3.1245 \(\int \cos ^{\frac{9}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=226 \[ \frac{2 a (16 A+18 B+21 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{8 a (16 A+18 B+21 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{315 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a (16 A+18 B+21 C) \sin (c+d x)}{315 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 a (A+9 B) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{63 d \sqrt{a \sec (c+d x)+a}}+\frac{2 A \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}{9 d} \]

[Out]

(16*a*(16*A + 18*B + 21*C)*Sin[c + d*x])/(315*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (8*a*(16*A + 18
*B + 21*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(16*A + 18*B + 21*C)*Cos[c
+ d*x]^(3/2)*Sin[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(A + 9*B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])
/(63*d*Sqrt[a + a*Sec[c + d*x]]) + (2*A*Cos[c + d*x]^(7/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(9*d)

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Rubi [A]  time = 0.628749, antiderivative size = 226, normalized size of antiderivative = 1., 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 = {4265, 4086, 4015, 3805, 3804} \[ \frac{2 a (16 A+18 B+21 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{8 a (16 A+18 B+21 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{315 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a (16 A+18 B+21 C) \sin (c+d x)}{315 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 a (A+9 B) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{63 d \sqrt{a \sec (c+d x)+a}}+\frac{2 A \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(9/2)*Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(16*a*(16*A + 18*B + 21*C)*Sin[c + d*x])/(315*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (8*a*(16*A + 18
*B + 21*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(16*A + 18*B + 21*C)*Cos[c
+ d*x]^(3/2)*Sin[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(A + 9*B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])
/(63*d*Sqrt[a + a*Sec[c + d*x]]) + (2*A*Cos[c + d*x]^(7/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(9*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 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{9}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \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{\sqrt{a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac{9}{2}}(c+d x)} \, dx\\ &=\frac{2 A \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{9 d}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \sec (c+d x)} \left (\frac{1}{2} a (A+9 B)+\frac{3}{2} a (2 A+3 C) \sec (c+d x)\right )}{\sec ^{\frac{7}{2}}(c+d x)} \, dx}{9 a}\\ &=\frac{2 a (A+9 B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 A \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{9 d}+\frac{1}{21} \left ((16 A+18 B+21 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 (16 A+18 B+21 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (A+9 B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 A \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{9 d}+\frac{1}{105} \left (4 (16 A+18 B+21 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 (16 A+18 B+21 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{315 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (16 A+18 B+21 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (A+9 B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 A \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{9 d}+\frac{1}{315} \left (8 (16 A+18 B+21 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\\ &=\frac{16 a (16 A+18 B+21 C) \sin (c+d x)}{315 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{8 a (16 A+18 B+21 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{315 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (16 A+18 B+21 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (A+9 B) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 A \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{9 d}\\ \end{align*}

Mathematica [A]  time = 0.566825, size = 127, normalized size = 0.56 \[ \frac{\sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a (\sec (c+d x)+1)} ((752 A+846 B+672 C) \cos (c+d x)+4 (83 A+54 B+63 C) \cos (2 (c+d x))+80 A \cos (3 (c+d x))+35 A \cos (4 (c+d x))+1321 A+90 B \cos (3 (c+d x))+1368 B+1596 C)}{1260 d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^(9/2)*Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(Sqrt[Cos[c + d*x]]*(1321*A + 1368*B + 1596*C + (752*A + 846*B + 672*C)*Cos[c + d*x] + 4*(83*A + 54*B + 63*C)*
Cos[2*(c + d*x)] + 80*A*Cos[3*(c + d*x)] + 90*B*Cos[3*(c + d*x)] + 35*A*Cos[4*(c + d*x)])*Sqrt[a*(1 + Sec[c +
d*x])]*Sin[c + d*x])/(1260*d*(1 + Cos[c + d*x]))

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Maple [A]  time = 0.401, size = 153, normalized size = 0.7 \begin{align*} -{\frac{ \left ( -2+2\,\cos \left ( dx+c \right ) \right ) \left ( 35\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+40\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+45\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+48\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+54\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+63\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+64\,A\cos \left ( dx+c \right ) +72\,B\cos \left ( dx+c \right ) +84\,C\cos \left ( dx+c \right ) +128\,A+144\,B+168\,C \right ) }{315\,d\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}\sqrt{\cos \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(9/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x)

[Out]

-2/315/d*(-1+cos(d*x+c))*(35*A*cos(d*x+c)^4+40*A*cos(d*x+c)^3+45*B*cos(d*x+c)^3+48*A*cos(d*x+c)^2+54*B*cos(d*x
+c)^2+63*C*cos(d*x+c)^2+64*A*cos(d*x+c)+72*B*cos(d*x+c)+84*C*cos(d*x+c)+128*A+144*B+168*C)*(a*(cos(d*x+c)+1)/c
os(d*x+c))^(1/2)*cos(d*x+c)^(1/2)/sin(d*x+c)

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Maxima [B]  time = 2.42619, size = 902, normalized size = 3.99 \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)^(9/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/5040*(sqrt(2)*(1890*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) + 420*
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) + 252*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) + 45*cos(2/9*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) - 1890*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))) - 420*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))) - 252*
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))) - 45*cos(9/2*d*x + 9/2*c)*si
n(2/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 70*sin(9/2*d*x + 9/2*c) + 45*sin(7/9*arctan2(sin(
9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 252*sin(5/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) +
420*sin(1/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 1890*sin(1/9*arctan2(sin(9/2*d*x + 9/2*c),
cos(9/2*d*x + 9/2*c))))*A*sqrt(a) - 18*sqrt(2)*(7*(15*sin(3*d*x + 3*c) + 5*sin(2*d*x + 2*c) + sin(d*x + c))*co
s(7/2*arctan2(sin(d*x + c), cos(d*x + c))) - (105*cos(3*d*x + 3*c) + 35*cos(2*d*x + 2*c) + 7*cos(d*x + c) + 10
)*sin(7/2*arctan2(sin(d*x + c), cos(d*x + c))) - 7*sin(5/2*arctan2(sin(d*x + c), cos(d*x + c))) - 35*sin(3/2*a
rctan2(sin(d*x + c), cos(d*x + c))) - 105*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))))*B*sqrt(a) - 84*sqrt(2)
*(5*(6*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5/2*arctan2(sin(d*x + c), cos(d*x + c))) - (30*cos(2*d*x + 2*c) +
5*cos(d*x + c) + 6)*sin(5/2*arctan2(sin(d*x + c), cos(d*x + c))) - 5*sin(3/2*arctan2(sin(d*x + c), cos(d*x + c
))) - 30*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))))*C*sqrt(a))/d

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Fricas [A]  time = 0.495573, size = 344, normalized size = 1.52 \begin{align*} \frac{2 \,{\left (35 \, A \cos \left (d x + c\right )^{4} + 5 \,{\left (8 \, A + 9 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (16 \, A + 18 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (16 \, A + 18 \, B + 21 \, C\right )} \cos \left (d x + c\right ) + 128 \, A + 144 \, B + 168 \, C\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 )}{315 \,{\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)^(9/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/315*(35*A*cos(d*x + c)^4 + 5*(8*A + 9*B)*cos(d*x + c)^3 + 3*(16*A + 18*B + 21*C)*cos(d*x + c)^2 + 4*(16*A +
18*B + 21*C)*cos(d*x + c) + 128*A + 144*B + 168*C)*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)**(9/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)*(a+a*sec(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [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)^(9/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Timed out