3.1144 \(\int \cos ^{\frac{13}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=313 \[ \frac{2 a^2 (136 A+143 C) \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}{1287 d}+\frac{2 a^3 (2224 A+2717 C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{9009 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^3 (8368 A+10439 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{15015 d \sqrt{a \sec (c+d x)+a}}+\frac{8 a^3 (8368 A+10439 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{45045 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a^3 (8368 A+10439 C) \sin (c+d x)}{45045 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 A \sin (c+d x) \cos ^{\frac{11}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}+\frac{10 a A \sin (c+d x) \cos ^{\frac{9}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{143 d} \]

[Out]

(16*a^3*(8368*A + 10439*C)*Sin[c + d*x])/(45045*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (8*a^3*(8368*
A + 10439*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(45045*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(8368*A + 10439*C)*C
os[c + d*x]^(3/2)*Sin[c + d*x])/(15015*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(2224*A + 2717*C)*Cos[c + d*x]^(5/
2)*Sin[c + d*x])/(9009*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(136*A + 143*C)*Cos[c + d*x]^(7/2)*Sqrt[a + a*Sec[
c + d*x]]*Sin[c + d*x])/(1287*d) + (10*a*A*Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(143*d)
+ (2*A*Cos[c + d*x]^(11/2)*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(13*d)

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Rubi [A]  time = 1.01576, antiderivative size = 313, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.162, Rules used = {4265, 4087, 4017, 4015, 3805, 3804} \[ \frac{2 a^2 (136 A+143 C) \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}{1287 d}+\frac{2 a^3 (2224 A+2717 C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{9009 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^3 (8368 A+10439 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{15015 d \sqrt{a \sec (c+d x)+a}}+\frac{8 a^3 (8368 A+10439 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{45045 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a^3 (8368 A+10439 C) \sin (c+d x)}{45045 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 A \sin (c+d x) \cos ^{\frac{11}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d}+\frac{10 a A \sin (c+d x) \cos ^{\frac{9}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{143 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(13/2)*(a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2),x]

[Out]

(16*a^3*(8368*A + 10439*C)*Sin[c + d*x])/(45045*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (8*a^3*(8368*
A + 10439*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(45045*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(8368*A + 10439*C)*C
os[c + d*x]^(3/2)*Sin[c + d*x])/(15015*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(2224*A + 2717*C)*Cos[c + d*x]^(5/
2)*Sin[c + d*x])/(9009*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(136*A + 143*C)*Cos[c + d*x]^(7/2)*Sqrt[a + a*Sec[
c + d*x]]*Sin[c + d*x])/(1287*d) + (10*a*A*Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(143*d)
+ (2*A*Cos[c + d*x]^(11/2)*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(13*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 4087

Int[((A_.) + 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] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, 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{13}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+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))^{5/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac{13}{2}}(c+d x)} \, dx\\ &=\frac{2 A \cos ^{\frac{11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{13 d}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \sec (c+d x))^{5/2} \left (\frac{5 a A}{2}+\frac{1}{2} a (6 A+13 C) \sec (c+d x)\right )}{\sec ^{\frac{11}{2}}(c+d x)} \, dx}{13 a}\\ &=\frac{10 a A \cos ^{\frac{9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{143 d}+\frac{2 A \cos ^{\frac{11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{13 d}+\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \sec (c+d x))^{3/2} \left (\frac{1}{4} a^2 (136 A+143 C)+\frac{1}{4} a^2 (96 A+143 C) \sec (c+d x)\right )}{\sec ^{\frac{9}{2}}(c+d x)} \, dx}{143 a}\\ &=\frac{2 a^2 (136 A+143 C) \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{1287 d}+\frac{10 a A \cos ^{\frac{9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{143 d}+\frac{2 A \cos ^{\frac{11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{13 d}+\frac{\left (8 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \sec (c+d x)} \left (\frac{1}{8} a^3 (2224 A+2717 C)+\frac{15}{8} a^3 (112 A+143 C) \sec (c+d x)\right )}{\sec ^{\frac{7}{2}}(c+d x)} \, dx}{1287 a}\\ &=\frac{2 a^3 (2224 A+2717 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{9009 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (136 A+143 C) \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{1287 d}+\frac{10 a A \cos ^{\frac{9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{143 d}+\frac{2 A \cos ^{\frac{11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{13 d}+\frac{\left (a^2 (8368 A+10439 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}{3003}\\ &=\frac{2 a^3 (8368 A+10439 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15015 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^3 (2224 A+2717 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{9009 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (136 A+143 C) \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{1287 d}+\frac{10 a A \cos ^{\frac{9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{143 d}+\frac{2 A \cos ^{\frac{11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{13 d}+\frac{\left (4 a^2 (8368 A+10439 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}{15015}\\ &=\frac{8 a^3 (8368 A+10439 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{45045 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^3 (8368 A+10439 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15015 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^3 (2224 A+2717 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{9009 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (136 A+143 C) \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{1287 d}+\frac{10 a A \cos ^{\frac{9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{143 d}+\frac{2 A \cos ^{\frac{11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{13 d}+\frac{\left (8 a^2 (8368 A+10439 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}{45045}\\ &=\frac{16 a^3 (8368 A+10439 C) \sin (c+d x)}{45045 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{8 a^3 (8368 A+10439 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{45045 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^3 (8368 A+10439 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15015 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^3 (2224 A+2717 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{9009 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (136 A+143 C) \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{1287 d}+\frac{10 a A \cos ^{\frac{9}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{143 d}+\frac{2 A \cos ^{\frac{11}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{13 d}\\ \end{align*}

Mathematica [A]  time = 3.51184, size = 148, normalized size = 0.47 \[ \frac{a^2 \sqrt{\cos (c+d x)} \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)} (8 (226573 A+222794 C) \cos (c+d x)+(746519 A+581152 C) \cos (2 (c+d x))+287060 A \cos (3 (c+d x))+94010 A \cos (4 (c+d x))+23940 A \cos (5 (c+d x))+3465 A \cos (6 (c+d x))+2798182 A+148720 C \cos (3 (c+d x))+20020 C \cos (4 (c+d x))+3233516 C)}{720720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^(13/2)*(a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^2*Sqrt[Cos[c + d*x]]*(2798182*A + 3233516*C + 8*(226573*A + 222794*C)*Cos[c + d*x] + (746519*A + 581152*C)*
Cos[2*(c + d*x)] + 287060*A*Cos[3*(c + d*x)] + 148720*C*Cos[3*(c + d*x)] + 94010*A*Cos[4*(c + d*x)] + 20020*C*
Cos[4*(c + d*x)] + 23940*A*Cos[5*(c + d*x)] + 3465*A*Cos[6*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x
)/2])/(720720*d)

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Maple [A]  time = 0.368, size = 166, normalized size = 0.5 \begin{align*} -{\frac{2\,{a}^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 3465\,A \left ( \cos \left ( dx+c \right ) \right ) ^{6}+11970\,A \left ( \cos \left ( dx+c \right ) \right ) ^{5}+18305\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+5005\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}+20920\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+18590\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+25104\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+31317\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+33472\,A\cos \left ( dx+c \right ) +41756\,C\cos \left ( dx+c \right ) +66944\,A+83512\,C \right ) }{45045\,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)^(13/2)*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x)

[Out]

-2/45045/d*a^2*(-1+cos(d*x+c))*(3465*A*cos(d*x+c)^6+11970*A*cos(d*x+c)^5+18305*A*cos(d*x+c)^4+5005*C*cos(d*x+c
)^4+20920*A*cos(d*x+c)^3+18590*C*cos(d*x+c)^3+25104*A*cos(d*x+c)^2+31317*C*cos(d*x+c)^2+33472*A*cos(d*x+c)+417
56*C*cos(d*x+c)+66944*A+83512*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 = 2.2674, size = 1150, normalized size = 3.67 \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)^(13/2)*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/2882880*(sqrt(2)*(3783780*a^2*cos(12/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c)))*sin(13/2*d*
x + 13/2*c) + 1066065*a^2*cos(10/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c)))*sin(13/2*d*x + 13
/2*c) + 459459*a^2*cos(8/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c)))*sin(13/2*d*x + 13/2*c) +
193050*a^2*cos(6/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c)))*sin(13/2*d*x + 13/2*c) + 70070*a^
2*cos(4/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c)))*sin(13/2*d*x + 13/2*c) + 20475*a^2*cos(2/1
3*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c)))*sin(13/2*d*x + 13/2*c) - 3783780*a^2*cos(13/2*d*x +
13/2*c)*sin(12/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) - 1066065*a^2*cos(13/2*d*x + 13/2*
c)*sin(10/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) - 459459*a^2*cos(13/2*d*x + 13/2*c)*sin(
8/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) - 193050*a^2*cos(13/2*d*x + 13/2*c)*sin(6/13*arc
tan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) - 70070*a^2*cos(13/2*d*x + 13/2*c)*sin(4/13*arctan2(sin(
13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) - 20475*a^2*cos(13/2*d*x + 13/2*c)*sin(2/13*arctan2(sin(13/2*d*x
+ 13/2*c), cos(13/2*d*x + 13/2*c))) + 6930*a^2*sin(13/2*d*x + 13/2*c) + 20475*a^2*sin(11/13*arctan2(sin(13/2*d
*x + 13/2*c), cos(13/2*d*x + 13/2*c))) + 70070*a^2*sin(9/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/
2*c))) + 193050*a^2*sin(7/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) + 459459*a^2*sin(5/13*ar
ctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) + 1066065*a^2*sin(3/13*arctan2(sin(13/2*d*x + 13/2*c),
cos(13/2*d*x + 13/2*c))) + 3783780*a^2*sin(1/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))))*A*sq
rt(a) + 1144*sqrt(2)*(225*a^2*sin(7/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 378*a^2*sin(5/4*arctan2(s
in(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2100*a^2*sin(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4095*a^2
*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 63*(65*a^2*sin(4*d*x + 4*c) + 6*a^2*sin(2*d*x + 2*c))*
cos(9/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 7*(585*a^2*cos(4*d*x + 4*c) + 54*a^2*cos(2*d*x + 2*c) +
5*a^2)*sin(9/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*C*sqrt(a))/d

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Fricas [A]  time = 0.510444, size = 473, normalized size = 1.51 \begin{align*} \frac{2 \,{\left (3465 \, A a^{2} \cos \left (d x + c\right )^{6} + 11970 \, A a^{2} \cos \left (d x + c\right )^{5} + 35 \,{\left (523 \, A + 143 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 10 \,{\left (2092 \, A + 1859 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \,{\left (8368 \, A + 10439 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 4 \,{\left (8368 \, A + 10439 \, C\right )} a^{2} \cos \left (d x + c\right ) + 8 \,{\left (8368 \, A + 10439 \, C\right )} a^{2}\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 )}{45045 \,{\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)^(13/2)*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

2/45045*(3465*A*a^2*cos(d*x + c)^6 + 11970*A*a^2*cos(d*x + c)^5 + 35*(523*A + 143*C)*a^2*cos(d*x + c)^4 + 10*(
2092*A + 1859*C)*a^2*cos(d*x + c)^3 + 3*(8368*A + 10439*C)*a^2*cos(d*x + c)^2 + 4*(8368*A + 10439*C)*a^2*cos(d
*x + c) + 8*(8368*A + 10439*C)*a^2)*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)**(13/2)*(a+a*sec(d*x+c))**(5/2)*(A+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} + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )^{\frac{13}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(13/2)*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(a*sec(d*x + c) + a)^(5/2)*cos(d*x + c)^(13/2), x)