\(\int \frac {A+C \cos ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx\) [33]
Optimal result
Integrand size = 25, antiderivative size = 115 \[
\int \frac {A+C \cos ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\frac {2 (9 A+7 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 (9 A+7 C) \sin (c+d x)}{45 b d (b \sec (c+d x))^{3/2}}+\frac {2 b^2 C \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}}
\]
[Out]
2/45*(9*A+7*C)*sin(d*x+c)/b/d/(b*sec(d*x+c))^(3/2)+2/15*(9*A+7*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))/b^2/d/cos(d*x+c)^(1/2)/(b*sec(d*x+c))^(1/2)+2/9*b^2*C*tan(d*x+c)/d/(
b*sec(d*x+c))^(9/2)
Rubi [A] (verified)
Time = 0.15 (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, 4130, 3854, 3856, 2719}
\[
\int \frac {A+C \cos ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\frac {2 (9 A+7 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 (9 A+7 C) \sin (c+d x)}{45 b d (b \sec (c+d x))^{3/2}}+\frac {2 b^2 C \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}}
\]
[In]
Int[(A + C*Cos[c + d*x]^2)/(b*Sec[c + d*x])^(5/2),x]
[Out]
(2*(9*A + 7*C)*EllipticE[(c + d*x)/2, 2])/(15*b^2*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*(9*A + 7*C)*
Sin[c + d*x])/(45*b*d*(b*Sec[c + d*x])^(3/2)) + (2*b^2*C*Tan[c + d*x])/(9*d*(b*Sec[c + d*x])^(9/2))
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 3854
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
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[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 4130
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e
+ f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
Rubi steps \begin{align*}
\text {integral}& = b^2 \int \frac {C+A \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx \\ & = \frac {2 b^2 C \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}}+\frac {1}{9} (9 A+7 C) \int \frac {1}{(b \sec (c+d x))^{5/2}} \, dx \\ & = \frac {2 (9 A+7 C) \sin (c+d x)}{45 b d (b \sec (c+d x))^{3/2}}+\frac {2 b^2 C \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}}+\frac {(9 A+7 C) \int \frac {1}{\sqrt {b \sec (c+d x)}} \, dx}{15 b^2} \\ & = \frac {2 (9 A+7 C) \sin (c+d x)}{45 b d (b \sec (c+d x))^{3/2}}+\frac {2 b^2 C \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}}+\frac {(9 A+7 C) \int \sqrt {\cos (c+d x)} \, dx}{15 b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}} \\ & = \frac {2 (9 A+7 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 (9 A+7 C) \sin (c+d x)}{45 b d (b \sec (c+d x))^{3/2}}+\frac {2 b^2 C \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}} \\
\end{align*}
Mathematica [A] (verified)
Time = 1.00 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.70
\[
\int \frac {A+C \cos ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\frac {\frac {48 (9 A+7 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{\sqrt {\cos (c+d x)}}+4 (18 A+19 C+5 C \cos (2 (c+d x))) \sin (2 (c+d x))}{360 b^2 d \sqrt {b \sec (c+d x)}}
\]
[In]
Integrate[(A + C*Cos[c + d*x]^2)/(b*Sec[c + d*x])^(5/2),x]
[Out]
((48*(9*A + 7*C)*EllipticE[(c + d*x)/2, 2])/Sqrt[Cos[c + d*x]] + 4*(18*A + 19*C + 5*C*Cos[2*(c + d*x)])*Sin[2*
(c + d*x)])/(360*b^2*d*Sqrt[b*Sec[c + d*x]])
Maple [C] (verified)
Result contains complex when optimal does not.
Time = 23.00 (sec) , antiderivative size = 866, normalized size of antiderivative =
7.53
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\(\text {Expression too large to display}\) |
\(866\) |
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\(\text {Expression too large to display}\) |
\(876\) |
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[In]
int((A+C*cos(d*x+c)^2)/(b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
[Out]
-2/45/b^2/d/(1+cos(d*x+c))/(b*sec(d*x+c))^(1/2)*(21*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)-21*I*sec(d*x+c)*C*EllipticE(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)-27*I*sec(d*x+c)*A*EllipticE(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)+54*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)-5*C*sin(d*x+c)*cos(d*x+c)^4+27*I*sec(d*x+c)*A*El
lipticF(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)+42*I*C*Ellipti
cF(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)+21*I*sec(d*x+c)*C*E
llipticF(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)-21*I*cos(d*x+
c)*C*EllipticE(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)-5*C*cos
(d*x+c)^3*sin(d*x+c)-42*I*C*EllipticE(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)+27*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)-9*A*sin(d*x+c)*cos(d*x+c)^2-27*I*cos(d*x+c)*A*EllipticE(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)-54*I*A*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(1+co
s(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-7*C*cos(d*x+c)^2*sin(d*x+c)-9*A*sin(d*x+c)*cos(d*x+c)-7*C*c
os(d*x+c)*sin(d*x+c)-27*A*sin(d*x+c)-21*sin(d*x+c)*C)
Fricas [C] (verification not implemented)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.12
\[
\int \frac {A+C \cos ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=-\frac {3 \, \sqrt {2} {\left (-9 i \, A - 7 i \, C\right )} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (9 i \, A + 7 i \, C\right )} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (5 \, C \cos \left (d x + c\right )^{4} + {\left (9 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{45 \, b^{3} d}
\]
[In]
integrate((A+C*cos(d*x+c)^2)/(b*sec(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
-1/45*(3*sqrt(2)*(-9*I*A - 7*I*C)*sqrt(b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*s
in(d*x + c))) + 3*sqrt(2)*(9*I*A + 7*I*C)*sqrt(b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x +
c) - I*sin(d*x + c))) - 2*(5*C*cos(d*x + c)^4 + (9*A + 7*C)*cos(d*x + c)^2)*sqrt(b/cos(d*x + c))*sin(d*x + c))
/(b^3*d)
Sympy [F]
\[
\int \frac {A+C \cos ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int \frac {A + C \cos ^{2}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx
\]
[In]
integrate((A+C*cos(d*x+c)**2)/(b*sec(d*x+c))**(5/2),x)
[Out]
Integral((A + C*cos(c + d*x)**2)/(b*sec(c + d*x))**(5/2), x)
Maxima [F]
\[
\int \frac {A+C \cos ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x }
\]
[In]
integrate((A+C*cos(d*x+c)^2)/(b*sec(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
integrate((C*cos(d*x + c)^2 + A)/(b*sec(d*x + c))^(5/2), x)
Giac [F]
\[
\int \frac {A+C \cos ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x }
\]
[In]
integrate((A+C*cos(d*x+c)^2)/(b*sec(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + A)/(b*sec(d*x + c))^(5/2), x)
Mupad [F(-1)]
Timed out. \[
\int \frac {A+C \cos ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x
\]
[In]
int((A + C*cos(c + d*x)^2)/(b/cos(c + d*x))^(5/2),x)
[Out]
int((A + C*cos(c + d*x)^2)/(b/cos(c + d*x))^(5/2), x)