3.4.79 \(\int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx\) [379]

Optimal. Leaf size=96 \[ -\frac {3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}-\frac {F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {3 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))} \]

[Out]

-3*(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))/a/d-(cos(1/2*d*x+1/2*
c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a/d-sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*se
c(d*x+c))+3*sin(d*x+c)/a/d/cos(d*x+c)^(1/2)

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Rubi [A]
time = 0.13, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4349, 3903, 3872, 3856, 2720, 3853, 2719} \begin {gather*} -\frac {F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}-\frac {3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {3 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])),x]

[Out]

(-3*EllipticE[(c + d*x)/2, 2])/(a*d) - EllipticF[(c + d*x)/2, 2]/(a*d) + (3*Sin[c + d*x])/(a*d*Sqrt[Cos[c + d*
x]]) - Sin[c + d*x]/(d*Cos[c + d*x]^(3/2)*(a + a*Sec[c + d*x]))

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 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

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 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3903

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[d^2*Cot[e +
f*x]*((d*Csc[e + f*x])^(n - 2)/(f*(a + b*Csc[e + f*x]))), x] - Dist[d^2/(a*b), Int[(d*Csc[e + f*x])^(n - 2)*(
b*(n - 2) - a*(n - 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1]

Rule 4349

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps

\begin {align*} \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{a+a \sec (c+d x)} \, dx\\ &=-\frac {\sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \left (\frac {a}{2}-\frac {3}{2} a \sec (c+d x)\right ) \, dx}{a^2}\\ &=-\frac {\sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \, dx}{2 a}+\frac {\left (3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {3}{2}}(c+d x) \, dx}{2 a}\\ &=\frac {3 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))}-\frac {\int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{2 a}-\frac {\left (3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 a}\\ &=-\frac {F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {3 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))}-\frac {3 \int \sqrt {\cos (c+d x)} \, dx}{2 a}\\ &=-\frac {3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}-\frac {F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {3 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 1.73, size = 303, normalized size = 3.16 \begin {gather*} \frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\left (2 \cos \left (\frac {1}{2} (c-d x)\right )+\cos \left (\frac {1}{2} (3 c+d x)\right )+3 \cos \left (\frac {1}{2} (c+3 d x)\right )\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right )}{2 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {2 i \sqrt {2} e^{-i (c+d x)} \left (3 \left (1+e^{2 i (c+d x)}\right )+3 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i (c+d x)}\right )-e^{i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-e^{2 i (c+d x)}\right )\right ) \sec (c+d x)}{d \left (-1+e^{2 i c}\right ) \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}}\right )}{a (1+\sec (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])),x]

[Out]

(Cos[(c + d*x)/2]^2*(((2*Cos[(c - d*x)/2] + Cos[(3*c + d*x)/2] + 3*Cos[(c + 3*d*x)/2])*Csc[c/2]*Sec[c/2]*Sec[(
c + d*x)/2])/(2*d*Cos[c + d*x]^(3/2)) - ((2*I)*Sqrt[2]*(3*(1 + E^((2*I)*(c + d*x))) + 3*(-1 + E^((2*I)*c))*Sqr
t[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] - E^(I*(c + d*x))*(-1 + E^(
(2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))])*Sec[c + d*x])/
(d*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))])))/(a*(1 + Sec[c + d*x])
)

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Maple [A]
time = 0.10, size = 253, normalized size = 2.64

method	result	size
default	\(-\frac {-\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (\EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )+6 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(253\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-(-cos(1/2*d*x+1/2*c)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+s
in(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+6*(
-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-5*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*
x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2)/a/cos(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1
/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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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(1/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

integrate(1/((a*sec(d*x + c) + a)*cos(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 = 0.53, size = 236, normalized size = 2.46 \begin {gather*} \frac {2 \, {\left (3 \, \cos \left (d x + c\right ) + 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} + i \, \sqrt {2} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} - i \, \sqrt {2} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} + i \, \sqrt {2} \cos \left (d x + c\right )\right )} {\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 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} - i \, \sqrt {2} \cos \left (d x + c\right )\right )} {\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 (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/2*(2*(3*cos(d*x + c) + 2)*sqrt(cos(d*x + c))*sin(d*x + c) + (I*sqrt(2)*cos(d*x + c)^2 + I*sqrt(2)*cos(d*x +
c))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (-I*sqrt(2)*cos(d*x + c)^2 - I*sqrt(2)*cos(d*x
+ c))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*(I*sqrt(2)*cos(d*x + c)^2 + I*sqrt(2)*cos
(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*(-I*sqrt(2)*c
os(d*x + c)^2 - I*sqrt(2)*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin
(d*x + c))))/(a*d*cos(d*x + c)^2 + a*d*cos(d*x + c))

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)**(5/2)/(a+a*sec(d*x+c)),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3007 deep

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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(1/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate(1/((a*sec(d*x + c) + a)*cos(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 {1}{{\cos \left (c+d\,x\right )}^{5/2}\,\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^(5/2)*(a + a/cos(c + d*x))),x)

[Out]

int(1/(cos(c + d*x)^(5/2)*(a + a/cos(c + d*x))), x)

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