3.1253 \(\int \cos ^{\frac{11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=284 \[ \frac{2 a^2 (84 A+110 B+99 C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{693 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (336 A+374 B+429 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{1155 d \sqrt{a \sec (c+d x)+a}}+\frac{8 a^2 (336 A+374 B+429 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{3465 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a^2 (336 A+374 B+429 C) \sin (c+d x)}{3465 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 a (3 A+11 B) \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}{99 d}+\frac{2 A \sin (c+d x) \cos ^{\frac{9}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d} \]

[Out]

(16*a^2*(336*A + 374*B + 429*C)*Sin[c + d*x])/(3465*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (8*a^2*(3
36*A + 374*B + 429*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3465*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(336*A + 374
*B + 429*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(1155*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(84*A + 110*B + 99*C)*
Cos[c + d*x]^(5/2)*Sin[c + d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(3*A + 11*B)*Cos[c + d*x]^(7/2)*Sqrt[
a + a*Sec[c + d*x]]*Sin[c + d*x])/(99*d) + (2*A*Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(1
1*d)

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Rubi [A]  time = 0.885488, antiderivative size = 284, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {4265, 4086, 4017, 4015, 3805, 3804} \[ \frac{2 a^2 (84 A+110 B+99 C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{693 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (336 A+374 B+429 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{1155 d \sqrt{a \sec (c+d x)+a}}+\frac{8 a^2 (336 A+374 B+429 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{3465 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a^2 (336 A+374 B+429 C) \sin (c+d x)}{3465 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 a (3 A+11 B) \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}{99 d}+\frac{2 A \sin (c+d x) \cos ^{\frac{9}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*a^2*(336*A + 374*B + 429*C)*Sin[c + d*x])/(3465*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (8*a^2*(3
36*A + 374*B + 429*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3465*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(336*A + 374
*B + 429*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(1155*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(84*A + 110*B + 99*C)*
Cos[c + d*x]^(5/2)*Sin[c + d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(3*A + 11*B)*Cos[c + d*x]^(7/2)*Sqrt[
a + a*Sec[c + d*x]]*Sin[c + d*x])/(99*d) + (2*A*Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(1
1*d)

Rule 4265

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rule 4086

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*(d*
Csc[e + f*x])^n)/(f*n), x] - Dist[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 -
b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])

Rule 4017

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*n), x]
- Dist[b/(a*d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*(m - n - 1) - b*B*n - (a*
B*n + 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] && GtQ[m, 1/2] && LtQ[n, -1]

Rule 4015

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

Rule 3805

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

Rule 3804

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Simp[(-2*a*Co
t[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]]), x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^
2, 0]

Rubi steps

\begin{align*} \int \cos ^{\frac{11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac{11}{2}}(c+d x)} \, dx\\ &=\frac{2 A \cos ^{\frac{9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \sec (c+d x))^{3/2} \left (\frac{1}{2} a (3 A+11 B)+\frac{1}{2} a (6 A+11 C) \sec (c+d x)\right )}{\sec ^{\frac{9}{2}}(c+d x)} \, dx}{11 a}\\ &=\frac{2 a (3 A+11 B) \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{99 d}+\frac{2 A \cos ^{\frac{9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d}+\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \sec (c+d x)} \left (\frac{1}{4} a^2 (84 A+110 B+99 C)+\frac{3}{4} a^2 (24 A+22 B+33 C) \sec (c+d x)\right )}{\sec ^{\frac{7}{2}}(c+d x)} \, dx}{99 a}\\ &=\frac{2 a^2 (84 A+110 B+99 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (3 A+11 B) \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{99 d}+\frac{2 A \cos ^{\frac{9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d}+\frac{1}{231} \left (a (336 A+374 B+429 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \sec (c+d x)}}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 (336 A+374 B+429 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (84 A+110 B+99 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (3 A+11 B) \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{99 d}+\frac{2 A \cos ^{\frac{9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d}+\frac{\left (4 a (336 A+374 B+429 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \sec (c+d x)}}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{1155}\\ &=\frac{8 a^2 (336 A+374 B+429 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{3465 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (336 A+374 B+429 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (84 A+110 B+99 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (3 A+11 B) \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{99 d}+\frac{2 A \cos ^{\frac{9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d}+\frac{\left (8 a (336 A+374 B+429 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx}{3465}\\ &=\frac{16 a^2 (336 A+374 B+429 C) \sin (c+d x)}{3465 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{8 a^2 (336 A+374 B+429 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{3465 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (336 A+374 B+429 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{1155 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (84 A+110 B+99 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (3 A+11 B) \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{99 d}+\frac{2 A \cos ^{\frac{9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d}\\ \end{align*}

Mathematica [A]  time = 2.10568, size = 158, normalized size = 0.56 \[ \frac{a \sqrt{\cos (c+d x)} \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)} ((34734 A+44 (799 B+759 C)) \cos (c+d x)+8 (1743 A+1507 B+1287 C) \cos (2 (c+d x))+4935 A \cos (3 (c+d x))+1470 A \cos (4 (c+d x))+315 A \cos (5 (c+d x))+55482 A+3740 B \cos (3 (c+d x))+770 B \cos (4 (c+d x))+59158 B+1980 C \cos (3 (c+d x))+65208 C)}{27720 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[Cos[c + d*x]]*(55482*A + 59158*B + 65208*C + (34734*A + 44*(799*B + 759*C))*Cos[c + d*x] + 8*(1743*A +
1507*B + 1287*C)*Cos[2*(c + d*x)] + 4935*A*Cos[3*(c + d*x)] + 3740*B*Cos[3*(c + d*x)] + 1980*C*Cos[3*(c + d*x
)] + 1470*A*Cos[4*(c + d*x)] + 770*B*Cos[4*(c + d*x)] + 315*A*Cos[5*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])]*Tan
[(c + d*x)/2])/(27720*d)

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Maple [A]  time = 0.284, size = 187, normalized size = 0.7 \begin{align*} -{\frac{2\,a \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 315\,A \left ( \cos \left ( dx+c \right ) \right ) ^{5}+735\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+385\,B \left ( \cos \left ( dx+c \right ) \right ) ^{4}+840\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+935\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+495\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+1008\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+1122\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+1287\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+1344\,A\cos \left ( dx+c \right ) +1496\,B\cos \left ( dx+c \right ) +1716\,C\cos \left ( dx+c \right ) +2688\,A+2992\,B+3432\,C \right ) }{3465\,d\sin \left ( dx+c \right ) }\sqrt{\cos \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-2/3465/d*a*(-1+cos(d*x+c))*(315*A*cos(d*x+c)^5+735*A*cos(d*x+c)^4+385*B*cos(d*x+c)^4+840*A*cos(d*x+c)^3+935*B
*cos(d*x+c)^3+495*C*cos(d*x+c)^3+1008*A*cos(d*x+c)^2+1122*B*cos(d*x+c)^2+1287*C*cos(d*x+c)^2+1344*A*cos(d*x+c)
+1496*B*cos(d*x+c)+1716*C*cos(d*x+c)+2688*A+2992*B+3432*C)*cos(d*x+c)^(1/2)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2
)/sin(d*x+c)

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Maxima [B]  time = 2.49258, size = 1164, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/110880*(21*sqrt(2)*(3630*a*cos(10/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x +
11/2*c) + 990*a*cos(8/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 42
9*a*cos(6/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 165*a*cos(4/11*
arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 55*a*cos(2/11*arctan2(sin(11
/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) - 3630*a*cos(11/2*d*x + 11/2*c)*sin(10/11*ar
ctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 990*a*cos(11/2*d*x + 11/2*c)*sin(8/11*arctan2(sin(11/
2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 429*a*cos(11/2*d*x + 11/2*c)*sin(6/11*arctan2(sin(11/2*d*x + 11/2*
c), cos(11/2*d*x + 11/2*c))) - 165*a*cos(11/2*d*x + 11/2*c)*sin(4/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*
d*x + 11/2*c))) - 55*a*cos(11/2*d*x + 11/2*c)*sin(2/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))
) + 30*a*sin(11/2*d*x + 11/2*c) + 55*a*sin(9/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 165
*a*sin(7/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 429*a*sin(5/11*arctan2(sin(11/2*d*x + 1
1/2*c), cos(11/2*d*x + 11/2*c))) + 990*a*sin(3/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 3
630*a*sin(1/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))))*A*sqrt(a) - 44*sqrt(2)*(189*(10*a*sin
(4*d*x + 4*c) + a*sin(2*d*x + 2*c))*cos(9/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 7*(270*a*cos(4*d*x
+ 4*c) + 27*a*cos(2*d*x + 2*c) + 5*a)*sin(9/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 135*a*sin(7/4*arc
tan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 189*a*sin(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 1050*
a*sin(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 1890*a*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x +
2*c))))*B*sqrt(a) - 132*sqrt(2)*(175*a*cos(7/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2*d*x + 2*c) -
5*(35*a*cos(2*d*x + 2*c) + 6*a)*sin(7/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 126*a*sin(5/4*arctan2(
sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 175*a*sin(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 1470*a*sin
(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*C*sqrt(a))/d

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Fricas [A]  time = 0.511192, size = 437, normalized size = 1.54 \begin{align*} \frac{2 \,{\left (315 \, A a \cos \left (d x + c\right )^{5} + 35 \,{\left (21 \, A + 11 \, B\right )} a \cos \left (d x + c\right )^{4} + 5 \,{\left (168 \, A + 187 \, B + 99 \, C\right )} a \cos \left (d x + c\right )^{3} + 3 \,{\left (336 \, A + 374 \, B + 429 \, C\right )} a \cos \left (d x + c\right )^{2} + 4 \,{\left (336 \, A + 374 \, B + 429 \, C\right )} a \cos \left (d x + c\right ) + 8 \,{\left (336 \, A + 374 \, B + 429 \, C\right )} a\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3465 \,{\left (d \cos \left (d x + c\right ) + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*(315*A*a*cos(d*x + c)^5 + 35*(21*A + 11*B)*a*cos(d*x + c)^4 + 5*(168*A + 187*B + 99*C)*a*cos(d*x + c)^3
+ 3*(336*A + 374*B + 429*C)*a*cos(d*x + c)^2 + 4*(336*A + 374*B + 429*C)*a*cos(d*x + c) + 8*(336*A + 374*B +
429*C)*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c) + d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**(11/2)*(a+a*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cos \left (d x + c\right )^{\frac{11}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

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