\(\int (A+C \cos ^2(c+d x)) (b \sec (c+d x))^{7/2} \, dx\) [27]
Optimal result
Integrand size = 25, antiderivative size = 115 \[
\int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{7/2} \, dx=-\frac {2 b^4 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 b^3 (3 A+5 C) \sqrt {b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 A b^2 (b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}
\]
[Out]
-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)*EllipticE(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
Rubi [A] (verified)
Time = 0.14 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of
steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3317, 4131, 3853, 3856, 2719}
\[
\int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{7/2} \, dx=-\frac {2 b^4 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 b^3 (3 A+5 C) \sin (c+d x) \sqrt {b \sec (c+d x)}}{5 d}+\frac {2 A b^2 \tan (c+d x) (b \sec (c+d x))^{3/2}}{5 d}
\]
[In]
Int[(A + C*Cos[c + d*x]^2)*(b*Sec[c + d*x])^(7/2),x]
[Out]
(-2*b^4*(3*A + 5*C)*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*b^3*(3*A + 5
*C)*Sqrt[b*Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*A*b^2*(b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d)
Rule 2719
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]
Rule 3317
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[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]
Rule 3853
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] + Dist[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]
Rule 3856
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(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]
Rule 4131
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] + Dist[(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]
Rubi steps \begin{align*}
\text {integral}& = b^2 \int (b \sec (c+d x))^{3/2} \left (C+A \sec ^2(c+d x)\right ) \, dx \\ & = \frac {2 A b^2 (b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {1}{5} \left (b^2 (3 A+5 C)\right ) \int (b \sec (c+d x))^{3/2} \, dx \\ & = \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}-\frac {1}{5} \left (b^4 (3 A+5 C)\right ) \int \frac {1}{\sqrt {b \sec (c+d x)}} \, dx \\ & = \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}-\frac {\left (b^4 (3 A+5 C)\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}} \\ & = -\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} \\
\end{align*}
Mathematica [A] (verified)
Time = 1.38 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.69
\[
\int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{7/2} \, dx=-\frac {b^2 (b \sec (c+d x))^{3/2} \left (2 (3 A+5 C) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )-(3 A+5 C) \sin (2 (c+d x))-2 A \tan (c+d x)\right )}{5 d}
\]
[In]
Integrate[(A + C*Cos[c + d*x]^2)*(b*Sec[c + d*x])^(7/2),x]
[Out]
-1/5*(b^2*(b*Sec[c + d*x])^(3/2)*(2*(3*A + 5*C)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] - (3*A + 5*C)*Sin
[2*(c + d*x)] - 2*A*Tan[c + d*x]))/d
Maple [C] (verified)
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 |
\(\text {Expression too large to display}\) |
\(811\) |
| | |
|
|
|
|
[In]
int((A+C*cos(d*x+c)^2)*(b*sec(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
[Out]
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)-c
ot(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*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)-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)
Fricas [C] (verification not implemented)
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
\[
\int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{7/2} \, dx=\frac {-i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} b^{\frac {7}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} b^{\frac {7}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left ({\left (3 \, A + 5 \, C\right )} b^{3} \cos \left (d x + c\right )^{2} + A b^{3}\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{5 \, d \cos \left (d x + c\right )^{2}}
\]
[In]
integrate((A+C*cos(d*x+c)^2)*(b*sec(d*x+c))^(7/2),x, algorithm="fricas")
[Out]
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, weierstrassPInver
se(-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))*si
n(d*x + c))/(d*cos(d*x + c)^2)
Sympy [F(-1)]
Timed out. \[
\int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{7/2} \, dx=\text {Timed out}
\]
[In]
integrate((A+C*cos(d*x+c)**2)*(b*sec(d*x+c))**(7/2),x)
[Out]
Timed out
Maxima [F]
\[
\int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{7/2} \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {7}{2}} \,d x }
\]
[In]
integrate((A+C*cos(d*x+c)^2)*(b*sec(d*x+c))^(7/2),x, algorithm="maxima")
[Out]
integrate((C*cos(d*x + c)^2 + A)*(b*sec(d*x + c))^(7/2), x)
Giac [F]
\[
\int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{7/2} \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {7}{2}} \,d x }
\]
[In]
integrate((A+C*cos(d*x+c)^2)*(b*sec(d*x+c))^(7/2),x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + A)*(b*sec(d*x + c))^(7/2), x)
Mupad [F(-1)]
Timed out. \[
\int \left (A+C \cos ^2(c+d x)\right ) (b \sec (c+d x))^{7/2} \, dx=\int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{7/2} \,d x
\]
[In]
int((A + C*cos(c + d*x)^2)*(b/cos(c + d*x))^(7/2),x)
[Out]
int((A + C*cos(c + d*x)^2)*(b/cos(c + d*x))^(7/2), x)