3.7.3 \(\int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {11}{2}}(c+d x)} \, dx\) [603]

Optimal. Leaf size=284 \[ \frac {2 a^3 (1160 A+1364 B+1485 C) \sin (c+d x)}{3465 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (2840 A+3212 B+3795 C) \sin (c+d x)}{3465 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {4 a^3 (2840 A+3212 B+3795 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3465 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (32 A+44 B+33 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a (5 A+11 B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)} \]

[Out]

2/99*a*(5*A+11*B)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(7/2)+2/11*A*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c
)/d/sec(d*x+c)^(9/2)+2/3465*a^3*(1160*A+1364*B+1485*C)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+2/
3465*a^3*(2840*A+3212*B+3795*C)*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+4/3465*a^3*(2840*A+3212*B
+3795*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/231*a^2*(32*A+44*B+33*C)*sin(d*x+c)*(a+a*sec(d
*x+c))^(1/2)/d/sec(d*x+c)^(5/2)

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Rubi [A]
time = 0.59, antiderivative size = 284, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4171, 4102, 4100, 3890, 3889} \begin {gather*} \frac {2 a^3 (1160 A+1364 B+1485 C) \sin (c+d x)}{3465 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {4 a^3 (2840 A+3212 B+3795 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3465 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^3 (2840 A+3212 B+3795 C) \sin (c+d x)}{3465 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (32 A+44 B+33 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a (5 A+11 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(11/2),x]

[Out]

(2*a^3*(1160*A + 1364*B + 1485*C)*Sin[c + d*x])/(3465*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*
(2840*A + 3212*B + 3795*C)*Sin[c + d*x])/(3465*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (4*a^3*(2840*A
+ 3212*B + 3795*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3465*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(32*A + 44*B +
33*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(231*d*Sec[c + d*x]^(5/2)) + (2*a*(5*A + 11*B)*(a + a*Sec[c + d*
x])^(3/2)*Sin[c + d*x])/(99*d*Sec[c + d*x]^(7/2)) + (2*A*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(11*d*Sec[c
+ d*x]^(9/2))

Rule 3889

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]

Rule 3890

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 4100

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 4102

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 4171

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])

Rubi steps

\begin {align*} \int \frac {(a+a \sec (c+d x))^{5/2} \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 (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 \int \frac {(a+a \sec (c+d x))^{5/2} \left (\frac {1}{2} a (5 A+11 B)+\frac {1}{2} a (4 A+11 C) \sec (c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx}{11 a}\\ &=\frac {2 a (5 A+11 B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {4 \int \frac {(a+a \sec (c+d x))^{3/2} \left (\frac {3}{4} a^2 (32 A+44 B+33 C)+\frac {1}{4} a^2 (56 A+44 B+99 C) \sec (c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx}{99 a}\\ &=\frac {2 a^2 (32 A+44 B+33 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a (5 A+11 B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {8 \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {1}{8} a^3 (1160 A+1364 B+1485 C)+\frac {1}{8} a^3 (776 A+836 B+1089 C) \sec (c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{693 a}\\ &=\frac {2 a^3 (1160 A+1364 B+1485 C) \sin (c+d x)}{3465 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (32 A+44 B+33 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a (5 A+11 B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {\left (a^2 (2840 A+3212 B+3795 C)\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{1155}\\ &=\frac {2 a^3 (1160 A+1364 B+1485 C) \sin (c+d x)}{3465 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (2840 A+3212 B+3795 C) \sin (c+d x)}{3465 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (32 A+44 B+33 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a (5 A+11 B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {\left (2 a^2 (2840 A+3212 B+3795 C)\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx}{3465}\\ &=\frac {2 a^3 (1160 A+1364 B+1485 C) \sin (c+d x)}{3465 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (2840 A+3212 B+3795 C) \sin (c+d x)}{3465 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {4 a^3 (2840 A+3212 B+3795 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3465 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (32 A+44 B+33 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a (5 A+11 B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]
time = 1.70, size = 157, normalized size = 0.55 \begin {gather*} \frac {a^2 (114640 A+124366 B+137280 C+(69890 A+68552 B+66660 C) \cos (c+d x)+16 (1625 A+1397 B+990 C) \cos (2 (c+d x))+8675 A \cos (3 (c+d x))+5720 B \cos (3 (c+d x))+1980 C \cos (3 (c+d x))+2240 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 \left (\frac {1}{2} (c+d x)\right )}{27720 d \sqrt {\sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(11/2),x]

[Out]

(a^2*(114640*A + 124366*B + 137280*C + (69890*A + 68552*B + 66660*C)*Cos[c + d*x] + 16*(1625*A + 1397*B + 990*
C)*Cos[2*(c + d*x)] + 8675*A*Cos[3*(c + d*x)] + 5720*B*Cos[3*(c + d*x)] + 1980*C*Cos[3*(c + d*x)] + 2240*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*Sqrt[Sec[c + d*x]])

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Maple [A]
time = 0.24, size = 199, normalized size = 0.70

method	result	size
default	\(-\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (315 A \left (\cos ^{5}\left (d x +c \right )\right )+1120 A \left (\cos ^{4}\left (d x +c \right )\right )+385 B \left (\cos ^{4}\left (d x +c \right )\right )+1775 A \left (\cos ^{3}\left (d x +c \right )\right )+1430 B \left (\cos ^{3}\left (d x +c \right )\right )+495 C \left (\cos ^{3}\left (d x +c \right )\right )+2130 A \left (\cos ^{2}\left (d x +c \right )\right )+2409 B \left (\cos ^{2}\left (d x +c \right )\right )+1980 C \left (\cos ^{2}\left (d x +c \right )\right )+2840 A \cos \left (d x +c \right )+3212 B \cos \left (d x +c \right )+3795 C \cos \left (d x +c \right )+5680 A +6424 B +7590 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\cos ^{6}\left (d x +c \right )\right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {11}{2}} a^{2}}{3465 d \sin \left (d x +c \right )}\)	\(199\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x,method=_RETURNVERBOSE)

[Out]

-2/3465/d*(-1+cos(d*x+c))*(315*A*cos(d*x+c)^5+1120*A*cos(d*x+c)^4+385*B*cos(d*x+c)^4+1775*A*cos(d*x+c)^3+1430*
B*cos(d*x+c)^3+495*C*cos(d*x+c)^3+2130*A*cos(d*x+c)^2+2409*B*cos(d*x+c)^2+1980*C*cos(d*x+c)^2+2840*A*cos(d*x+c
)+3212*B*cos(d*x+c)+3795*C*cos(d*x+c)+5680*A+6424*B+7590*C)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^6*(
1/cos(d*x+c))^(11/2)/sin(d*x+c)*a^2

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 1266 vs. \(2 (248) = 496\).
time = 0.75, size = 1266, normalized size = 4.46 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

1/110880*(5*sqrt(2)*(31878*a^2*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) + 8778*a^2*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)
+ 3465*a^2*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) + 1287*a^
2*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) + 385*a^2*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) - 31878*a^2*cos(11/2*d*x + 11/
2*c)*sin(10/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 8778*a^2*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))) - 3465*a^2*cos(11/2*d*x + 11/2*c)*sin(6/11*arcta
n2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 1287*a^2*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))) - 385*a^2*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))) + 126*a^2*sin(11/2*d*x + 11/2*c) + 385*a^2*sin(9/11*arctan2(sin(11/2*d*x + 11/2
*c), cos(11/2*d*x + 11/2*c))) + 1287*a^2*sin(7/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 3
465*a^2*sin(5/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 8778*a^2*sin(3/11*arctan2(sin(11/2
*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 31878*a^2*sin(1/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 1
1/2*c))))*A*sqrt(a) + 22*sqrt(2)*(8190*a^2*cos(8/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/
2*d*x + 9/2*c) + 2100*a^2*cos(2/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) +
756*a^2*cos(4/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) + 225*a^2*cos(2/9*ar
ctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) - 8190*a^2*cos(9/2*d*x + 9/2*c)*sin(8/
9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) - 2100*a^2*cos(9/2*d*x + 9/2*c)*sin(2/3*arctan2(sin(9/2
*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) - 756*a^2*cos(9/2*d*x + 9/2*c)*sin(4/9*arctan2(sin(9/2*d*x + 9/2*c), cos
(9/2*d*x + 9/2*c))) - 225*a^2*cos(9/2*d*x + 9/2*c)*sin(2/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))
) + 70*a^2*sin(9/2*d*x + 9/2*c) + 225*a^2*sin(7/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 756*a
^2*sin(5/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 2100*a^2*sin(1/3*arctan2(sin(9/2*d*x + 9/2*c
), cos(9/2*d*x + 9/2*c))) + 8190*a^2*sin(1/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))))*B*sqrt(a) +
660*sqrt(2)*(315*a^2*cos(6/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 77*a
^2*cos(4/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 21*a^2*cos(2/7*arctan2(
sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) - 315*a^2*cos(7/2*d*x + 7/2*c)*sin(6/7*arcta
n2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 77*a^2*cos(7/2*d*x + 7/2*c)*sin(4/7*arctan2(sin(7/2*d*x + 7/
2*c), cos(7/2*d*x + 7/2*c))) - 21*a^2*cos(7/2*d*x + 7/2*c)*sin(2/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x +
7/2*c))) + 6*a^2*sin(7/2*d*x + 7/2*c) + 21*a^2*sin(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) +
77*a^2*sin(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 315*a^2*sin(1/7*arctan2(sin(7/2*d*x + 7
/2*c), cos(7/2*d*x + 7/2*c))))*C*sqrt(a))/d

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Fricas [A]
time = 4.21, size = 173, normalized size = 0.61 \begin {gather*} \frac {2 \, {\left (315 \, A a^{2} \cos \left (d x + c\right )^{6} + 35 \, {\left (32 \, A + 11 \, B\right )} a^{2} \cos \left (d x + c\right )^{5} + 5 \, {\left (355 \, A + 286 \, B + 99 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 3 \, {\left (710 \, A + 803 \, B + 660 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (2840 \, A + 3212 \, B + 3795 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (2840 \, A + 3212 \, B + 3795 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

2/3465*(315*A*a^2*cos(d*x + c)^6 + 35*(32*A + 11*B)*a^2*cos(d*x + c)^5 + 5*(355*A + 286*B + 99*C)*a^2*cos(d*x
+ c)^4 + 3*(710*A + 803*B + 660*C)*a^2*cos(d*x + c)^3 + (2840*A + 3212*B + 3795*C)*a^2*cos(d*x + c)^2 + 2*(284
0*A + 3212*B + 3795*C)*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/((d*cos(d*x + c)
+ d)*sqrt(cos(d*x + c)))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(11/2),x)

[Out]

Timed out

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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((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^(5/2)/sec(d*x + c)^(11/2), x)

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Mupad [B]
time = 11.31, size = 404, normalized size = 1.42 \begin {gather*} \frac {\sqrt {a-\frac {a}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}}\,\left (2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}-1\right )\,\left (\frac {A\,a^2\,\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{88\,d}+\frac {a^2\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )\,\left (13\,A+10\,B+4\,C\right )}{56\,d}+\frac {a^2\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )\,\left (19\,A+20\,B+22\,C\right )}{12\,d}+\frac {a^2\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )\,\left (25\,A+24\,B+20\,C\right )}{40\,d}+\frac {a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )\,\left (23\,A+26\,B+30\,C\right )}{4\,d}+\frac {a^2\,\sin \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )\,\left (5\,A+2\,B\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{72\,d}\right )}{2\,\sqrt {-\frac {1}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}}\,\left (2\,{\sin \left (\frac {c}{4}+\frac {d\,x}{4}\right )}^2-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(11/2),x)

[Out]

((a - a/(2*sin(c/2 + (d*x)/2)^2 - 1))^(1/2)*(sin((11*c)/2 + (11*d*x)/2)*1i + 2*sin((11*c)/4 + (11*d*x)/4)^2 -
1)*((A*a^2*sin((11*c)/2 + (11*d*x)/2)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11*c)/4 + (11*d*x)/4)^2 + 1))/(8
8*d) + (a^2*sin((7*c)/2 + (7*d*x)/2)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11*c)/4 + (11*d*x)/4)^2 + 1)*(13*
A + 10*B + 4*C))/(56*d) + (a^2*sin((3*c)/2 + (3*d*x)/2)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11*c)/4 + (11*
d*x)/4)^2 + 1)*(19*A + 20*B + 22*C))/(12*d) + (a^2*sin((5*c)/2 + (5*d*x)/2)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2
*sin((11*c)/4 + (11*d*x)/4)^2 + 1)*(25*A + 24*B + 20*C))/(40*d) + (a^2*sin(c/2 + (d*x)/2)*(sin((11*c)/2 + (11*
d*x)/2)*1i - 2*sin((11*c)/4 + (11*d*x)/4)^2 + 1)*(23*A + 26*B + 30*C))/(4*d) + (a^2*sin((9*c)/2 + (9*d*x)/2)*(
5*A + 2*B)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11*c)/4 + (11*d*x)/4)^2 + 1))/(72*d)))/(2*(-1/(2*sin(c/2 +
(d*x)/2)^2 - 1))^(1/2)*(2*sin(c/4 + (d*x)/4)^2 - 1))

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