3.3.17 \(\int \frac {1}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx\) [217]

Optimal. Leaf size=247 \[ \frac {231 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}-\frac {21 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{2 a^3 d}+\frac {77 \sin (c+d x)}{10 a^3 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {21 \sin (c+d x)}{2 a^3 d \sqrt {\sec (c+d x)}}-\frac {\sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {4 \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {63 \sin (c+d x)}{10 d \sec ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

77/10*sin(d*x+c)/a^3/d/sec(d*x+c)^(3/2)-1/5*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^3-4/5*sin(d*x+c)/a/
d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^2-63/10*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a^3+a^3*sec(d*x+c))-21/2*sin(d*x+c)
/a^3/d/sec(d*x+c)^(1/2)+231/10*(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-21/2*(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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^3/d

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Rubi [A]
time = 0.26, 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 = {3902, 4105, 3872, 3854, 3856, 2719, 2720} \begin {gather*} -\frac {63 \sin (c+d x)}{10 d \sec ^{\frac {3}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}+\frac {77 \sin (c+d x)}{10 a^3 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {21 \sin (c+d x)}{2 a^3 d \sqrt {\sec (c+d x)}}-\frac {21 \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 {231 \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 {4 \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}-\frac {\sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^3),x]

[Out]

(231*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) - (21*Sqrt[Cos[c + d*x]]*Elli
pticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(2*a^3*d) + (77*Sin[c + d*x])/(10*a^3*d*Sec[c + d*x]^(3/2)) - (21*Si
n[c + d*x])/(2*a^3*d*Sqrt[Sec[c + d*x]]) - Sin[c + d*x]/(5*d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^3) - (4*S
in[c + d*x])/(5*a*d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2) - (63*Sin[c + d*x])/(10*d*Sec[c + d*x]^(3/2)*(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 3854

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Csc[c + d*x])^(n + 1)/(b*d*n)), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[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 3902

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

Rule 4105

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(
2*m + 1))), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*
(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, 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 {1}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx &=-\frac {\sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {\int \frac {-\frac {15 a}{2}+\frac {9}{2} a \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {\sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {4 \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {\int \frac {-\frac {105 a^2}{2}+42 a^2 \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx}{15 a^4}\\ &=-\frac {\sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {4 \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {63 \sin (c+d x)}{10 d \sec ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right )}-\frac {\int \frac {-\frac {1155 a^3}{4}+\frac {945}{4} a^3 \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{15 a^6}\\ &=-\frac {\sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {4 \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {63 \sin (c+d x)}{10 d \sec ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right )}-\frac {63 \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{4 a^3}+\frac {77 \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{4 a^3}\\ &=\frac {77 \sin (c+d x)}{10 a^3 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {21 \sin (c+d x)}{2 a^3 d \sqrt {\sec (c+d x)}}-\frac {\sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {4 \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {63 \sin (c+d x)}{10 d \sec ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right )}-\frac {21 \int \sqrt {\sec (c+d x)} \, dx}{4 a^3}+\frac {231 \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{20 a^3}\\ &=\frac {77 \sin (c+d x)}{10 a^3 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {21 \sin (c+d x)}{2 a^3 d \sqrt {\sec (c+d x)}}-\frac {\sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {4 \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {63 \sin (c+d x)}{10 d \sec ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right )}-\frac {\left (21 \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 (231 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{20 a^3}\\ &=\frac {231 \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 {21 \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 {77 \sin (c+d x)}{10 a^3 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {21 \sin (c+d x)}{2 a^3 d \sqrt {\sec (c+d x)}}-\frac {\sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {4 \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {63 \sin (c+d x)}{10 d \sec ^{\frac {3}{2}}(c+d x) \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 = 2.72, size = 297, normalized size = 1.20 \begin {gather*} -\frac {e^{-i d x} \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (-3465 i \cos \left (\frac {1}{2} (c+d x)\right )-2541 i \cos \left (\frac {3}{2} (c+d x)\right )-1155 i \cos \left (\frac {5}{2} (c+d x)\right )-231 i \cos \left (\frac {7}{2} (c+d x)\right )+3360 \cos ^5\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 )+77 i e^{-\frac {3}{2} i (c+d x)} \left (1+e^{i (c+d x)}\right )^5 \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 )+125 \sin \left (\frac {1}{2} (c+d x)\right )+359 \sin \left (\frac {3}{2} (c+d x)\right )+350 \sin \left (\frac {5}{2} (c+d x)\right )+138 \sin \left (\frac {7}{2} (c+d x)\right )+5 \sin \left (\frac {9}{2} (c+d x)\right )-\sin \left (\frac {11}{2} (c+d x)\right )\right )}{40 a^3 d (1+\sec (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^3),x]

[Out]

-1/40*(Cos[(c + d*x)/2]*Sec[c + d*x]^(7/2)*(Cos[d*x] + I*Sin[d*x])*((-3465*I)*Cos[(c + d*x)/2] - (2541*I)*Cos[
(3*(c + d*x))/2] - (1155*I)*Cos[(5*(c + d*x))/2] - (231*I)*Cos[(7*(c + d*x))/2] + 3360*Cos[(c + d*x)/2]^5*Sqrt
[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + ((77*I)*(1 + E^(I*(c + d*x)))^5*Sqrt[1 + E^((2*I)*(c + d*x))]*Hyper
geometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^(((3*I)/2)*(c + d*x)) + 125*Sin[(c + d*x)/2] + 359*Sin[(3
*(c + d*x))/2] + 350*Sin[(5*(c + d*x))/2] + 138*Sin[(7*(c + d*x))/2] + 5*Sin[(9*(c + d*x))/2] - Sin[(11*(c + d
*x))/2]))/(a^3*d*E^(I*d*x)*(1 + Sec[c + d*x])^3)

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Maple [A]
time = 0.10, size = 296, normalized size = 1.20

method	result	size
default	\(-\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (64 \left (\cos ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-288 \left (\cos ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-76 \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-210 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-462 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+530 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-248 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+19 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}{20 a^{3} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(296\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

-1/20/a^3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(64*cos(1/2*d*x+1/2*c)^12-288*cos(1/2*d*x+1/
2*c)^10-76*cos(1/2*d*x+1/2*c)^8-210*(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(c
os(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5-462*(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)^5*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+530*cos(1/2*d*x+1/2*c)^6-248*cos(1/2*d*x+1/2*c
)^4+19*cos(1/2*d*x+1/2*c)^2-1)/(-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/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/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate(1/((a*sec(d*x + c) + a)^3*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 = 0.97, size = 373, normalized size = 1.51 \begin {gather*} -\frac {105 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{3} - 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 3 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 ) + 105 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{3} + 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 3 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 ) + 231 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{3} - 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 3 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 ) + 231 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{3} + 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 3 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 (4 \, \cos \left (d x + c\right )^{5} - 8 \, \cos \left (d x + c\right )^{4} - 147 \, \cos \left (d x + c\right )^{3} - 238 \, \cos \left (d x + c\right )^{2} - 105 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{20 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/20*(105*(-I*sqrt(2)*cos(d*x + c)^3 - 3*I*sqrt(2)*cos(d*x + c)^2 - 3*I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*wei
erstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 105*(I*sqrt(2)*cos(d*x + c)^3 + 3*I*sqrt(2)*cos(d*x +
c)^2 + 3*I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 231*
(-I*sqrt(2)*cos(d*x + c)^3 - 3*I*sqrt(2)*cos(d*x + c)^2 - 3*I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*weierstrassZet
a(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 231*(I*sqrt(2)*cos(d*x + c)^3 + 3*I*sqrt
(2)*cos(d*x + c)^2 + 3*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))) - 2*(4*cos(d*x + c)^5 - 8*cos(d*x + c)^4 - 147*cos(d*x + c)^3 - 238*cos(d*x + c
)^2 - 105*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^
3*d*cos(d*x + c) + a^3*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {1}{\sec ^{\frac {11}{2}}{\left (c + d x \right )} + 3 \sec ^{\frac {9}{2}}{\left (c + d x \right )} + 3 \sec ^{\frac {7}{2}}{\left (c + d x \right )} + \sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)**(5/2)/(a+a*sec(d*x+c))**3,x)

[Out]

Integral(1/(sec(c + d*x)**(11/2) + 3*sec(c + d*x)**(9/2) + 3*sec(c + d*x)**(7/2) + sec(c + d*x)**(5/2)), x)/a*
*3

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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/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(1/((a*sec(d*x + c) + a)^3*sec(d*x + c)^(5/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3\,{\left (\frac {1}{\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(1/((a + a/cos(c + d*x))^3*(1/cos(c + d*x))^(5/2)),x)

[Out]

int(1/((a + a/cos(c + d*x))^3*(1/cos(c + d*x))^(5/2)), x)

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