3.3.2 \(\int \frac {\sec ^{\frac {7}{2}}(c+d x)}{(a+a \sec (c+d x))^2} \, dx\) [202]

Optimal. Leaf size=176 \[ -\frac {4 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}-\frac {5 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}+\frac {4 \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 d}-\frac {5 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2} \]

[Out]

-5/3*sec(d*x+c)^(3/2)*sin(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^2+4*s
in(d*x+c)*sec(d*x+c)^(1/2)/a^2/d-4*(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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^2/d-5/3*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipti
cF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^2/d

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Rubi [A]
time = 0.17, antiderivative size = 176, 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 = {3901, 4104, 3872, 3856, 2720, 3853, 2719} \begin {gather*} -\frac {5 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac {4 \sin (c+d x) \sqrt {\sec (c+d x)}}{a^2 d}-\frac {5 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac {4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^(7/2)/(a + a*Sec[c + d*x])^2,x]

[Out]

(-4*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d) - (5*Sqrt[Cos[c + d*x]]*EllipticF
[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) + (4*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(a^2*d) - (5*Sec[c + d*x]
^(3/2)*Sin[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - (Sec[c + d*x]^(5/2)*Sin[c + d*x])/(3*d*(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 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 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]

Rubi steps

\begin {align*} \int \frac {\sec ^{\frac {7}{2}}(c+d x)}{(a+a \sec (c+d x))^2} \, dx &=-\frac {\sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (\frac {3 a}{2}-\frac {7}{2} a \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac {5 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \sqrt {\sec (c+d x)} \left (\frac {5 a^2}{2}-6 a^2 \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac {5 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {5 \int \sqrt {\sec (c+d x)} \, dx}{6 a^2}+\frac {2 \int \sec ^{\frac {3}{2}}(c+d x) \, dx}{a^2}\\ &=\frac {4 \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 d}-\frac {5 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {2 \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{a^2}-\frac {\left (5 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a^2}\\ &=-\frac {5 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}+\frac {4 \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 d}-\frac {5 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{a^2}\\ &=-\frac {4 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}-\frac {5 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}+\frac {4 \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 d}-\frac {5 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 d (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 = 1.31, size = 252, normalized size = 1.43 \begin {gather*} -\frac {e^{-i d x} \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \left (-4 i e^{-i (c+d x)} \left (1+e^{i (c+d x)}\right )^3 \sqrt {1+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 )+40 \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-i \sin \left (\frac {1}{2} (c+d x)\right )\right )+i (29+50 \cos (c+d x)+17 \cos (2 (c+d x))+12 i \sin (c+d x)+7 i \sin (2 (c+d x)))\right ) \left (\cos \left (\frac {1}{2} (c+3 d x)\right )+i \sin \left (\frac {1}{2} (c+3 d x)\right )\right )}{6 a^2 d (1+\sec (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^(7/2)/(a + a*Sec[c + d*x])^2,x]

[Out]

-1/6*(Cos[(c + d*x)/2]*Sec[c + d*x]^(5/2)*(((-4*I)*(1 + E^(I*(c + d*x)))^3*Sqrt[1 + E^((2*I)*(c + d*x))]*Hyper
geometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^(I*(c + d*x)) + 40*Cos[(c + d*x)/2]^3*Sqrt[Cos[c + d*x]]*
EllipticF[(c + d*x)/2, 2]*(Cos[(c + d*x)/2] - I*Sin[(c + d*x)/2]) + I*(29 + 50*Cos[c + d*x] + 17*Cos[2*(c + d*
x)] + (12*I)*Sin[c + d*x] + (7*I)*Sin[2*(c + d*x)]))*(Cos[(c + 3*d*x)/2] + I*Sin[(c + 3*d*x)/2]))/(a^2*d*E^(I*
d*x)*(1 + Sec[c + d*x])^2)

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Maple [A]
time = 0.07, size = 405, normalized size = 2.30

method	result	size
default	\(-\frac {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 )}\, \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}}\, \left (5 \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-12 \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\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 )}\, \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}}\, \left (5 \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-12 \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-48 \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 ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+86 \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 )-37 \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 )}{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}\)	\(405\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(7/2)/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-1/6*(2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2
*c)^2)^(1/2)*(5*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/
2*c)*sin(1/2*d*x+1/2*c)^2-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1
/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(5*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-12*EllipticE(cos(1/2*d*x+1/2*c),2^(1
/2)))*cos(1/2*d*x+1/2*c)-48*(-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)^6+86*(-2*s
in(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)^4-37*(-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^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(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(7/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.54, size = 278, normalized size = 1.58 \begin {gather*} -\frac {5 \, {\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 ) + 5 \, {\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 ) + 12 \, {\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 ) + 12 \, {\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 ) - \frac {2 \, {\left (12 \, \cos \left (d x + c\right )^{2} + 19 \, \cos \left (d x + c\right ) + 6\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\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(sec(d*x+c)^(7/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/6*(5*(-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)) + 5*(I*sqrt(2)*cos(d*x + c)^2 + 2*I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*weierstrassPInver
se(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 12*(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, cos(d*x + c) + I*sin(d*x + c))) + 12*(-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, cos(d*x + c) -
I*sin(d*x + c))) - 2*(12*cos(d*x + c)^2 + 19*cos(d*x + c) + 6)*sin(d*x + c)/sqrt(cos(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(sec(d*x+c)**(7/2)/(a+a*sec(d*x+c))**2,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3063 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(sec(d*x+c)^(7/2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^(7/2)/(a*sec(d*x + c) + a)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{{\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))^(7/2)/(a + a/cos(c + d*x))^2,x)

[Out]

int((1/cos(c + d*x))^(7/2)/(a + a/cos(c + d*x))^2, x)

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