3.4.90 \(\int \frac {\cos ^{\frac {3}{2}}(c+d x)}{(a+a \sec (c+d x))^3} \, dx\) [390]

Optimal. Leaf size=181 \[ -\frac {119 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {11 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{2 a^3 d}+\frac {11 \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a^3 d}-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {2 \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d (a+a \sec (c+d x))^2}-\frac {119 \sqrt {\cos (c+d x)} \sin (c+d x)}{30 d \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

-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))/a^3/d+11/2*(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))/a^3/d+11/2*sin(d*x+c)*cos(d*x
+c)^(1/2)/a^3/d-1/5*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^3-2/3*sin(d*x+c)*cos(d*x+c)^(1/2)/a/d/(a+a*
sec(d*x+c))^2-119/30*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a^3+a^3*sec(d*x+c))

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Rubi [A]
time = 0.28, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {4349, 3902, 4105, 3872, 3854, 3856, 2720, 2719} \begin {gather*} \frac {11 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{2 a^3 d}-\frac {119 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {11 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a^3 d}-\frac {119 \sin (c+d x) \sqrt {\cos (c+d x)}}{30 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d (a \sec (c+d x)+a)^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \sec (c+d x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-119*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) + (11*EllipticF[(c + d*x)/2, 2])/(2*a^3*d) + (11*Sqrt[Cos[c + d*x]
]*Sin[c + d*x])/(2*a^3*d) - (Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - (2*Sqrt[Cos[c + d
*x]]*Sin[c + d*x])/(3*a*d*(a + a*Sec[c + d*x])^2) - (119*Sqrt[Cos[c + d*x]]*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 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]

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 {\cos ^{\frac {3}{2}}(c+d x)}{(a+a \sec (c+d x))^3} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx\\ &=-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {13 a}{2}+\frac {7}{2} a \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {2 \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d (a+a \sec (c+d x))^2}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {69 a^2}{2}+25 a^2 \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx}{15 a^4}\\ &=-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {2 \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d (a+a \sec (c+d x))^2}-\frac {119 \sqrt {\cos (c+d x)} \sin (c+d x)}{30 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {495 a^3}{4}+\frac {357}{4} a^3 \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{15 a^6}\\ &=-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {2 \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d (a+a \sec (c+d x))^2}-\frac {119 \sqrt {\cos (c+d x)} \sin (c+d x)}{30 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {\left (119 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{20 a^3}+\frac {\left (33 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{4 a^3}\\ &=\frac {11 \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a^3 d}-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {2 \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d (a+a \sec (c+d x))^2}-\frac {119 \sqrt {\cos (c+d x)} \sin (c+d x)}{30 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {119 \int \sqrt {\cos (c+d x)} \, dx}{20 a^3}+\frac {\left (11 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \, dx}{4 a^3}\\ &=-\frac {119 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {11 \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a^3 d}-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {2 \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d (a+a \sec (c+d x))^2}-\frac {119 \sqrt {\cos (c+d x)} \sin (c+d x)}{30 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {11 \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{4 a^3}\\ &=-\frac {119 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {11 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{2 a^3 d}+\frac {11 \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a^3 d}-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {2 \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d (a+a \sec (c+d x))^2}-\frac {119 \sqrt {\cos (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 = 2.18, size = 375, normalized size = 2.07 \begin {gather*} \frac {\cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (-\frac {4 i \sqrt {2} e^{-i (c+d x)} \left (119 \left (1+e^{2 i (c+d x)}\right )+119 \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 )+55 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 ^3(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 )}}+\frac {720 \cot (c)+708 \csc (c)+\frac {1}{4} \sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (1061 \sin \left (\frac {d x}{2}\right )-709 \sin \left (c+\frac {d x}{2}\right )+715 \sin \left (c+\frac {3 d x}{2}\right )-170 \sin \left (2 c+\frac {3 d x}{2}\right )+202 \sin \left (2 c+\frac {5 d x}{2}\right )+25 \sin \left (3 c+\frac {5 d x}{2}\right )+5 \sin \left (3 c+\frac {7 d x}{2}\right )+5 \sin \left (4 c+\frac {7 d x}{2}\right )\right )}{3 d \cos ^{\frac {5}{2}}(c+d x)}\right )}{5 a^3 (1+\sec (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^6*(((-4*I)*Sqrt[2]*(119*(1 + E^((2*I)*(c + d*x))) + 119*(-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))] + 55*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*S
qrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))])*Sec[c + d*x]^3)/(d*E^(I*(
c + d*x))*(-1 + E^((2*I)*c))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]) + (720*Cot[c] + 708*Csc[c] + (Se
c[c/2]*Sec[(c + d*x)/2]^5*(1061*Sin[(d*x)/2] - 709*Sin[c + (d*x)/2] + 715*Sin[c + (3*d*x)/2] - 170*Sin[2*c + (
3*d*x)/2] + 202*Sin[2*c + (5*d*x)/2] + 25*Sin[3*c + (5*d*x)/2] + 5*Sin[3*c + (7*d*x)/2] + 5*Sin[4*c + (7*d*x)/
2]))/4)/(3*d*Cos[c + d*x]^(5/2))))/(5*a^3*(1 + Sec[c + d*x])^3)

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Maple [A]
time = 0.11, size = 283, normalized size = 1.56

method	result	size
default	\(-\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (160 \left (\cos ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+468 \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+330 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+714 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-1058 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+474 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-47 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3\right )}{60 a^{3} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(283\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(160*cos(1/2*d*x+1/2*c)^10+468*cos(1/2*d*x+1/2*c
)^8+330*(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))*c
os(1/2*d*x+1/2*c)^5+714*(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*El
lipticE(cos(1/2*d*x+1/2*c),2^(1/2))-1058*cos(1/2*d*x+1/2*c)^6+474*cos(1/2*d*x+1/2*c)^4-47*cos(1/2*d*x+1/2*c)^2
+3)/a^3/(-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(cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.70, size = 354, normalized size = 1.96 \begin {gather*} \frac {2 \, {\left (20 \, \cos \left (d x + c\right )^{3} + 237 \, \cos \left (d x + c\right )^{2} + 376 \, \cos \left (d x + c\right ) + 165\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 165 \, {\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 ) - 165 \, {\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 ) - 357 \, {\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 ) - 357 \, {\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 )}{60 \, {\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(cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/60*(2*(20*cos(d*x + c)^3 + 237*cos(d*x + c)^2 + 376*cos(d*x + c) + 165)*sqrt(cos(d*x + c))*sin(d*x + c) - 16
5*(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))*weierstrassPI
nverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 165*(-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)) - 357*(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, w
eierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 357*(-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, cos(d*x +
c) - I*sin(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 {\cos ^{\frac {3}{2}}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cos(c + d*x)**(3/2)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), 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(cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^{3/2}}{{\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(cos(c + d*x)^(3/2)/(a + a/cos(c + d*x))^3,x)

[Out]

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

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