Integrand size = 25, antiderivative size = 176 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{9/2}} \, dx=\frac {2 a \left (7 a^2+6 b^2\right ) E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{15 d^4 f \sqrt {d \sec (e+f x)}}-\frac {2 \cos ^4(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{9 d^4 f \sqrt {d \sec (e+f x)}}-\frac {2 \cos ^2(e+f x) \left (2 b \left (5 a^2+2 b^2\right )-a \left (7 a^2+b^2\right ) \tan (e+f x)\right )}{45 d^4 f \sqrt {d \sec (e+f x)}} \]
2/15*a*(7*a^2+6*b^2)*(cos(1/2*arctan(tan(f*x+e)))^2)^(1/2)/cos(1/2*arctan( tan(f*x+e)))*EllipticE(sin(1/2*arctan(tan(f*x+e))),2^(1/2))*(sec(f*x+e)^2) ^(1/4)/d^4/f/(d*sec(f*x+e))^(1/2)-2/9*cos(f*x+e)^4*(b-a*tan(f*x+e))*(a+b*t an(f*x+e))^2/d^4/f/(d*sec(f*x+e))^(1/2)-2/45*cos(f*x+e)^2*(2*b*(5*a^2+2*b^ 2)-a*(7*a^2+b^2)*tan(f*x+e))/d^4/f/(d*sec(f*x+e))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(372\) vs. \(2(176)=352\).
Time = 9.41 (sec) , antiderivative size = 372, normalized size of antiderivative = 2.11 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{9/2}} \, dx=\frac {\sec ^{\frac {3}{2}}(e+f x) \left (\frac {2 \left (56 a^3+48 a b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{\sqrt {\cos (e+f x)} \sqrt {\sec (e+f x)}}-\frac {2 \left (15 a^2 b+7 b^3\right ) \sin ^2(e+f x)}{\sqrt {1-\cos ^2(e+f x)} \sqrt {\sec (e+f x)} \sqrt {\cos ^2(e+f x) \left (-1+\sec ^2(e+f x)\right )}}\right ) (a+b \tan (e+f x))^3}{120 f (d \sec (e+f x))^{9/2} (a \cos (e+f x)+b \sin (e+f x))^3}+\frac {\sec ^2(e+f x) \left (-\frac {1}{90} b \left (15 a^2+4 b^2\right ) \cos (e+f x)-\frac {1}{360} b \left (75 a^2+11 b^2\right ) \cos (3 (e+f x))-\frac {1}{72} b \left (3 a^2-b^2\right ) \cos (5 (e+f x))+\frac {1}{180} a \left (19 a^2-3 b^2\right ) \sin (e+f x)+\frac {1}{360} a \left (43 a^2-21 b^2\right ) \sin (3 (e+f x))+\frac {1}{72} a \left (a^2-3 b^2\right ) \sin (5 (e+f x))\right ) (a+b \tan (e+f x))^3}{f (d \sec (e+f x))^{9/2} (a \cos (e+f x)+b \sin (e+f x))^3} \]
(Sec[e + f*x]^(3/2)*((2*(56*a^3 + 48*a*b^2)*EllipticE[(e + f*x)/2, 2])/(Sq rt[Cos[e + f*x]]*Sqrt[Sec[e + f*x]]) - (2*(15*a^2*b + 7*b^3)*Sin[e + f*x]^ 2)/(Sqrt[1 - Cos[e + f*x]^2]*Sqrt[Sec[e + f*x]]*Sqrt[Cos[e + f*x]^2*(-1 + Sec[e + f*x]^2)]))*(a + b*Tan[e + f*x])^3)/(120*f*(d*Sec[e + f*x])^(9/2)*( a*Cos[e + f*x] + b*Sin[e + f*x])^3) + (Sec[e + f*x]^2*(-1/90*(b*(15*a^2 + 4*b^2)*Cos[e + f*x]) - (b*(75*a^2 + 11*b^2)*Cos[3*(e + f*x)])/360 - (b*(3* a^2 - b^2)*Cos[5*(e + f*x)])/72 + (a*(19*a^2 - 3*b^2)*Sin[e + f*x])/180 + (a*(43*a^2 - 21*b^2)*Sin[3*(e + f*x)])/360 + (a*(a^2 - 3*b^2)*Sin[5*(e + f *x)])/72)*(a + b*Tan[e + f*x])^3)/(f*(d*Sec[e + f*x])^(9/2)*(a*Cos[e + f*x ] + b*Sin[e + f*x])^3)
Time = 0.37 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3994, 495, 27, 675, 212}
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.
| \(\displaystyle \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{9/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{9/2}}dx\) |
| \(\Big \downarrow \) 3994 |
| \(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \int \frac {(a+b \tan (e+f x))^3}{\left (\tan ^2(e+f x)+1\right )^{13/4}}d(b \tan (e+f x))}{b d^4 f \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 495 |
| \(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \left (\frac {2}{9} b^2 \int \frac {(a+b \tan (e+f x)) \left (\left (\frac {7 a^2}{b^2}+4\right ) b^2+3 a \tan (e+f x) b\right )}{2 b^2 \left (\tan ^2(e+f x)+1\right )^{9/4}}d(b \tan (e+f x))-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{9 \left (\tan ^2(e+f x)+1\right )^{9/4}}\right )}{b d^4 f \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \left (\frac {1}{9} \int \frac {(a+b \tan (e+f x)) \left (7 a^2+3 b \tan (e+f x) a+4 b^2\right )}{\left (\tan ^2(e+f x)+1\right )^{9/4}}d(b \tan (e+f x))-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{9 \left (\tan ^2(e+f x)+1\right )^{9/4}}\right )}{b d^4 f \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 675 |
| \(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \left (\frac {1}{9} \left (\frac {3}{5} a \left (7 a^2+6 b^2\right ) \int \frac {1}{\left (\tan ^2(e+f x)+1\right )^{5/4}}d(b \tan (e+f x))-\frac {4 b^2 \left (5 a^2+2 b^2\right )}{5 \left (\tan ^2(e+f x)+1\right )^{5/4}}+\frac {2 a b \left (7 a^2+b^2\right ) \tan (e+f x)}{5 \left (\tan ^2(e+f x)+1\right )^{5/4}}\right )-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{9 \left (\tan ^2(e+f x)+1\right )^{9/4}}\right )}{b d^4 f \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 212 |
| \(\displaystyle \frac {\sqrt [4]{\sec ^2(e+f x)} \left (\frac {1}{9} \left (\frac {6}{5} a b \left (7 a^2+6 b^2\right ) E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right )-\frac {4 b^2 \left (5 a^2+2 b^2\right )}{5 \left (\tan ^2(e+f x)+1\right )^{5/4}}+\frac {2 a b \left (7 a^2+b^2\right ) \tan (e+f x)}{5 \left (\tan ^2(e+f x)+1\right )^{5/4}}\right )-\frac {2 (a+b \tan (e+f x))^2 \left (b^2-a b \tan (e+f x)\right )}{9 \left (\tan ^2(e+f x)+1\right )^{9/4}}\right )}{b d^4 f \sqrt {d \sec (e+f x)}}\) |
((Sec[e + f*x]^2)^(1/4)*((-2*(a + b*Tan[e + f*x])^2*(b^2 - a*b*Tan[e + f*x ]))/(9*(1 + Tan[e + f*x]^2)^(9/4)) + ((6*a*b*(7*a^2 + 6*b^2)*EllipticE[Arc Tan[Tan[e + f*x]]/2, 2])/5 - (4*b^2*(5*a^2 + 2*b^2))/(5*(1 + Tan[e + f*x]^ 2)^(5/4)) + (2*a*b*(7*a^2 + b^2)*Tan[e + f*x])/(5*(1 + Tan[e + f*x]^2)^(5/ 4)))/9))/(b*d^4*f*Sqrt[d*Sec[e + f*x]])
3.7.1.3.1 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[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) )*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a , 0] && PosQ[b/a]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (a*d - b*c*x)*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(c + d*x)^(n - 2)*(a + b*x^2)^(p + 1)*Simp[a* d^2*(n - 1) - b*c^2*(2*p + 3) - b*c*d*(n + 2*p + 2)*x, x], x], x] /; FreeQ[ {a, b, c, d}, x] && LtQ[p, -1] && GtQ[n, 1] && IntQuadraticQ[a, 0, b, c, d, n, p, x]
Int[((d_) + (e_.)*(x_))*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[a*(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + (- Simp[(c*d*f - a*e*g)*x*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] - Simp[(a* e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)) Int[(a + c*x^2)^(p + 1), x], x]) / ; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1] && !(IntegerQ[p] && NiceSqrtQ [(-a)*c])
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_), x_Symbol] :> Simp[d^(2*IntPart[m/2])*((d*Sec[e + f*x])^(2*FracP art[m/2])/(b*f*(Sec[e + f*x]^2)^FracPart[m/2])) Subst[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 + b^2, 0] && !IntegerQ[m] && IntegerQ[n]
Result contains complex when optimal does not.
Time = 32.98 (sec) , antiderivative size = 976, normalized size of antiderivative = 5.55
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(976\) |
| default | \(\text {Expression too large to display}\) | \(1035\) |
2/45*a^3/f/(cos(f*x+e)+1)/(d*sec(f*x+e))^(1/2)/d^4*(21*I*cos(f*x+e)*Ellipt icE(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)-21*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)+5*cos(f*x+e)^4*s in(f*x+e)+42*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)-42*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)+5*cos( f*x+e)^3*sin(f*x+e)+21*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)-21*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/(cos(f*x+e)+1))^(1/2)+7*sin(f*x+e)*cos(f*x+e)^2+7*sin(f*x+e)*cos(f*x+e )+21*sin(f*x+e))+2/45*b^3/f/(d*sec(f*x+e))^(1/2)/d^4*(5*cos(f*x+e)^4-9*cos (f*x+e)^2)+2/15*a*b^2/f/(cos(f*x+e)+1)/(d*sec(f*x+e))^(1/2)/d^4*(6*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)-6*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)-5*cos (f*x+e)^4*sin(f*x+e)+12*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)-12*I*(1/(cos(f*x+e)+1))^ (1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e) ),I)-5*cos(f*x+e)^3*sin(f*x+e)+6*I*sec(f*x+e)*(1/(cos(f*x+e)+1))^(1/2)*...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.12 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.09 \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{9/2}} \, dx=-\frac {3 \, \sqrt {2} {\left (-7 i \, a^{3} - 6 i \, a b^{2}\right )} \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 \, \sqrt {2} {\left (7 i \, a^{3} + 6 i \, a b^{2}\right )} \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) + 2 \, {\left (9 \, b^{3} \cos \left (f x + e\right )^{3} + 5 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (f x + e\right )^{5} - {\left (5 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (f x + e\right )^{4} + {\left (7 \, a^{3} + 6 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{45 \, d^{5} f} \]
-1/45*(3*sqrt(2)*(-7*I*a^3 - 6*I*a*b^2)*sqrt(d)*weierstrassZeta(-4, 0, wei erstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) + 3*sqrt(2)*(7*I*a ^3 + 6*I*a*b^2)*sqrt(d)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) + 2*(9*b^3*cos(f*x + e)^3 + 5*(3*a^2*b - b ^3)*cos(f*x + e)^5 - (5*(a^3 - 3*a*b^2)*cos(f*x + e)^4 + (7*a^3 + 6*a*b^2) *cos(f*x + e)^2)*sin(f*x + e))*sqrt(d/cos(f*x + e)))/(d^5*f)
Timed out. \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{9/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{9/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\left (d \sec \left (f x + e\right )\right )^{\frac {9}{2}}} \,d x } \]
\[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{9/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\left (d \sec \left (f x + e\right )\right )^{\frac {9}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{9/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{9/2}} \,d x \]