Integrand size = 25, antiderivative size = 145 \[ \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{5/2}} \, dx=-\frac {2 a b}{15 f (d \sec (e+f x))^{5/2}}+\frac {2 \left (3 a^2+2 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 d^2 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 \left (3 a^2+2 b^2\right ) \sin (e+f x)}{15 d f (d \sec (e+f x))^{3/2}}-\frac {2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}} \]
-2/15*a*b/f/(d*sec(f*x+e))^(5/2)+2/15*(3*a^2+2*b^2)*sin(f*x+e)/d/f/(d*sec( f*x+e))^(3/2)+2/5*(3*a^2+2*b^2)*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1 /2*e)*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))/d^2/f/cos(f*x+e)^(1/2)/(d*sec( f*x+e))^(1/2)-2/3*b*(a+b*tan(f*x+e))/f/(d*sec(f*x+e))^(5/2)
Time = 2.83 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.63 \[ \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{5/2}} \, dx=\frac {\left (6 a^2+4 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+2 \cos ^{\frac {3}{2}}(e+f x) \left (-2 a b \cos (e+f x)+\left (a^2-b^2\right ) \sin (e+f x)\right )}{5 f \cos ^{\frac {5}{2}}(e+f x) (d \sec (e+f x))^{5/2}} \]
((6*a^2 + 4*b^2)*EllipticE[(e + f*x)/2, 2] + 2*Cos[e + f*x]^(3/2)*(-2*a*b* Cos[e + f*x] + (a^2 - b^2)*Sin[e + f*x]))/(5*f*Cos[e + f*x]^(5/2)*(d*Sec[e + f*x])^(5/2))
Time = 0.70 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.97, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 3993, 27, 3042, 3967, 3042, 4256, 3042, 4258, 3042, 3119}
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.
| \(\displaystyle \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3993 |
| \(\displaystyle -\frac {2}{3} \int -\frac {3 a^2+b \tan (e+f x) a+2 b^2}{2 (d \sec (e+f x))^{5/2}}dx-\frac {2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \int \frac {3 a^2+b \tan (e+f x) a+2 b^2}{(d \sec (e+f x))^{5/2}}dx-\frac {2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {3 a^2+b \tan (e+f x) a+2 b^2}{(d \sec (e+f x))^{5/2}}dx-\frac {2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3967 |
| \(\displaystyle \frac {1}{3} \left (\left (3 a^2+2 b^2\right ) \int \frac {1}{(d \sec (e+f x))^{5/2}}dx-\frac {2 a b}{5 f (d \sec (e+f x))^{5/2}}\right )-\frac {2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\left (3 a^2+2 b^2\right ) \int \frac {1}{\left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{5/2}}dx-\frac {2 a b}{5 f (d \sec (e+f x))^{5/2}}\right )-\frac {2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 4256 |
| \(\displaystyle \frac {1}{3} \left (\left (3 a^2+2 b^2\right ) \left (\frac {3 \int \frac {1}{\sqrt {d \sec (e+f x)}}dx}{5 d^2}+\frac {2 \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}\right )-\frac {2 a b}{5 f (d \sec (e+f x))^{5/2}}\right )-\frac {2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\left (3 a^2+2 b^2\right ) \left (\frac {3 \int \frac {1}{\sqrt {d \csc \left (e+f x+\frac {\pi }{2}\right )}}dx}{5 d^2}+\frac {2 \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}\right )-\frac {2 a b}{5 f (d \sec (e+f x))^{5/2}}\right )-\frac {2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{3} \left (\left (3 a^2+2 b^2\right ) \left (\frac {3 \int \sqrt {\cos (e+f x)}dx}{5 d^2 \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}\right )-\frac {2 a b}{5 f (d \sec (e+f x))^{5/2}}\right )-\frac {2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\left (3 a^2+2 b^2\right ) \left (\frac {3 \int \sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}dx}{5 d^2 \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}\right )-\frac {2 a b}{5 f (d \sec (e+f x))^{5/2}}\right )-\frac {2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{3} \left (\left (3 a^2+2 b^2\right ) \left (\frac {6 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 d^2 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}\right )-\frac {2 a b}{5 f (d \sec (e+f x))^{5/2}}\right )-\frac {2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}}\) |
((-2*a*b)/(5*f*(d*Sec[e + f*x])^(5/2)) + (3*a^2 + 2*b^2)*((6*EllipticE[(e + f*x)/2, 2])/(5*d^2*f*Sqrt[Cos[e + f*x]]*Sqrt[d*Sec[e + f*x]]) + (2*Sin[e + f*x])/(5*d*f*(d*Sec[e + f*x])^(3/2))))/3 - (2*b*(a + b*Tan[e + f*x]))/( 3*f*(d*Sec[e + f*x])^(5/2))
3.6.91.3.1 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[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[b*((d*Sec[e + f*x])^m/(f*m)), x] + Simp[a Int[(d *Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2*m] || NeQ[a^2 + b^2, 0])
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^2, x_Symbol] :> Simp[b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])/(f*(m + 1))), x] + Simp[1/(m + 1) Int[(d*Sec[e + f*x])^m*(a^2*(m + 1) - b^2 + a*b*(m + 2)*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^ 2 + b^2, 0] && !IntegerQ[m]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n) Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2* n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Result contains complex when optimal does not.
Time = 23.41 (sec) , antiderivative size = 863, normalized size of antiderivative = 5.95
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(863\) |
| default | \(\text {Expression too large to display}\) | \(890\) |
2/5*a^2/f/(cos(f*x+e)+1)/(d*sec(f*x+e))^(1/2)/d^2*(3*I*cos(f*x+e)*Elliptic E(I*(csc(f*x+e)-cot(f*x+e)),I)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f *x+e)+1))^(1/2)-3*I*cos(f*x+e)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(1/( cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+6*I*(1/(cos(f*x+e)+ 1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(csc(f*x+e)-cot(f* x+e)),I)-6*I*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(csc(f*x+e)-cot (f*x+e)),I)*(1/(cos(f*x+e)+1))^(1/2)+3*I*sec(f*x+e)*EllipticE(I*(csc(f*x+e )-cot(f*x+e)),I)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2 )-3*I*sec(f*x+e)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(1/(cos(f*x+e)+1)) ^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+sin(f*x+e)*cos(f*x+e)^2+sin(f*x+e )*cos(f*x+e)+3*sin(f*x+e))-2/5*b^2/f/(cos(f*x+e)+1)/(d*sec(f*x+e))^(1/2)/d ^2*(-2*I*cos(f*x+e)*EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I)*(1/(cos(f*x+e)+ 1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+2*I*EllipticF(I*(csc(f*x+e)-co t(f*x+e)),I)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*co s(f*x+e)-4*I*EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I)*(1/(cos(f*x+e)+1))^(1/ 2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+4*I*EllipticF(I*(csc(f*x+e)-cot(f*x+e )),I)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+sin(f*x+e )*cos(f*x+e)^2-2*I*sec(f*x+e)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f* x+e)+1))^(1/2)*EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I)+2*I*sec(f*x+e)*(cos( f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(1/...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.96 \[ \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{5/2}} \, dx=\frac {\sqrt {2} {\left (3 i \, a^{2} + 2 i \, b^{2}\right )} \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + \sqrt {2} {\left (-3 i \, a^{2} - 2 i \, b^{2}\right )} \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) - 2 \, {\left (2 \, a b \cos \left (f x + e\right )^{3} - {\left (a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{5 \, d^{3} f} \]
1/5*(sqrt(2)*(3*I*a^2 + 2*I*b^2)*sqrt(d)*weierstrassZeta(-4, 0, weierstras sPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) + sqrt(2)*(-3*I*a^2 - 2*I *b^2)*sqrt(d)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) - 2*(2*a*b*cos(f*x + e)^3 - (a^2 - b^2)*cos(f*x + e) ^2*sin(f*x + e))*sqrt(d/cos(f*x + e)))/(d^3*f)
\[ \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{5/2}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{2}}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{5/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{5/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{5/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]