Integrand size = 25, antiderivative size = 143 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^2 \, dx=\frac {2 \left (7 a^2-2 b^2\right ) d^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{21 f}+\frac {18 a b (d \sec (e+f x))^{5/2}}{35 f}+\frac {2 \left (7 a^2-2 b^2\right ) d (d \sec (e+f x))^{3/2} \sin (e+f x)}{21 f}+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f} \]
18/35*a*b*(d*sec(f*x+e))^(5/2)/f+2/21*(7*a^2-2*b^2)*d*(d*sec(f*x+e))^(3/2) *sin(f*x+e)/f+2/21*(7*a^2-2*b^2)*d^2*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2* f*x+1/2*e)*EllipticF(sin(1/2*f*x+1/2*e),2^(1/2))*cos(f*x+e)^(1/2)*(d*sec(f *x+e))^(1/2)/f+2/7*b*(d*sec(f*x+e))^(5/2)*(a+b*tan(f*x+e))/f
Time = 2.44 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.89 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^2 \, dx=\frac {2 d^2 \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2 \left (5 \left (7 a^2-2 b^2\right ) \cos ^{\frac {5}{2}}(e+f x) \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )+\frac {5}{2} \left (7 a^2-2 b^2\right ) \sin (2 (e+f x))+3 b (14 a+5 b \tan (e+f x))\right )}{105 f (a \cos (e+f x)+b \sin (e+f x))^2} \]
(2*d^2*Sqrt[d*Sec[e + f*x]]*(a + b*Tan[e + f*x])^2*(5*(7*a^2 - 2*b^2)*Cos[ e + f*x]^(5/2)*EllipticF[(e + f*x)/2, 2] + (5*(7*a^2 - 2*b^2)*Sin[2*(e + f *x)])/2 + 3*b*(14*a + 5*b*Tan[e + f*x])))/(105*f*(a*Cos[e + f*x] + b*Sin[e + f*x])^2)
Time = 0.69 (sec) , antiderivative size = 139, 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, 4255, 3042, 4258, 3042, 3120}
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.
| \(\displaystyle \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^2dx\) |
| \(\Big \downarrow \) 3993 |
| \(\displaystyle \frac {2}{7} \int \frac {1}{2} (d \sec (e+f x))^{5/2} \left (7 a^2+9 b \tan (e+f x) a-2 b^2\right )dx+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} \int (d \sec (e+f x))^{5/2} \left (7 a^2+9 b \tan (e+f x) a-2 b^2\right )dx+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \int (d \sec (e+f x))^{5/2} \left (7 a^2+9 b \tan (e+f x) a-2 b^2\right )dx+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}\) |
| \(\Big \downarrow \) 3967 |
| \(\displaystyle \frac {1}{7} \left (\left (7 a^2-2 b^2\right ) \int (d \sec (e+f x))^{5/2}dx+\frac {18 a b (d \sec (e+f x))^{5/2}}{5 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\left (7 a^2-2 b^2\right ) \int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{5/2}dx+\frac {18 a b (d \sec (e+f x))^{5/2}}{5 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{7} \left (\left (7 a^2-2 b^2\right ) \left (\frac {1}{3} d^2 \int \sqrt {d \sec (e+f x)}dx+\frac {2 d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {18 a b (d \sec (e+f x))^{5/2}}{5 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\left (7 a^2-2 b^2\right ) \left (\frac {1}{3} d^2 \int \sqrt {d \csc \left (e+f x+\frac {\pi }{2}\right )}dx+\frac {2 d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {18 a b (d \sec (e+f x))^{5/2}}{5 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{7} \left (\left (7 a^2-2 b^2\right ) \left (\frac {1}{3} d^2 \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx+\frac {2 d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {18 a b (d \sec (e+f x))^{5/2}}{5 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\left (7 a^2-2 b^2\right ) \left (\frac {1}{3} d^2 \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx+\frac {2 d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {18 a b (d \sec (e+f x))^{5/2}}{5 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{7} \left (\left (7 a^2-2 b^2\right ) \left (\frac {2 d^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{3 f}+\frac {2 d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {18 a b (d \sec (e+f x))^{5/2}}{5 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}\) |
((18*a*b*(d*Sec[e + f*x])^(5/2))/(5*f) + (7*a^2 - 2*b^2)*((2*d^2*Sqrt[Cos[ e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[d*Sec[e + f*x]])/(3*f) + (2*d*(d* Sec[e + f*x])^(3/2)*Sin[e + f*x])/(3*f)))/7 + (2*b*(d*Sec[e + f*x])^(5/2)* (a + b*Tan[e + f*x]))/(7*f)
3.6.86.3.1 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[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[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 = 146.29 (sec) , antiderivative size = 317, normalized size of antiderivative = 2.22
| method | result | size |
| default | \(-\frac {2 d^{2} \sqrt {d \sec \left (f x +e \right )}\, \left (35 i \cos \left (f x +e \right ) F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, a^{2}-10 i \cos \left (f x +e \right ) F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, b^{2}+35 i \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) a^{2}-10 i \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) b^{2}-35 \tan \left (f x +e \right ) a^{2}+10 \tan \left (f x +e \right ) b^{2}-42 \left (\sec ^{2}\left (f x +e \right )\right ) a b -15 \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right ) b^{2}\right )}{105 f}\) | \(317\) |
| parts | \(-\frac {2 a^{2} \sqrt {d \sec \left (f x +e \right )}\, d^{2} \left (i \cos \left (f x +e \right ) F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+i \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}-\tan \left (f x +e \right )\right )}{3 f}+\frac {2 i b^{2} \sqrt {d \sec \left (f x +e \right )}\, d^{2} \left (2 F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \cos \left (f x +e \right )+2 \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}+2 i \tan \left (f x +e \right )-3 i \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right )\right )}{21 f}+\frac {4 a b \left (d \sec \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f}\) | \(328\) |
-2/105*d^2/f*(d*sec(f*x+e))^(1/2)*(35*I*cos(f*x+e)*EllipticF(I*(csc(f*x+e) -cot(f*x+e)),I)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2) *a^2-10*I*cos(f*x+e)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(cos(f*x+e)/(c os(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*b^2+35*I*(cos(f*x+e)/(cos(f*x +e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)) ,I)*a^2-10*I*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*El lipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*b^2-35*tan(f*x+e)*a^2+10*tan(f*x+e)*b ^2-42*sec(f*x+e)^2*a*b-15*tan(f*x+e)*sec(f*x+e)^2*b^2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.20 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^2 \, dx=\frac {-5 i \, \sqrt {2} {\left (7 \, a^{2} - 2 \, b^{2}\right )} d^{\frac {5}{2}} \cos \left (f x + e\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 5 i \, \sqrt {2} {\left (7 \, a^{2} - 2 \, b^{2}\right )} d^{\frac {5}{2}} \cos \left (f x + e\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, {\left (42 \, a b d^{2} \cos \left (f x + e\right ) + 5 \, {\left ({\left (7 \, a^{2} - 2 \, b^{2}\right )} d^{2} \cos \left (f x + e\right )^{2} + 3 \, b^{2} d^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{105 \, f \cos \left (f x + e\right )^{3}} \]
1/105*(-5*I*sqrt(2)*(7*a^2 - 2*b^2)*d^(5/2)*cos(f*x + e)^3*weierstrassPInv erse(-4, 0, cos(f*x + e) + I*sin(f*x + e)) + 5*I*sqrt(2)*(7*a^2 - 2*b^2)*d ^(5/2)*cos(f*x + e)^3*weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) + 2*(42*a*b*d^2*cos(f*x + e) + 5*((7*a^2 - 2*b^2)*d^2*cos(f*x + e)^2 + 3*b^2*d^2)*sin(f*x + e))*sqrt(d/cos(f*x + e)))/(f*cos(f*x + e)^3)
\[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^2 \, dx=\int \left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{2}} \left (a + b \tan {\left (e + f x \right )}\right )^{2}\, dx \]
\[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^2 \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}} {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \,d x } \]
\[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^2 \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}} {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \,d x } \]
Timed out. \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^2 \, dx=\int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2 \,d x \]