3.4.86 \(\int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx\) [386]

Optimal. Leaf size=109 \[ \frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac {\sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)} (1+\sec (c+d x))}-\frac {\sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \]

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 109, 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, 3901, 4104, 3872, 3856, 2719, 2720} \begin {gather*} \frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac {\sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)} (\sec (c+d x)+1)}-\frac {\sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

EllipticE[(c + d*x)/2, 2]/(a^2*d) + (2*EllipticF[(c + d*x)/2, 2])/(3*a^2*d) - Sin[c + d*x]/(a^2*d*Sqrt[Cos[c +
d*x]]*(1 + Sec[c + d*x])) - Sin[c + d*x]/(3*d*Cos[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^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 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 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 3901

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

Rule 4104

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

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))^2} \, 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))^2} \, dx\\ &=-\frac {\sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)} \left (\frac {a}{2}-\frac {5}{2} a \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac {\sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)} (1+\sec (c+d x))}-\frac {\sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {3 a^2}{2}-a^2 \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{3 a^4}\\ &=-\frac {\sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)} (1+\sec (c+d x))}-\frac {\sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \, dx}{3 a^2}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {\sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)} (1+\sec (c+d x))}-\frac {\sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}+\frac {\int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 a^2}+\frac {\int \sqrt {\cos (c+d x)} \, dx}{2 a^2}\\ &=\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac {\sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)} (1+\sec (c+d x))}-\frac {\sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^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 = 4.98, size = 312, normalized size = 2.86 \begin {gather*} \frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (-\frac {\left (7 \cos \left (\frac {1}{2} (c-d x)\right )+2 \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 ^3\left (\frac {1}{2} (c+d x)\right )}{2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 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 )-2 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 ^2(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 )}{3 a^2 (1+\sec (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^4*(-1/2*((7*Cos[(c - d*x)/2] + 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]^3)/(d*Cos[c + d*x]^(3/2)) + ((4*I)*Sqrt[2]*(3*(1 + E^((2*I)*(c + d*x))) + 3*(-1 + E^((2*I)*
c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] - 2*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]^2)/(d*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))])))/(3*a^2*(1
+ Sec[c + d*x])^2)

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Maple [A]
time = 0.10, size = 257, normalized size = 2.36

method	result	size
default	\(\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (12 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-16 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right )}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(257\)

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))^2,x,method=_RETURNVERBOSE)

[Out]

1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(12*cos(1/2*d*x+1/2*c)^6-4*(sin(1/2*d*x+1/2*c)^2)^
(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^3+6*(sin(1/2*
d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*cos(1/2*d*x+1/2*c)^3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2
))-16*cos(1/2*d*x+1/2*c)^4+3*cos(1/2*d*x+1/2*c)^2+1)/a^2/cos(1/2*d*x+1/2*c)^3/(-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))^2,x, algorithm="maxima")

[Out]

integrate(1/((a*sec(d*x + c) + a)^2*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 = 0.50, size = 268, normalized size = 2.46 \begin {gather*} -\frac {2 \, {\left (3 \, \cos \left (d x + c\right ) + 4\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 2 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\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} - 2 i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\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} + 2 i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\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 )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\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))^2,x, algorithm="fricas")

[Out]

-1/6*(2*(3*cos(d*x + c) + 4)*sqrt(cos(d*x + c))*sin(d*x + c) + 2*(I*sqrt(2)*cos(d*x + c)^2 + 2*I*sqrt(2)*cos(d
*x + c) + I*sqrt(2))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 2*(-I*sqrt(2)*cos(d*x + c)^2
- 2*I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*(-I*sqrt
(2)*cos(d*x + c)^2 - 2*I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, c
os(d*x + c) + I*sin(d*x + c))) + 3*(I*sqrt(2)*cos(d*x + c)^2 + 2*I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*weierstra
ssZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(
d*x + c) + a^2*d)

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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))**2,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 5009 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))^2,x, algorithm="giac")

[Out]

integrate(1/((a*sec(d*x + c) + a)^2*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 )}^2} \,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))^2),x)

[Out]

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

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