Integrand size = 25, antiderivative size = 143 \[ \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2 \, dx=-\frac {2 \left (5 a^2-2 b^2\right ) d^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {14 a b (d \sec (e+f x))^{3/2}}{15 f}+\frac {2 \left (5 a^2-2 b^2\right ) d \sqrt {d \sec (e+f x)} \sin (e+f x)}{5 f}+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f} \] Output:
-2/5*(5*a^2-2*b^2)*d^2*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))/f/cos(f*x+e)^ (1/2)/(d*sec(f*x+e))^(1/2)+14/15*a*b*(d*sec(f*x+e))^(3/2)/f+2/5*(5*a^2-2*b ^2)*d*(d*sec(f*x+e))^(1/2)*sin(f*x+e)/f+2/5*b*(d*sec(f*x+e))^(3/2)*(a+b*ta n(f*x+e))/f
Time = 0.95 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.88 \[ \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2 \, dx=-\frac {2 d^2 (a+b \tan (e+f x))^2 \left (3 \left (5 a^2-2 b^2\right ) \cos ^{\frac {3}{2}}(e+f x) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+\left (-\frac {15 a^2}{2}+3 b^2\right ) \sin (2 (e+f x))-b (10 a+3 b \tan (e+f x))\right )}{15 f \sqrt {d \sec (e+f x)} (a \cos (e+f x)+b \sin (e+f x))^2} \] Input:
Integrate[(d*Sec[e + f*x])^(3/2)*(a + b*Tan[e + f*x])^2,x]
Output:
(-2*d^2*(a + b*Tan[e + f*x])^2*(3*(5*a^2 - 2*b^2)*Cos[e + f*x]^(3/2)*Ellip ticE[(e + f*x)/2, 2] + ((-15*a^2)/2 + 3*b^2)*Sin[2*(e + f*x)] - b*(10*a + 3*b*Tan[e + f*x])))/(15*f*Sqrt[d*Sec[e + f*x]]*(a*Cos[e + f*x] + b*Sin[e + f*x])^2)
Time = 0.72 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.94, 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, 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.
\(\displaystyle \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2dx\) |
\(\Big \downarrow \) 3993 |
\(\displaystyle \frac {2}{5} \int \frac {1}{2} (d \sec (e+f x))^{3/2} \left (5 a^2+7 b \tan (e+f x) a-2 b^2\right )dx+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int (d \sec (e+f x))^{3/2} \left (5 a^2+7 b \tan (e+f x) a-2 b^2\right )dx+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \int (d \sec (e+f x))^{3/2} \left (5 a^2+7 b \tan (e+f x) a-2 b^2\right )dx+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f}\) |
\(\Big \downarrow \) 3967 |
\(\displaystyle \frac {1}{5} \left (\left (5 a^2-2 b^2\right ) \int (d \sec (e+f x))^{3/2}dx+\frac {14 a b (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\left (5 a^2-2 b^2\right ) \int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {14 a b (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {1}{5} \left (\left (5 a^2-2 b^2\right ) \left (\frac {2 d \sin (e+f x) \sqrt {d \sec (e+f x)}}{f}-d^2 \int \frac {1}{\sqrt {d \sec (e+f x)}}dx\right )+\frac {14 a b (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\left (5 a^2-2 b^2\right ) \left (\frac {2 d \sin (e+f x) \sqrt {d \sec (e+f x)}}{f}-d^2 \int \frac {1}{\sqrt {d \csc \left (e+f x+\frac {\pi }{2}\right )}}dx\right )+\frac {14 a b (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {1}{5} \left (\left (5 a^2-2 b^2\right ) \left (\frac {2 d \sin (e+f x) \sqrt {d \sec (e+f x)}}{f}-\frac {d^2 \int \sqrt {\cos (e+f x)}dx}{\sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}\right )+\frac {14 a b (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\left (5 a^2-2 b^2\right ) \left (\frac {2 d \sin (e+f x) \sqrt {d \sec (e+f x)}}{f}-\frac {d^2 \int \sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}dx}{\sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}\right )+\frac {14 a b (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {1}{5} \left (\left (5 a^2-2 b^2\right ) \left (\frac {2 d \sin (e+f x) \sqrt {d \sec (e+f x)}}{f}-\frac {2 d^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}\right )+\frac {14 a b (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f}\) |
Input:
Int[(d*Sec[e + f*x])^(3/2)*(a + b*Tan[e + f*x])^2,x]
Output:
((14*a*b*(d*Sec[e + f*x])^(3/2))/(3*f) + (5*a^2 - 2*b^2)*((-2*d^2*Elliptic E[(e + f*x)/2, 2])/(f*Sqrt[Cos[e + f*x]]*Sqrt[d*Sec[e + f*x]]) + (2*d*Sqrt [d*Sec[e + f*x]]*Sin[e + f*x])/f))/5 + (2*b*(d*Sec[e + f*x])^(3/2)*(a + b* Tan[e + f*x]))/(5*f)
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[(-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 = 23.29 (sec) , antiderivative size = 392, normalized size of antiderivative = 2.74
method | result | size |
default | \(\frac {2 \sqrt {d \sec \left (f x +e \right )}\, \left (15 i \left (-\cos \left (f x +e \right )^{2}-2 \cos \left (f x +e \right )-1\right ) \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, a^{2} \operatorname {EllipticE}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )+6 i \left (\cos \left (f x +e \right )^{2}+2 \cos \left (f x +e \right )+1\right ) \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, b^{2} \operatorname {EllipticE}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )+15 i \left (\cos \left (f x +e \right )^{2}+2 \cos \left (f x +e \right )+1\right ) \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, a^{2} \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )+6 i \left (-\cos \left (f x +e \right )^{2}-2 \cos \left (f x +e \right )-1\right ) \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, b^{2} \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )+15 a^{2} \sin \left (f x +e \right )+10 b a \left (1+\sec \left (f x +e \right )\right )+3 b^{2} \left (-2 \sin \left (f x +e \right )+\tan \left (f x +e \right )+\sec \left (f x +e \right ) \tan \left (f x +e \right )\right )\right ) d}{15 f \left (1+\cos \left (f x +e \right )\right )}\) | \(392\) |
parts | \(\frac {2 a^{2} \left (i \left (\cos \left (f x +e \right )^{2}+2 \cos \left (f x +e \right )+1\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )+i \left (-\cos \left (f x +e \right )^{2}-2 \cos \left (f x +e \right )-1\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticE}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )+\sin \left (f x +e \right )\right ) \sqrt {d \sec \left (f x +e \right )}\, d}{f \left (1+\cos \left (f x +e \right )\right )}-\frac {2 b^{2} \sqrt {d \sec \left (f x +e \right )}\, d \left (2 \sin \left (f x +e \right )-\tan \left (f x +e \right )-\sec \left (f x +e \right ) \tan \left (f x +e \right )+i \left (2 \cos \left (f x +e \right )^{2}+4 \cos \left (f x +e \right )+2\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )+i \left (-2 \cos \left (f x +e \right )^{2}-4 \cos \left (f x +e \right )-2\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticE}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )\right )}{5 f \left (1+\cos \left (f x +e \right )\right )}+\frac {4 a b \left (d \sec \left (f x +e \right )\right )^{\frac {3}{2}}}{3 f}\) | \(413\) |
Input:
int((d*sec(f*x+e))^(3/2)*(a+b*tan(f*x+e))^2,x,method=_RETURNVERBOSE)
Output:
2/15/f*(d*sec(f*x+e))^(1/2)/(1+cos(f*x+e))*(15*I*(-cos(f*x+e)^2-2*cos(f*x+ e)-1)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(1+cos(f*x+e)))^(1/2)*a^2*Ellip ticE(I*(csc(f*x+e)-cot(f*x+e)),I)+6*I*(cos(f*x+e)^2+2*cos(f*x+e)+1)*(cos(f *x+e)/(1+cos(f*x+e)))^(1/2)*(1/(1+cos(f*x+e)))^(1/2)*b^2*EllipticE(I*(csc( f*x+e)-cot(f*x+e)),I)+15*I*(cos(f*x+e)^2+2*cos(f*x+e)+1)*(cos(f*x+e)/(1+co s(f*x+e)))^(1/2)*(1/(1+cos(f*x+e)))^(1/2)*a^2*EllipticF(I*(csc(f*x+e)-cot( f*x+e)),I)+6*I*(-cos(f*x+e)^2-2*cos(f*x+e)-1)*(cos(f*x+e)/(1+cos(f*x+e)))^ (1/2)*(1/(1+cos(f*x+e)))^(1/2)*b^2*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)+ 15*a^2*sin(f*x+e)+10*b*a*(1+sec(f*x+e))+3*b^2*(-2*sin(f*x+e)+tan(f*x+e)+se c(f*x+e)*tan(f*x+e)))*d
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.20 \[ \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2 \, dx=\frac {-3 i \, \sqrt {2} {\left (5 \, a^{2} - 2 \, b^{2}\right )} d^{\frac {3}{2}} \cos \left (f x + e\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 i \, \sqrt {2} {\left (5 \, a^{2} - 2 \, b^{2}\right )} d^{\frac {3}{2}} \cos \left (f x + e\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) + 2 \, {\left (10 \, a b d \cos \left (f x + e\right ) + 3 \, {\left ({\left (5 \, a^{2} - 2 \, b^{2}\right )} d \cos \left (f x + e\right )^{2} + b^{2} d\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{15 \, f \cos \left (f x + e\right )^{2}} \] Input:
integrate((d*sec(f*x+e))^(3/2)*(a+b*tan(f*x+e))^2,x, algorithm="fricas")
Output:
1/15*(-3*I*sqrt(2)*(5*a^2 - 2*b^2)*d^(3/2)*cos(f*x + e)^2*weierstrassZeta( -4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) + 3*I*sq rt(2)*(5*a^2 - 2*b^2)*d^(3/2)*cos(f*x + e)^2*weierstrassZeta(-4, 0, weiers trassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) + 2*(10*a*b*d*cos(f*x + e) + 3*((5*a^2 - 2*b^2)*d*cos(f*x + e)^2 + b^2*d)*sin(f*x + e))*sqrt(d/ cos(f*x + e)))/(f*cos(f*x + e)^2)
\[ \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2 \, dx=\int \left (d \sec {\left (e + f x \right )}\right )^{\frac {3}{2}} \left (a + b \tan {\left (e + f x \right )}\right )^{2}\, dx \] Input:
integrate((d*sec(f*x+e))**(3/2)*(a+b*tan(f*x+e))**2,x)
Output:
Integral((d*sec(e + f*x))**(3/2)*(a + b*tan(e + f*x))**2, x)
\[ \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2 \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}} {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \,d x } \] Input:
integrate((d*sec(f*x+e))^(3/2)*(a+b*tan(f*x+e))^2,x, algorithm="maxima")
Output:
integrate((d*sec(f*x + e))^(3/2)*(b*tan(f*x + e) + a)^2, x)
\[ \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2 \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}} {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \,d x } \] Input:
integrate((d*sec(f*x+e))^(3/2)*(a+b*tan(f*x+e))^2,x, algorithm="giac")
Output:
integrate((d*sec(f*x + e))^(3/2)*(b*tan(f*x + e) + a)^2, x)
Timed out. \[ \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2 \, dx=\int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{3/2}\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2 \,d x \] Input:
int((d/cos(e + f*x))^(3/2)*(a + b*tan(e + f*x))^2,x)
Output:
int((d/cos(e + f*x))^(3/2)*(a + b*tan(e + f*x))^2, x)
\[ \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2 \, dx=\frac {\sqrt {d}\, d \left (4 \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right ) a b +3 \left (\int \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right ) \tan \left (f x +e \right )^{2}d x \right ) b^{2} f +3 \left (\int \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )d x \right ) a^{2} f \right )}{3 f} \] Input:
int((d*sec(f*x+e))^(3/2)*(a+b*tan(f*x+e))^2,x)
Output:
(sqrt(d)*d*(4*sqrt(sec(e + f*x))*sec(e + f*x)*a*b + 3*int(sqrt(sec(e + f*x ))*sec(e + f*x)*tan(e + f*x)**2,x)*b**2*f + 3*int(sqrt(sec(e + f*x))*sec(e + f*x),x)*a**2*f))/(3*f)