3.830 \(\int \frac {\cos ^{\frac {3}{2}}(c+d x)}{(a+b \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=346 \[ \frac {b^2 \left (13 a^2-7 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 a^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {b \left (24 a^4-65 a^2 b^2+35 b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^4 d \left (a^2-b^2\right )^2}-\frac {b^3 \left (63 a^4-86 a^2 b^2+35 b^4\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^5 d (a-b)^2 (a+b)^3}+\frac {\left (8 a^4-61 a^2 b^2+35 b^4\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{12 a^3 d \left (a^2-b^2\right )^2}+\frac {\left (8 a^6+128 a^4 b^2-223 a^2 b^4+105 b^6\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{12 a^5 d \left (a^2-b^2\right )^2} \]

[Out]

-1/4*b*(24*a^4-65*a^2*b^2+35*b^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))/a^4/(a^2-b^2)^2/d+1/12*(8*a^6+128*a^4*b^2-223*a^2*b^4+105*b^6)*(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^5/(a^2-b^2)^2/d-1/4*b^3*(63*a^4-86*a^2*b^2+35*b^4)*(cos(1/2
*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))/a^5/(a-b)^2/(a+b)^3/d
+1/12*(8*a^4-61*a^2*b^2+35*b^4)*sin(d*x+c)*cos(d*x+c)^(1/2)/a^3/(a^2-b^2)^2/d+1/2*b^2*sin(d*x+c)*cos(d*x+c)^(1
/2)/a/(a^2-b^2)/d/(a+b*sec(d*x+c))^2+1/4*b^2*(13*a^2-7*b^2)*sin(d*x+c)*cos(d*x+c)^(1/2)/a^2/(a^2-b^2)^2/d/(a+b
*sec(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 1.06, antiderivative size = 346, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {4264, 3847, 4100, 4104, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ \frac {\left (128 a^4 b^2-223 a^2 b^4+8 a^6+105 b^6\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{12 a^5 d \left (a^2-b^2\right )^2}-\frac {b \left (-65 a^2 b^2+24 a^4+35 b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^4 d \left (a^2-b^2\right )^2}-\frac {b^3 \left (-86 a^2 b^2+63 a^4+35 b^4\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^5 d (a-b)^2 (a+b)^3}+\frac {\left (-61 a^2 b^2+8 a^4+35 b^4\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{12 a^3 d \left (a^2-b^2\right )^2}+\frac {b^2 \left (13 a^2-7 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 a^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(24*a^4 - 65*a^2*b^2 + 35*b^4)*EllipticE[(c + d*x)/2, 2])/(4*a^4*(a^2 - b^2)^2*d) + ((8*a^6 + 128*a^4*b^2
- 223*a^2*b^4 + 105*b^6)*EllipticF[(c + d*x)/2, 2])/(12*a^5*(a^2 - b^2)^2*d) - (b^3*(63*a^4 - 86*a^2*b^2 + 35*
b^4)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(4*a^5*(a - b)^2*(a + b)^3*d) + ((8*a^4 - 61*a^2*b^2 + 35*b^4)
*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(12*a^3*(a^2 - b^2)^2*d) + (b^2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*a*(a^2 -
b^2)*d*(a + b*Sec[c + d*x])^2) + (b^2*(13*a^2 - 7*b^2)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*a^2*(a^2 - b^2)^2*
d*(a + b*Sec[c + d*x]))

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rule 3771

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 3787

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 3847

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

Rule 3849

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[d*Sqrt[d*S
in[e + f*x]]*Sqrt[d*Csc[e + f*x]], Int[1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d
, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4100

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

Rule 4104

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

Rule 4106

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))), x_Symbol] :> Dist[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2), Int[(d*Csc[
e + f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Dist[1/a^2, Int[(a*A - (A*b - a*B)*Csc[e + f*x])/Sqrt[d*Csc[e +
f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]

Rule 4264

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+b \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+b \sec (c+d x))^3} \, dx\\ &=\frac {b^2 \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-2 a^2+\frac {7 b^2}{2}+2 a b \sec (c+d x)-\frac {5}{2} b^2 \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac {b^2 \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {b^2 \left (13 a^2-7 b^2\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{4} \left (8 a^4-61 a^2 b^2+35 b^4\right )-a b \left (4 a^2-b^2\right ) \sec (c+d x)+\frac {3}{4} b^2 \left (13 a^2-7 b^2\right ) \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac {\left (8 a^4-61 a^2 b^2+35 b^4\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{12 a^3 \left (a^2-b^2\right )^2 d}+\frac {b^2 \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {b^2 \left (13 a^2-7 b^2\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {3}{8} b \left (24 a^4-65 a^2 b^2+35 b^4\right )-\frac {1}{2} a \left (2 a^4+14 a^2 b^2-7 b^4\right ) \sec (c+d x)-\frac {1}{8} b \left (8 a^4-61 a^2 b^2+35 b^4\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{3 a^3 \left (a^2-b^2\right )^2}\\ &=\frac {\left (8 a^4-61 a^2 b^2+35 b^4\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{12 a^3 \left (a^2-b^2\right )^2 d}+\frac {b^2 \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {b^2 \left (13 a^2-7 b^2\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {3}{8} a b \left (24 a^4-65 a^2 b^2+35 b^4\right )-\left (\frac {1}{2} a^2 \left (2 a^4+14 a^2 b^2-7 b^4\right )+\frac {3}{8} b^2 \left (24 a^4-65 a^2 b^2+35 b^4\right )\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{3 a^5 \left (a^2-b^2\right )^2}-\frac {\left (b^3 \left (63 a^4-86 a^2 b^2+35 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{8 a^5 \left (a^2-b^2\right )^2}\\ &=\frac {\left (8 a^4-61 a^2 b^2+35 b^4\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{12 a^3 \left (a^2-b^2\right )^2 d}+\frac {b^2 \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {b^2 \left (13 a^2-7 b^2\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac {\left (b^3 \left (63 a^4-86 a^2 b^2+35 b^4\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{8 a^5 \left (a^2-b^2\right )^2}-\frac {\left (b \left (24 a^4-65 a^2 b^2+35 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{8 a^4 \left (a^2-b^2\right )^2}+\frac {\left (\left (8 a^6+128 a^4 b^2-223 a^2 b^4+105 b^6\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \, dx}{24 a^5 \left (a^2-b^2\right )^2}\\ &=-\frac {b^3 \left (63 a^4-86 a^2 b^2+35 b^4\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^5 (a-b)^2 (a+b)^3 d}+\frac {\left (8 a^4-61 a^2 b^2+35 b^4\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{12 a^3 \left (a^2-b^2\right )^2 d}+\frac {b^2 \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {b^2 \left (13 a^2-7 b^2\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac {\left (b \left (24 a^4-65 a^2 b^2+35 b^4\right )\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 a^4 \left (a^2-b^2\right )^2}+\frac {\left (8 a^6+128 a^4 b^2-223 a^2 b^4+105 b^6\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{24 a^5 \left (a^2-b^2\right )^2}\\ &=-\frac {b \left (24 a^4-65 a^2 b^2+35 b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^4 \left (a^2-b^2\right )^2 d}+\frac {\left (8 a^6+128 a^4 b^2-223 a^2 b^4+105 b^6\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{12 a^5 \left (a^2-b^2\right )^2 d}-\frac {b^3 \left (63 a^4-86 a^2 b^2+35 b^4\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^5 (a-b)^2 (a+b)^3 d}+\frac {\left (8 a^4-61 a^2 b^2+35 b^4\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{12 a^3 \left (a^2-b^2\right )^2 d}+\frac {b^2 \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {b^2 \left (13 a^2-7 b^2\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 4.35, size = 353, normalized size = 1.02 \[ \frac {\frac {-\frac {2 \left (56 a^4 b-73 a^2 b^3+35 b^5\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+\frac {16 \left (2 a^4+14 a^2 b^2-7 b^4\right ) \left ((a+b) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-b \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{a+b}-\frac {6 \left (24 a^4-65 a^2 b^2+35 b^4\right ) \sin (c+d x) \left (\left (a^2-2 b^2\right ) \Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{a^2 \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}+\frac {4 \sin (c+d x) \sqrt {\cos (c+d x)} \left (4 a^6+4 \left (a^3-a b^2\right )^2 \cos (2 (c+d x))-57 a^2 b^4+a b \left (16 a^4-83 a^2 b^2+49 b^4\right ) \cos (c+d x)+35 b^6\right )}{\left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}}{48 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*Sqrt[Cos[c + d*x]]*(4*a^6 - 57*a^2*b^4 + 35*b^6 + a*b*(16*a^4 - 83*a^2*b^2 + 49*b^4)*Cos[c + d*x] + 4*(a^3
- a*b^2)^2*Cos[2*(c + d*x)])*Sin[c + d*x])/((a^2 - b^2)^2*(b + a*Cos[c + d*x])^2) + ((-2*(56*a^4*b - 73*a^2*b
^3 + 35*b^5)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + (16*(2*a^4 + 14*a^2*b^2 - 7*b^4)*((a + b)*El
lipticF[(c + d*x)/2, 2] - b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]))/(a + b) - (6*(24*a^4 - 65*a^2*b^2 + 35
*b^4)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1
] + (a^2 - 2*b^2)*EllipticPi[-(a/b), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a^2*Sqrt[Sin[c + d*x]^2])
)/((a - b)^2*(a + b)^2))/(48*a^3*d)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 17.27, size = 2216, normalized size = 6.40 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4/3/a^3*(2*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)
+2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*(sin(
1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(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*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-2/a^4*(2*a+3*b)*(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)*(Elliptic
F(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2*(a^2+3*a*b+6*b^2)/a^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(c
os(1/2*d*x+1/2*c),2^(1/2))+20*b^3/a^4/(a^2-a*b)*(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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))-2/a^5*b
^5*(1/2*a^2/b/(a^2-b^2)*cos(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)/(2*a*cos(1/2*d
*x+1/2*c)^2-a+b)^2+3/4*a^2*(a^2-3*b^2)/b^2/(a^2-b^2)^2*cos(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)/(2*a*cos(1/2*d*x+1/2*c)^2-a+b)-3/8/(a+b)/(a^2-b^2)/b^2*(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))*a^2-1/4/(a+b)/(a^2-b^2)/b*(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))*a+7/8/(a+b)/(a^2-b^2)*(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)*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2))+3/8*a^3/b^2/(a^2-b^2)^2*(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))-9/8*a/(a^2
-b^2)^2*(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))-3/8*a^3/b^2/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*c
os(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)*EllipticE(cos(1/2*d*x+1/2*c)
,2^(1/2))+9/8*a/(a^2-b^2)^2*(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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3/8/(a-b)/(a+b)/(a^2-b^2)/b^2/(a^2-a*b
)*a^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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+3/4/(a-b)/(a+b)/(a^2-b^2)/(a^2-a*b)*a^3*(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)*Ell
ipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))-15/8/(a-b)/(a+b)/(a^2-b^2)*b^2/(a^2-a*b)*a*(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)*EllipticPi(cos(
1/2*d*x+1/2*c),2*a/(a-b),2^(1/2)))+10/a^5*b^4*(a^2/b/(a^2-b^2)*cos(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)/(2*a*cos(1/2*d*x+1/2*c)^2-a+b)-1/2/(a+b)/b*(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)
)+1/2*a/b/(a^2-b^2)*(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+si
n(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2*a/b/(a^2-b^2)*(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)*EllipticE(cos(1/2*d*x+
1/2*c),2^(1/2))-1/2/b/(a^2-b^2)/(a^2-a*b)*a^3*(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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+3/2*b/(a^
2-b^2)/(a^2-a*b)*a*(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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),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)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)/(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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