3.1.21 \(\int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx\) [21]

Optimal. Leaf size=77 \[ \frac {2 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}} \]

[Out]

2/5*(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))/b^2/d/cos(
d*x+c)^(1/2)/(b*sec(d*x+c))^(1/2)+2/5*A*tan(d*x+c)/d/(b*sec(d*x+c))^(5/2)

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Rubi [A]
time = 0.05, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {4130, 3856, 2719} \begin {gather*} \frac {2 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(5/2),x]

[Out]

(2*(3*A + 5*C)*EllipticE[(c + d*x)/2, 2])/(5*b^2*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*A*Tan[c + d*x
])/(5*d*(b*Sec[c + d*x])^(5/2))

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 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*} \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx &=\frac {2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}+\frac {(3 A+5 C) \int \frac {1}{\sqrt {b \sec (c+d x)}} \, dx}{5 b^2}\\ &=\frac {2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}+\frac {(3 A+5 C) \int \sqrt {\cos (c+d x)} \, dx}{5 b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\\ &=\frac {2 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.77, size = 133, normalized size = 1.73 \begin {gather*} \frac {e^{-i d x} \sec ^2(c+d x) (\cos (d x)+i \sin (d x)) \left (12 i (3 A+5 C)-\frac {8 i (3 A+5 C) e^{2 i (c+d x)} \, _2F_1\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)}}}+6 A \sin (2 (c+d x))\right )}{30 d (b \sec (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(5/2),x]

[Out]

(Sec[c + d*x]^2*(Cos[d*x] + I*Sin[d*x])*((12*I)*(3*A + 5*C) - ((8*I)*(3*A + 5*C)*E^((2*I)*(c + d*x))*Hypergeom
etric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] + 6*A*Sin[2*(c + d*x)]))/(30*d*E^
(I*d*x)*(b*Sec[c + d*x])^(5/2))

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Maple [C] Result contains complex when optimal does not.
time = 22.27, size = 614, normalized size = 7.97

method	result	size
default	\(\frac {\frac {6 i A \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )}{5}-\frac {6 i A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )}{5}+2 i C \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-2 i C \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+\frac {6 i A \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{5}-\frac {6 i A \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{5}+2 i C \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-2 i C \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-\frac {2 A \left (\cos ^{4}\left (d x +c \right )\right )}{5}-\frac {4 A \left (\cos ^{2}\left (d x +c \right )\right )}{5}-2 C \left (\cos ^{2}\left (d x +c \right )\right )+\frac {6 A \cos \left (d x +c \right )}{5}+2 C \cos \left (d x +c \right )}{d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3} \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}}\)	\(614\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/5/d*(3*I*A*sin(d*x+c)*cos(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(-1+
cos(d*x+c))/sin(d*x+c),I)-3*I*A*sin(d*x+c)*cos(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(1+cos(d*x+
c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+5*I*C*sin(d*x+c)*cos(d*x+c)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c
),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-5*I*C*sin(d*x+c)*cos(d*x+c)*EllipticE(I*(-1+co
s(d*x+c))/sin(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*sin(d*x+c)*EllipticF(
I*(-1+cos(d*x+c))/sin(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*sin(d*x+c)*El
lipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+5*I*C*sin(d
*x+c)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-5*I
*C*sin(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(
1/2)-A*cos(d*x+c)^4-2*A*cos(d*x+c)^2-5*C*cos(d*x+c)^2+3*A*cos(d*x+c)+5*C*cos(d*x+c))/sin(d*x+c)/cos(d*x+c)^3/(
b/cos(d*x+c))^(5/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + A)/(b*sec(d*x + c))^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.75, size = 108, normalized size = 1.40 \begin {gather*} \frac {2 \, A \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + \sqrt {2} {\left (3 i \, A + 5 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 ) + \sqrt {2} {\left (-3 i \, A - 5 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 )}{5 \, b^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/5*(2*A*sqrt(b/cos(d*x + c))*cos(d*x + c)^2*sin(d*x + c) + sqrt(2)*(3*I*A + 5*I*C)*sqrt(b)*weierstrassZeta(-4
, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + sqrt(2)*(-3*I*A - 5*I*C)*sqrt(b)*weierstrass
Zeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(b^3*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + C \sec ^{2}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)**2)/(b*sec(d*x+c))**(5/2),x)

[Out]

Integral((A + C*sec(c + d*x)**2)/(b*sec(c + d*x))**(5/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)/(b*sec(d*x + c))^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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