3.1135 \(\int \cos ^{\frac{11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=266 \[ \frac{2 a^2 (28 A+33 C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{231 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (112 A+143 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{385 d \sqrt{a \sec (c+d x)+a}}+\frac{8 a^2 (112 A+143 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{1155 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a^2 (112 A+143 C) \sin (c+d x)}{1155 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 A \sin (c+d x) \cos ^{\frac{9}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}+\frac{2 a A \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}{33 d} \]

[Out]

(16*a^2*(112*A + 143*C)*Sin[c + d*x])/(1155*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (8*a^2*(112*A + 1
43*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(1155*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(112*A + 143*C)*Cos[c + d*x]
^(3/2)*Sin[c + d*x])/(385*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(28*A + 33*C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/
(231*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*A*Cos[c + d*x]^(7/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(33*d) + (
2*A*Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(11*d)

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Rubi [A]  time = 0.806646, antiderivative size = 266, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{2 a^2 (28 A+33 C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{231 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (112 A+143 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{385 d \sqrt{a \sec (c+d x)+a}}+\frac{8 a^2 (112 A+143 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{1155 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a^2 (112 A+143 C) \sin (c+d x)}{1155 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 A \sin (c+d x) \cos ^{\frac{9}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d}+\frac{2 a A \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}{33 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*a^2*(112*A + 143*C)*Sin[c + d*x])/(1155*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (8*a^2*(112*A + 1
43*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(1155*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(112*A + 143*C)*Cos[c + d*x]
^(3/2)*Sin[c + d*x])/(385*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(28*A + 33*C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/
(231*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*A*Cos[c + d*x]^(7/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(33*d) + (
2*A*Cos[c + d*x]^(9/2)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(11*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{11}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \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))^{3/2} \left (A+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{3 a A}{2}+\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 A \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{33 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{3}{4} a^2 (28 A+33 C)+\frac{9}{4} a^2 (8 A+11 C) \sec (c+d x)\right )}{\sec ^{\frac{7}{2}}(c+d x)} \, dx}{99 a}\\ &=\frac{2 a^2 (28 A+33 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{231 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a A \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{33 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}{77} \left (a (112 A+143 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 (112 A+143 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{385 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (28 A+33 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{231 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a A \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{33 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}{385} \left (4 a (112 A+143 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\\ &=\frac{8 a^2 (112 A+143 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{1155 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (112 A+143 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{385 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (28 A+33 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{231 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a A \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{33 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 (112 A+143 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}{1155}\\ &=\frac{16 a^2 (112 A+143 C) \sin (c+d x)}{1155 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{8 a^2 (112 A+143 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{1155 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (112 A+143 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{385 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (28 A+33 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{231 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a A \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{33 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 = 1.99648, size = 125, normalized size = 0.47 \[ \frac{a \sqrt{\cos (c+d x)} \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)} (2 (5789 A+5566 C) \cos (c+d x)+8 (581 A+429 C) \cos (2 (c+d x))+1645 A \cos (3 (c+d x))+490 A \cos (4 (c+d x))+105 A \cos (5 (c+d x))+18494 A+660 C \cos (3 (c+d x))+21736 C)}{9240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[Cos[c + d*x]]*(18494*A + 21736*C + 2*(5789*A + 5566*C)*Cos[c + d*x] + 8*(581*A + 429*C)*Cos[2*(c + d*x
)] + 1645*A*Cos[3*(c + d*x)] + 660*C*Cos[3*(c + d*x)] + 490*A*Cos[4*(c + d*x)] + 105*A*Cos[5*(c + d*x)])*Sqrt[
a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(9240*d)

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Maple [A]  time = 0.351, size = 142, normalized size = 0.5 \begin{align*} -{\frac{2\,a \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 105\,A \left ( \cos \left ( dx+c \right ) \right ) ^{5}+245\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+280\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+165\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+336\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+429\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+448\,A\cos \left ( dx+c \right ) +572\,C\cos \left ( dx+c \right ) +896\,A+1144\,C \right ) }{1155\,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+C*sec(d*x+c)^2),x)

[Out]

-2/1155/d*a*(-1+cos(d*x+c))*(105*A*cos(d*x+c)^5+245*A*cos(d*x+c)^4+280*A*cos(d*x+c)^3+165*C*cos(d*x+c)^3+336*A
*cos(d*x+c)^2+429*C*cos(d*x+c)^2+448*A*cos(d*x+c)+572*C*cos(d*x+c)+896*A+1144*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 = 2.24813, size = 880, normalized size = 3.31 \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+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/36960*(7*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 + 1
1/2*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) + 429*
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*ar
ctan2(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*arct
an2(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 + 11/
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))) + 363
0*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)*(175*a*cos(7/4*a
rctan2(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 = 0.499507, size = 378, normalized size = 1.42 \begin{align*} \frac{2 \,{\left (105 \, A a \cos \left (d x + c\right )^{5} + 245 \, A a \cos \left (d x + c\right )^{4} + 5 \,{\left (56 \, A + 33 \, C\right )} a \cos \left (d x + c\right )^{3} + 3 \,{\left (112 \, A + 143 \, C\right )} a \cos \left (d x + c\right )^{2} + 4 \,{\left (112 \, A + 143 \, C\right )} a \cos \left (d x + c\right ) + 8 \,{\left (112 \, A + 143 \, 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 )}{1155 \,{\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+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

2/1155*(105*A*a*cos(d*x + c)^5 + 245*A*a*cos(d*x + c)^4 + 5*(56*A + 33*C)*a*cos(d*x + c)^3 + 3*(112*A + 143*C)
*a*cos(d*x + c)^2 + 4*(112*A + 143*C)*a*cos(d*x + c) + 8*(112*A + 143*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+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{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+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

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