3.3.9 \(\int \frac {\sec ^{\frac {11}{2}}(c+d x)}{(a+a \sec (c+d x))^3} \, dx\) [209]

Optimal. Leaf size=247 \[ \frac {119 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}+\frac {11 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{2 a^3 d}-\frac {119 \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a^3 d}+\frac {11 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a^3 d}-\frac {\sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \sec (c+d x))^2}-\frac {119 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{30 d \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

11/2*sec(d*x+c)^(3/2)*sin(d*x+c)/a^3/d-1/5*sec(d*x+c)^(9/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^3-2/3*sec(d*x+c)^(7/
2)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^2-119/30*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a^3+a^3*sec(d*x+c))-119/10*sin(d*x+
c)*sec(d*x+c)^(1/2)/a^3/d+119/10*(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^3/d+11/2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elliptic
F(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^3/d

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Rubi [A]
time = 0.25, antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3901, 4104, 3872, 3853, 3856, 2719, 2720} \begin {gather*} -\frac {119 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{30 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {11 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 a^3 d}-\frac {119 \sin (c+d x) \sqrt {\sec (c+d x)}}{10 a^3 d}+\frac {11 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{2 a^3 d}+\frac {119 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {\sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 a d (a \sec (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^(11/2)/(a + a*Sec[c + d*x])^3,x]

[Out]

(119*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) + (11*Sqrt[Cos[c + d*x]]*Elli
pticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(2*a^3*d) - (119*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(10*a^3*d) + (11*S
ec[c + d*x]^(3/2)*Sin[c + d*x])/(2*a^3*d) - (Sec[c + d*x]^(9/2)*Sin[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - (
2*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(3*a*d*(a + a*Sec[c + d*x])^2) - (119*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(30*
d*(a^3 + a^3*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 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 {11}{2}}(c+d x)}{(a+a \sec (c+d x))^3} \, dx &=-\frac {\sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {\int \frac {\sec ^{\frac {7}{2}}(c+d x) \left (\frac {7 a}{2}-\frac {13}{2} a \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {\sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \sec (c+d x))^2}-\frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (25 a^2-\frac {69}{2} a^2 \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac {\sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \sec (c+d x))^2}-\frac {119 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{30 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {\int \sec ^{\frac {3}{2}}(c+d x) \left (\frac {357 a^3}{4}-\frac {495}{4} a^3 \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac {\sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \sec (c+d x))^2}-\frac {119 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{30 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {119 \int \sec ^{\frac {3}{2}}(c+d x) \, dx}{20 a^3}+\frac {33 \int \sec ^{\frac {5}{2}}(c+d x) \, dx}{4 a^3}\\ &=-\frac {119 \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a^3 d}+\frac {11 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a^3 d}-\frac {\sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \sec (c+d x))^2}-\frac {119 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{30 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {11 \int \sqrt {\sec (c+d x)} \, dx}{4 a^3}+\frac {119 \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{20 a^3}\\ &=-\frac {119 \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a^3 d}+\frac {11 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a^3 d}-\frac {\sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \sec (c+d x))^2}-\frac {119 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{30 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {\left (11 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{4 a^3}+\frac {\left (119 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{20 a^3}\\ &=\frac {119 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}+\frac {11 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{2 a^3 d}-\frac {119 \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a^3 d}+\frac {11 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a^3 d}-\frac {\sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 a d (a+a \sec (c+d x))^2}-\frac {119 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{30 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 4.97, size = 378, normalized size = 1.53 \begin {gather*} \frac {e^{-i d x} \csc \left (\frac {c}{2}\right ) \left (-119 \sqrt {2} e^{2 i d x} \left (-1+e^{2 i c}\right ) \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \cos ^6\left (\frac {1}{2} (c+d x)\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right ) \sec \left (\frac {c}{2}\right ) \sec ^3(c+d x)+\frac {e^{-\frac {3}{2} i (2 c+d x)} \left (-1+e^{i c}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (165+944 e^{i (c+d x)}+2476 e^{2 i (c+d x)}+4148 e^{3 i (c+d x)}+5134 e^{4 i (c+d x)}+4664 e^{5 i (c+d x)}+3340 e^{6 i (c+d x)}+1620 e^{7 i (c+d x)}+357 e^{8 i (c+d x)}-165 i \left (1+e^{i (c+d x)}\right )^5 \left (1+e^{2 i (c+d x)}\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right ) \sec ^{\frac {7}{2}}(c+d x)}{16 \left (1+e^{2 i (c+d x)}\right )}\right )}{15 a^3 d (1+\sec (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^(11/2)/(a + a*Sec[c + d*x])^3,x]

[Out]

(Csc[c/2]*(-119*Sqrt[2]*E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[
1 + E^((2*I)*(c + d*x))]*Cos[(c + d*x)/2]^6*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]*Sec[c/2]*Se
c[c + d*x]^3 + ((-1 + E^(I*c))*Cos[(c + d*x)/2]*(165 + 944*E^(I*(c + d*x)) + 2476*E^((2*I)*(c + d*x)) + 4148*E
^((3*I)*(c + d*x)) + 5134*E^((4*I)*(c + d*x)) + 4664*E^((5*I)*(c + d*x)) + 3340*E^((6*I)*(c + d*x)) + 1620*E^(
(7*I)*(c + d*x)) + 357*E^((8*I)*(c + d*x)) - (165*I)*(1 + E^(I*(c + d*x)))^5*(1 + E^((2*I)*(c + d*x)))*Sqrt[Co
s[c + d*x]]*EllipticF[(c + d*x)/2, 2])*Sec[c + d*x]^(7/2))/(16*E^(((3*I)/2)*(2*c + d*x))*(1 + E^((2*I)*(c + d*
x))))))/(15*a^3*d*E^(I*d*x)*(1 + Sec[c + d*x])^3)

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Maple [A]
time = 0.10, size = 453, normalized size = 1.83

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 (\frac {\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 )}}{5 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}+\frac {32 \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 )}}{15 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {118 \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 )}}{5 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {128 \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 )}{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 )}}+\frac {238 \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 (\EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {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 )}}+\frac {48 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{\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 )}}-\frac {4 \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 )}}{3 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{2}}\right )}{4 a^{3} \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}\)	\(453\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(11/2)/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

-1/4*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a^3*(1/5*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1
/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)^5+32/15*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2
*c)^3+118/5*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)-128/5*(sin(1/2*d*x+1/2*c)^
2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(
1/2*d*x+1/2*c),2^(1/2))+238/5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1
/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2
)))+48*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)-4/3*c
os(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)/(cos(1/2*d*x+1/2*c)^2-1/2)^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)^(11/2)/(a+a*sec(d*x+c))^3,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 = 0.72, size = 404, normalized size = 1.64 \begin {gather*} -\frac {165 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{4} + 3 i \, \sqrt {2} \cos \left (d x + c\right )^{3} + 3 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 ) + 165 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{4} - 3 i \, \sqrt {2} \cos \left (d x + c\right )^{3} - 3 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 ) + 357 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{4} - 3 i \, \sqrt {2} \cos \left (d x + c\right )^{3} - 3 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 ) + 357 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{4} + 3 i \, \sqrt {2} \cos \left (d x + c\right )^{3} + 3 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 ) + \frac {2 \, {\left (357 \, \cos \left (d x + c\right )^{4} + 906 \, \cos \left (d x + c\right )^{3} + 695 \, \cos \left (d x + c\right )^{2} + 120 \, \cos \left (d x + c\right ) - 20\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(11/2)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/60*(165*(I*sqrt(2)*cos(d*x + c)^4 + 3*I*sqrt(2)*cos(d*x + c)^3 + 3*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)) + 165*(-I*sqrt(2)*cos(d*x + c)^4 - 3*I*sq
rt(2)*cos(d*x + c)^3 - 3*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)) + 357*(-I*sqrt(2)*cos(d*x + c)^4 - 3*I*sqrt(2)*cos(d*x + c)^3 - 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)
)) + 357*(I*sqrt(2)*cos(d*x + c)^4 + 3*I*sqrt(2)*cos(d*x + c)^3 + 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))) + 2*(357*cos(d*x +
c)^4 + 906*cos(d*x + c)^3 + 695*cos(d*x + c)^2 + 120*cos(d*x + c) - 20)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*
d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))

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Sympy [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)**(11/2)/(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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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)^(11/2)/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^(11/2)/(a*sec(d*x + c) + a)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{11/2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cos(c + d*x))^(11/2)/(a + a/cos(c + d*x))^3,x)

[Out]

int((1/cos(c + d*x))^(11/2)/(a + a/cos(c + d*x))^3, x)

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