(f *x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*(a-sin(f*x+e)*a)^(1/2)*a^3*cos(f*x+e)^2-44 8*sin(f*x+e)*a^(7/2)+132*(a-sin(f*x+e)*a)^(1/2)*arctanh(1/2*(a-sin(f*x+e)* a)^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*sin(f*x+e)*a^3-320*a^(7/2)+132*(a-sin(f* x+e)*a)^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)* a^3)/(1+sin(f*x+e))^2/cos(f*x+e)/(a+sin(f*x+e)*a)^(1/2)/f
Time = 0.16 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.67 \[ \int \frac {\tan ^2(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {33 \, \sqrt {2} {\left (3 \, \cos \left (f x + e\right )^{3} + {\left (\cos \left (f x + e\right )^{3} - 4 \, \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) - 4 \, \cos \left (f x + e\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 4 \, {\left (77 \, \cos \left (f x + e\right )^{2} + {\left (33 \, \cos \left (f x + e\right )^{2} - 224\right )} \sin \left (f x + e\right ) - 160\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{1536 \, {\left (3 \, a^{3} f \cos \left (f x + e\right )^{3} - 4 \, a^{3} f \cos \left (f x + e\right ) + {\left (a^{3} f \cos \left (f x + e\right )^{3} - 4 \, a^{3} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate(tan(f*x+e)^2/(a+a*sin(f*x+e))^(5/2),x, algorithm="fricas")
Output:
1/1536*(33*sqrt(2)*(3*cos(f*x + e)^3 + (cos(f*x + e)^3 - 4*cos(f*x + e))*s in(f*x + e) - 4*cos(f*x + e))*sqrt(a)*log(-(a*cos(f*x + e)^2 - 2*sqrt(2)*s qrt(a*sin(f*x + e) + a)*sqrt(a)*(cos(f*x + e) - sin(f*x + e) + 1) + 3*a*co s(f*x + e) - (a*cos(f*x + e) - 2*a)*sin(f*x + e) + 2*a)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) + 4*(77*cos(f*x + e)^ 2 + (33*cos(f*x + e)^2 - 224)*sin(f*x + e) - 160)*sqrt(a*sin(f*x + e) + a) )/(3*a^3*f*cos(f*x + e)^3 - 4*a^3*f*cos(f*x + e) + (a^3*f*cos(f*x + e)^3 - 4*a^3*f*cos(f*x + e))*sin(f*x + e))
\[ \int \frac {\tan ^2(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\int \frac {\tan ^{2}{\left (e + f x \right )}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(tan(f*x+e)**2/(a+a*sin(f*x+e))**(5/2),x)
Output:
Integral(tan(e + f*x)**2/(a*(sin(e + f*x) + 1))**(5/2), x)
\[ \int \frac {\tan ^2(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\int { \frac {\tan \left (f x + e\right )^{2}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(tan(f*x+e)^2/(a+a*sin(f*x+e))^(5/2),x, algorithm="maxima")
Output:
integrate(tan(f*x + e)^2/(a*sin(f*x + e) + a)^(5/2), x)
Time = 0.26 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.84 \[ \int \frac {\tan ^2(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\frac {48 \, \sqrt {2}}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )} - \frac {\sqrt {2} {\left (15 \, \sqrt {a} \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 56 \, \sqrt {a} \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 33 \, \sqrt {a} \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{3} a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{768 \, f} \] Input:
integrate(tan(f*x+e)^2/(a+a*sin(f*x+e))^(5/2),x, algorithm="giac")
Output:
1/768*(48*sqrt(2)/(a^(5/2)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(3/4*pi + 1/2*f*x + 1/2*e)) - sqrt(2)*(15*sqrt(a)*sin(3/4*pi + 1/2*f*x + 1/2*e)^5 - 56*sqrt(a)*sin(3/4*pi + 1/2*f*x + 1/2*e)^3 + 33*sqrt(a)*sin(3/4*pi + 1/2 *f*x + 1/2*e))/((sin(3/4*pi + 1/2*f*x + 1/2*e)^2 - 1)^3*a^3*sgn(cos(-1/4*p i + 1/2*f*x + 1/2*e))))/f
Timed out. \[ \int \frac {\tan ^2(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^2}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:
int(tan(e + f*x)^2/(a + a*sin(e + f*x))^(5/2),x)
Output:
int(tan(e + f*x)^2/(a + a*sin(e + f*x))^(5/2), x)
\[ \int \frac {\tan ^2(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \tan \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right )}{a^{3}} \] Input:
int(tan(f*x+e)^2/(a+a*sin(f*x+e))^(5/2),x)
Output:
(sqrt(a)*int((sqrt(sin(e + f*x) + 1)*tan(e + f*x)**2)/(sin(e + f*x)**3 + 3 *sin(e + f*x)**2 + 3*sin(e + f*x) + 1),x))/a**3