Integrand size = 25, antiderivative size = 115 \[ \int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{7/2} \, dx=-\frac {2 b^4 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 b^3 (3 A+5 C) \sqrt {b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 A b^2 (b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \]
-2/5*b^4*(3*A+5*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipti cE(sin(1/2*d*x+1/2*c),2^(1/2))/d/cos(d*x+c)^(1/2)/(b*sec(d*x+c))^(1/2)+2/5 *b^3*(3*A+5*C)*sin(d*x+c)*(b*sec(d*x+c))^(1/2)/d+2/5*A*b^2*(b*sec(d*x+c))^ (3/2)*tan(d*x+c)/d
Time = 1.38 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.69 \[ \int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{7/2} \, dx=-\frac {b^2 (b \sec (c+d x))^{3/2} \left (2 (3 A+5 C) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )-(3 A+5 C) \sin (2 (c+d x))-2 A \tan (c+d x)\right )}{5 d} \]
-1/5*(b^2*(b*Sec[c + d*x])^(3/2)*(2*(3*A + 5*C)*Cos[c + d*x]^(3/2)*Ellipti cE[(c + d*x)/2, 2] - (3*A + 5*C)*Sin[2*(c + d*x)] - 2*A*Tan[c + d*x]))/d
Time = 0.61 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.94, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3717, 3042, 4534, 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.
\(\displaystyle \int (b \sec (c+d x))^{7/2} \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 3717 |
\(\displaystyle b^2 \int (b \sec (c+d x))^{3/2} \left (A \sec ^2(c+d x)+C\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (A \csc \left (c+d x+\frac {\pi }{2}\right )^2+C\right )dx\) |
\(\Big \downarrow \) 4534 |
\(\displaystyle b^2 \left (\frac {1}{5} (3 A+5 C) \int (b \sec (c+d x))^{3/2}dx+\frac {2 A \tan (c+d x) (b \sec (c+d x))^{3/2}}{5 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \left (\frac {1}{5} (3 A+5 C) \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {2 A \tan (c+d x) (b \sec (c+d x))^{3/2}}{5 d}\right )\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle b^2 \left (\frac {1}{5} (3 A+5 C) \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-b^2 \int \frac {1}{\sqrt {b \sec (c+d x)}}dx\right )+\frac {2 A \tan (c+d x) (b \sec (c+d x))^{3/2}}{5 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \left (\frac {1}{5} (3 A+5 C) \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-b^2 \int \frac {1}{\sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 A \tan (c+d x) (b \sec (c+d x))^{3/2}}{5 d}\right )\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle b^2 \left (\frac {1}{5} (3 A+5 C) \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {b^2 \int \sqrt {\cos (c+d x)}dx}{\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\right )+\frac {2 A \tan (c+d x) (b \sec (c+d x))^{3/2}}{5 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \left (\frac {1}{5} (3 A+5 C) \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {b^2 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\right )+\frac {2 A \tan (c+d x) (b \sec (c+d x))^{3/2}}{5 d}\right )\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle b^2 \left (\frac {1}{5} (3 A+5 C) \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {2 b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\right )+\frac {2 A \tan (c+d x) (b \sec (c+d x))^{3/2}}{5 d}\right )\) |
b^2*(((3*A + 5*C)*((-2*b^2*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x] ]*Sqrt[b*Sec[c + d*x]]) + (2*b*Sqrt[b*Sec[c + d*x]]*Sin[c + d*x])/d))/5 + (2*A*(b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d))
3.1.27.3.1 Defintions of rubi rules used
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[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]^(n_.))^(p_.), x_Symbol] :> Simp[d^(n*p) Int[(d*Csc[e + f*x])^(m - n*p )*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && !IntegerQ[m] && IntegersQ[n, p]
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]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) )), x] + Simp[(C*m + A*(m + 1))/(m + 1) Int[(b*Csc[e + f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] && !LeQ[m, -1]
Result contains complex when optimal does not.
Time = 72.59 (sec) , antiderivative size = 798, normalized size of antiderivative = 6.94
method | result | size |
default | \(\text {Expression too large to display}\) | \(798\) |
parts | \(\text {Expression too large to display}\) | \(811\) |
2/5*b^3/d*(b*sec(d*x+c))^(1/2)/(1+cos(d*x+c))*(-5*I*C*(1/(1+cos(d*x+c)))^( 1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)) ,I)*cos(d*x+c)^2-6*I*cos(d*x+c)*A*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+ cos(d*x+c)))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)+3*I*A*(1/(1+cos( d*x+c)))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(cos(d*x+c)/(1+cos(d *x+c)))^(1/2)+5*I*C*(1/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d *x+c)),I)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-3*I*A*(1/(1+cos(d*x+c)))^(1/2) *(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)+ 6*I*cos(d*x+c)*A*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(1+cos(d*x+c))) ^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-10*I*cos(d*x+c)*C*(1/(1+cos(d*x+c )))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d* x+c)),I)+3*I*A*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(1+cos(d*x+c)))^( 1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2+10*I*cos(d*x+c)*C*Elli pticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1 +cos(d*x+c)))^(1/2)-3*I*A*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+ c)))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)*cos(d*x+c)^2-5*I*C*(1/(1 +cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(csc(d*x +c)-cot(d*x+c)),I)+5*I*C*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(1+cos( d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2+3*A*sin(d*x+ c)+5*sin(d*x+c)*C+A*tan(d*x+c)+tan(d*x+c)*sec(d*x+c)*A)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.28 \[ \int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{7/2} \, dx=\frac {-i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} b^{\frac {7}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} b^{\frac {7}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left ({\left (3 \, A + 5 \, C\right )} b^{3} \cos \left (d x + c\right )^{2} + A b^{3}\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{5 \, d \cos \left (d x + c\right )^{2}} \]
1/5*(-I*sqrt(2)*(3*A + 5*C)*b^(7/2)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + I*sqrt(2)*(3* A + 5*C)*b^(7/2)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse (-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*((3*A + 5*C)*b^3*cos(d*x + c)^ 2 + A*b^3)*sqrt(b/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2)
Timed out. \[ \int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{7/2} \, dx=\text {Timed out} \]
\[ \int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{7/2} \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {7}{2}} \,d x } \]
\[ \int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{7/2} \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {7}{2}} \,d x } \]
Timed out. \[ \int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{7/2} \, dx=\int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{7/2} \,d x \]