Integrand size = 25, antiderivative size = 184 \[ \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{7/2}} \, dx=-\frac {6 a b}{35 f (d \sec (e+f x))^{7/2}}+\frac {2 \left (5 a^2+2 b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{21 d^4 f}+\frac {2 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{35 d f (d \sec (e+f x))^{5/2}}+\frac {2 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{21 d^3 f \sqrt {d \sec (e+f x)}}-\frac {2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}} \] Output:
-6/35*a*b/f/(d*sec(f*x+e))^(7/2)+2/21*(5*a^2+2*b^2)*cos(f*x+e)^(1/2)*Inver seJacobiAM(1/2*f*x+1/2*e,2^(1/2))*(d*sec(f*x+e))^(1/2)/d^4/f+2/35*(5*a^2+2 *b^2)*sin(f*x+e)/d/f/(d*sec(f*x+e))^(5/2)+2/21*(5*a^2+2*b^2)*sin(f*x+e)/d^ 3/f/(d*sec(f*x+e))^(1/2)-2/5*b*(a+b*tan(f*x+e))/f/(d*sec(f*x+e))^(7/2)
Time = 2.85 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.69 \[ \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{7/2}} \, dx=\frac {-18 a b \cos (e+f x)-6 a b \cos (3 (e+f x))+\frac {4 \left (5 a^2+2 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{\sqrt {\cos (e+f x)}}+23 a^2 \sin (e+f x)+5 b^2 \sin (e+f x)+3 a^2 \sin (3 (e+f x))-3 b^2 \sin (3 (e+f x))}{42 d^3 f \sqrt {d \sec (e+f x)}} \] Input:
Integrate[(a + b*Tan[e + f*x])^2/(d*Sec[e + f*x])^(7/2),x]
Output:
(-18*a*b*Cos[e + f*x] - 6*a*b*Cos[3*(e + f*x)] + (4*(5*a^2 + 2*b^2)*Ellipt icF[(e + f*x)/2, 2])/Sqrt[Cos[e + f*x]] + 23*a^2*Sin[e + f*x] + 5*b^2*Sin[ e + f*x] + 3*a^2*Sin[3*(e + f*x)] - 3*b^2*Sin[3*(e + f*x)])/(42*d^3*f*Sqrt [d*Sec[e + f*x]])
Time = 0.90 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.96, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 3993, 27, 3042, 3967, 3042, 4256, 3042, 4256, 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.
\(\displaystyle \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{7/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{7/2}}dx\) |
\(\Big \downarrow \) 3993 |
\(\displaystyle -\frac {2}{5} \int -\frac {5 a^2+3 b \tan (e+f x) a+2 b^2}{2 (d \sec (e+f x))^{7/2}}dx-\frac {2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {5 a^2+3 b \tan (e+f x) a+2 b^2}{(d \sec (e+f x))^{7/2}}dx-\frac {2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \int \frac {5 a^2+3 b \tan (e+f x) a+2 b^2}{(d \sec (e+f x))^{7/2}}dx-\frac {2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}\) |
\(\Big \downarrow \) 3967 |
\(\displaystyle \frac {1}{5} \left (\left (5 a^2+2 b^2\right ) \int \frac {1}{(d \sec (e+f x))^{7/2}}dx-\frac {6 a b}{7 f (d \sec (e+f x))^{7/2}}\right )-\frac {2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\left (5 a^2+2 b^2\right ) \int \frac {1}{\left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{7/2}}dx-\frac {6 a b}{7 f (d \sec (e+f x))^{7/2}}\right )-\frac {2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}\) |
\(\Big \downarrow \) 4256 |
\(\displaystyle \frac {1}{5} \left (\left (5 a^2+2 b^2\right ) \left (\frac {5 \int \frac {1}{(d \sec (e+f x))^{3/2}}dx}{7 d^2}+\frac {2 \sin (e+f x)}{7 d f (d \sec (e+f x))^{5/2}}\right )-\frac {6 a b}{7 f (d \sec (e+f x))^{7/2}}\right )-\frac {2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\left (5 a^2+2 b^2\right ) \left (\frac {5 \int \frac {1}{\left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{7 d^2}+\frac {2 \sin (e+f x)}{7 d f (d \sec (e+f x))^{5/2}}\right )-\frac {6 a b}{7 f (d \sec (e+f x))^{7/2}}\right )-\frac {2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}\) |
\(\Big \downarrow \) 4256 |
\(\displaystyle \frac {1}{5} \left (\left (5 a^2+2 b^2\right ) \left (\frac {5 \left (\frac {\int \sqrt {d \sec (e+f x)}dx}{3 d^2}+\frac {2 \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}}\right )}{7 d^2}+\frac {2 \sin (e+f x)}{7 d f (d \sec (e+f x))^{5/2}}\right )-\frac {6 a b}{7 f (d \sec (e+f x))^{7/2}}\right )-\frac {2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\left (5 a^2+2 b^2\right ) \left (\frac {5 \left (\frac {\int \sqrt {d \csc \left (e+f x+\frac {\pi }{2}\right )}dx}{3 d^2}+\frac {2 \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}}\right )}{7 d^2}+\frac {2 \sin (e+f x)}{7 d f (d \sec (e+f x))^{5/2}}\right )-\frac {6 a b}{7 f (d \sec (e+f x))^{7/2}}\right )-\frac {2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {1}{5} \left (\left (5 a^2+2 b^2\right ) \left (\frac {5 \left (\frac {\sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx}{3 d^2}+\frac {2 \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}}\right )}{7 d^2}+\frac {2 \sin (e+f x)}{7 d f (d \sec (e+f x))^{5/2}}\right )-\frac {6 a b}{7 f (d \sec (e+f x))^{7/2}}\right )-\frac {2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\left (5 a^2+2 b^2\right ) \left (\frac {5 \left (\frac {\sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx}{3 d^2}+\frac {2 \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}}\right )}{7 d^2}+\frac {2 \sin (e+f x)}{7 d f (d \sec (e+f x))^{5/2}}\right )-\frac {6 a b}{7 f (d \sec (e+f x))^{7/2}}\right )-\frac {2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {1}{5} \left (\left (5 a^2+2 b^2\right ) \left (\frac {5 \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{3 d^2 f}+\frac {2 \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}}\right )}{7 d^2}+\frac {2 \sin (e+f x)}{7 d f (d \sec (e+f x))^{5/2}}\right )-\frac {6 a b}{7 f (d \sec (e+f x))^{7/2}}\right )-\frac {2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}\) |
Input:
Int[(a + b*Tan[e + f*x])^2/(d*Sec[e + f*x])^(7/2),x]
Output:
((-6*a*b)/(7*f*(d*Sec[e + f*x])^(7/2)) + (5*a^2 + 2*b^2)*((2*Sin[e + f*x]) /(7*d*f*(d*Sec[e + f*x])^(5/2)) + (5*((2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[d*Sec[e + f*x]])/(3*d^2*f) + (2*Sin[e + f*x])/(3*d*f*Sqrt [d*Sec[e + f*x]])))/(7*d^2)))/5 - (2*b*(a + b*Tan[e + f*x]))/(5*f*(d*Sec[e + f*x])^(7/2))
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[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 = 26.70 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.15
method | result | size |
default | \(-\frac {2 \left (i \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, a^{2} \operatorname {EllipticF}\left (i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right ), i\right ) \left (-5-5 \sec \left (f x +e \right )\right )+i \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, b^{2} \operatorname {EllipticF}\left (i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right ), i\right ) \left (-2-2 \sec \left (f x +e \right )\right )+\sin \left (f x +e \right ) \left (-3 \cos \left (f x +e \right )^{2}-5\right ) a^{2}+6 \cos \left (f x +e \right )^{3} a b +\sin \left (f x +e \right ) \left (3 \cos \left (f x +e \right )^{2}-2\right ) b^{2}\right )}{21 f \sqrt {d \sec \left (f x +e \right )}\, d^{3}}\) | \(212\) |
parts | \(-\frac {2 a^{2} \left (\sin \left (f x +e \right ) \left (-3 \cos \left (f x +e \right )^{2}-5\right )+i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \left (5+5 \sec \left (f x +e \right )\right )\right )}{21 f \sqrt {d \sec \left (f x +e \right )}\, d^{3}}-\frac {2 b^{2} \left (\sin \left (f x +e \right ) \left (3 \cos \left (f x +e \right )^{2}-2\right )+i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \left (2+2 \sec \left (f x +e \right )\right )\right )}{21 f \sqrt {d \sec \left (f x +e \right )}\, d^{3}}-\frac {4 a b}{7 f \left (d \sec \left (f x +e \right )\right )^{\frac {7}{2}}}\) | \(231\) |
Input:
int((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(7/2),x,method=_RETURNVERBOSE)
Output:
-2/21/f/(d*sec(f*x+e))^(1/2)/d^3*(I*(cos(f*x+e)/(1+cos(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)*(-5-5*sec (f*x+e))+I*(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)*(-2-2*sec(f*x+e))+sin(f*x+e)*(-3*c os(f*x+e)^2-5)*a^2+6*cos(f*x+e)^3*a*b+sin(f*x+e)*(3*cos(f*x+e)^2-2)*b^2)
Result contains complex when optimal does not.
Time = 0.12 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.84 \[ \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{7/2}} \, dx=\frac {\sqrt {2} {\left (-5 i \, a^{2} - 2 i \, b^{2}\right )} \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + \sqrt {2} {\left (5 i \, a^{2} + 2 i \, b^{2}\right )} \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 2 \, {\left (6 \, a b \cos \left (f x + e\right )^{4} - {\left (3 \, {\left (a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (5 \, a^{2} + 2 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{21 \, d^{4} f} \] Input:
integrate((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(7/2),x, algorithm="fricas")
Output:
1/21*(sqrt(2)*(-5*I*a^2 - 2*I*b^2)*sqrt(d)*weierstrassPInverse(-4, 0, cos( f*x + e) + I*sin(f*x + e)) + sqrt(2)*(5*I*a^2 + 2*I*b^2)*sqrt(d)*weierstra ssPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) - 2*(6*a*b*cos(f*x + e)^4 - (3*(a^2 - b^2)*cos(f*x + e)^3 + (5*a^2 + 2*b^2)*cos(f*x + e))*sin(f*x + e))*sqrt(d/cos(f*x + e)))/(d^4*f)
\[ \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{7/2}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{2}}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx \] Input:
integrate((a+b*tan(f*x+e))**2/(d*sec(f*x+e))**(7/2),x)
Output:
Integral((a + b*tan(e + f*x))**2/(d*sec(e + f*x))**(7/2), x)
\[ \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{7/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(7/2),x, algorithm="maxima")
Output:
integrate((b*tan(f*x + e) + a)^2/(d*sec(f*x + e))^(7/2), x)
\[ \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{7/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(7/2),x, algorithm="giac")
Output:
integrate((b*tan(f*x + e) + a)^2/(d*sec(f*x + e))^(7/2), x)
Timed out. \[ \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{7/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{7/2}} \,d x \] Input:
int((a + b*tan(e + f*x))^2/(d/cos(e + f*x))^(7/2),x)
Output:
int((a + b*tan(e + f*x))^2/(d/cos(e + f*x))^(7/2), x)
\[ \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{7/2}} \, dx=\frac {\sqrt {d}\, \left (\left (\int \frac {\sqrt {\sec \left (f x +e \right )}}{\sec \left (f x +e \right )^{4}}d x \right ) a^{2}+\left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \tan \left (f x +e \right )^{2}}{\sec \left (f x +e \right )^{4}}d x \right ) b^{2}+2 \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \tan \left (f x +e \right )}{\sec \left (f x +e \right )^{4}}d x \right ) a b \right )}{d^{4}} \] Input:
int((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(7/2),x)
Output:
(sqrt(d)*(int(sqrt(sec(e + f*x))/sec(e + f*x)**4,x)*a**2 + int((sqrt(sec(e + f*x))*tan(e + f*x)**2)/sec(e + f*x)**4,x)*b**2 + 2*int((sqrt(sec(e + f* x))*tan(e + f*x))/sec(e + f*x)**4,x)*a*b))/d**4