\(\int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx\) [23]

Optimal result

Integrand size = 25, antiderivative size = 112 \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx=\frac {2 (7 A+9 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 b^4 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 (7 A+9 C) \sin (c+d x)}{45 b^3 d (b \sec (c+d x))^{3/2}}+\frac {2 A \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}} \]

[Out]

2/45*(7*A+9*C)*sin(d*x+c)/b^3/d/(b*sec(d*x+c))^(3/2)+2/15*(7*A+9*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^4/d/cos(d*x+c)^(1/2)/(b*sec(d*x+c))^(1/2)+2/9*A*tan(d*x+c)/d/(b*
sec(d*x+c))^(9/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {4130, 3854, 3856, 2719} \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx=\frac {2 (7 A+9 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 b^4 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 (7 A+9 C) \sin (c+d x)}{45 b^3 d (b \sec (c+d x))^{3/2}}+\frac {2 A \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}} \]

[In]

Int[(A + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(9/2),x]

[Out]

(2*(7*A + 9*C)*EllipticE[(c + d*x)/2, 2])/(15*b^4*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*(7*A + 9*C)*
Sin[c + d*x])/(45*b^3*d*(b*Sec[c + d*x])^(3/2)) + (2*A*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 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}& = \frac {2 A \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}}+\frac {(7 A+9 C) \int \frac {1}{(b \sec (c+d x))^{5/2}} \, dx}{9 b^2} \\ & = \frac {2 (7 A+9 C) \sin (c+d x)}{45 b^3 d (b \sec (c+d x))^{3/2}}+\frac {2 A \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}}+\frac {(7 A+9 C) \int \frac {1}{\sqrt {b \sec (c+d x)}} \, dx}{15 b^4} \\ & = \frac {2 (7 A+9 C) \sin (c+d x)}{45 b^3 d (b \sec (c+d x))^{3/2}}+\frac {2 A \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}}+\frac {(7 A+9 C) \int \sqrt {\cos (c+d x)} \, dx}{15 b^4 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}} \\ & = \frac {2 (7 A+9 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 b^4 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 (7 A+9 C) \sin (c+d x)}{45 b^3 d (b \sec (c+d x))^{3/2}}+\frac {2 A \tan (c+d x)}{9 d (b \sec (c+d x))^{9/2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 1.32 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.28 \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx=\frac {e^{-i d x} (\cos (d x)+i \sin (d x)) \left (336 i A+432 i C-\frac {32 i (7 A+9 C) e^{2 i (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}+(76 A+72 C) \sin (2 (c+d x))+10 A \sin (4 (c+d x))\right )}{360 b^4 d \sqrt {b \sec (c+d x)}} \]

[In]

Integrate[(A + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(9/2),x]

[Out]

((Cos[d*x] + I*Sin[d*x])*((336*I)*A + (432*I)*C - ((32*I)*(7*A + 9*C)*E^((2*I)*(c + d*x))*Hypergeometric2F1[1/
2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] + (76*A + 72*C)*Sin[2*(c + d*x)] + 10*A*Sin[
4*(c + d*x)]))/(360*b^4*d*E^(I*d*x)*Sqrt[b*Sec[c + d*x]])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 7.76 (sec) , antiderivative size = 866, normalized size of antiderivative = 7.73

method	result	size
default	\(\text {Expression too large to display}\)	\(866\)
parts	\(\text {Expression too large to display}\)	\(876\)

[In]

int((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(9/2),x,method=_RETURNVERBOSE)

[Out]

-2/45/b^4/d/(cos(d*x+c)+1)/(b*sec(d*x+c))^(1/2)*(-27*I*C*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^
(1/2)*EllipticF(I*(cot(d*x+c)-csc(d*x+c)),I)*sec(d*x+c)+42*I*A*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c
)+1))^(1/2)*EllipticE(I*(cot(d*x+c)-csc(d*x+c)),I)-5*A*cos(d*x+c)^4*sin(d*x+c)-21*I*A*(1/(cos(d*x+c)+1))^(1/2)
*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cot(d*x+c)-csc(d*x+c)),I)*sec(d*x+c)+21*I*A*(1/(cos(d*x+c)+1))
^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cot(d*x+c)-csc(d*x+c)),I)*sec(d*x+c)+54*I*C*(1/(cos(d*x+
c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cot(d*x+c)-csc(d*x+c)),I)-42*I*A*(1/(cos(d*x+c)+1)
)^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cot(d*x+c)-csc(d*x+c)),I)-5*A*cos(d*x+c)^3*sin(d*x+c)+2
7*I*C*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cot(d*x+c)-csc(d*x+c)),I)*sec(d*
x+c)-54*I*C*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cot(d*x+c)-csc(d*x+c)),I)-
21*I*A*EllipticF(I*(cot(d*x+c)-csc(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d
*x+c)-27*I*C*EllipticF(I*(cot(d*x+c)-csc(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)
*cos(d*x+c)-7*A*cos(d*x+c)^2*sin(d*x+c)+21*I*A*EllipticE(I*(cot(d*x+c)-csc(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)
*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+27*I*C*EllipticE(I*(cot(d*x+c)-csc(d*x+c)),I)*(1/(cos(d*x+c)+1))
^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)-9*C*cos(d*x+c)^2*sin(d*x+c)-7*A*cos(d*x+c)*sin(d*x+c)-9*C*
cos(d*x+c)*sin(d*x+c)-21*A*sin(d*x+c)-27*C*sin(d*x+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 = 1.15 \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx=-\frac {3 \, \sqrt {2} {\left (-7 i \, A - 9 i \, C\right )} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (7 i \, A + 9 i \, C\right )} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (5 \, A \cos \left (d x + c\right )^{4} + {\left (7 \, A + 9 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{45 \, b^{5} d} \]

[In]

integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(9/2),x, algorithm="fricas")

[Out]

-1/45*(3*sqrt(2)*(-7*I*A - 9*I*C)*sqrt(b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*s
in(d*x + c))) + 3*sqrt(2)*(7*I*A + 9*I*C)*sqrt(b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x +
c) - I*sin(d*x + c))) - 2*(5*A*cos(d*x + c)^4 + (7*A + 9*C)*cos(d*x + c)^2)*sqrt(b/cos(d*x + c))*sin(d*x + c))
/(b^5*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx=\text {Timed out} \]

[In]

integrate((A+C*sec(d*x+c)**2)/(b*sec(d*x+c))**(9/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {9}{2}}} \,d x } \]

[In]

integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(9/2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + A)/(b*sec(d*x + c))^(9/2), x)

Giac [F]

\[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {9}{2}}} \,d x } \]

[In]

integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(9/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)/(b*sec(d*x + c))^(9/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{9/2}} \,d x \]

[In]

int((A + C/cos(c + d*x)^2)/(b/cos(c + d*x))^(9/2),x)

[Out]

int((A + C/cos(c + d*x)^2)/(b/cos(c + d*x))^(9/2), x)