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

Optimal. Leaf size=213 \[ \frac{2 a (16 A+21 C) \sin (c+d x)}{105 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}+\frac{16 a (16 A+21 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{315 d \sqrt{a \sec (c+d x)+a}}+\frac{8 a (16 A+21 C) \sin (c+d x)}{315 d \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 A \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 a A \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}} \]

[Out]

(2*a*A*Sin[c + d*x])/(63*d*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(16*A + 21*C)*Sin[c + d*x])/(10
5*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (8*a*(16*A + 21*C)*Sin[c + d*x])/(315*d*Sqrt[Sec[c + d*x]]*
Sqrt[a + a*Sec[c + d*x]]) + (16*a*(16*A + 21*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x
]]) + (2*A*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2))

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Rubi [A]  time = 0.458937, antiderivative size = 213, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.108, Rules used = {4087, 4015, 3805, 3804} \[ \frac{2 a (16 A+21 C) \sin (c+d x)}{105 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}+\frac{16 a (16 A+21 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{315 d \sqrt{a \sec (c+d x)+a}}+\frac{8 a (16 A+21 C) \sin (c+d x)}{315 d \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 A \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 a A \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*A*Sin[c + d*x])/(63*d*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(16*A + 21*C)*Sin[c + d*x])/(10
5*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (8*a*(16*A + 21*C)*Sin[c + d*x])/(315*d*Sqrt[Sec[c + d*x]]*
Sqrt[a + a*Sec[c + d*x]]) + (16*a*(16*A + 21*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x
]]) + (2*A*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2))

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

Mathematica [A]  time = 1.18383, size = 102, normalized size = 0.48 \[ \frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)} (16 (47 A+42 C) \cos (c+d x)+4 (83 A+63 C) \cos (2 (c+d x))+80 A \cos (3 (c+d x))+35 A \cos (4 (c+d x))+1321 A+1596 C)}{1260 d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1321*A + 1596*C + 16*(47*A + 42*C)*Cos[c + d*x] + 4*(83*A + 63*C)*Cos[2*(c + d*x)] + 80*A*Cos[3*(c + d*x)] +
35*A*Cos[4*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(1260*d*Sqrt[Sec[c + d*x]])

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Maple [A]  time = 0.394, size = 129, normalized size = 0.6 \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}+48\,A \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 ) +84\,C\cos \left ( dx+c \right ) +128\,A+168\,C \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{315\,d\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315/d*(-1+cos(d*x+c))*(35*A*cos(d*x+c)^4+40*A*cos(d*x+c)^3+48*A*cos(d*x+c)^2+63*C*cos(d*x+c)^2+64*A*cos(d*x
+c)+84*C*cos(d*x+c)+128*A+168*C)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*cos(d*x+c)^5*(1/cos(d*x+c))^(9/2)/sin(d*x
+c)

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Maxima [B]  time = 2.08835, size = 792, normalized size = 3.72 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(9/2),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) + 84*sqrt(2)*(30*cos(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))
)*sin(5/2*d*x + 5/2*c) + 5*cos(2/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) -
30*cos(5/2*d*x + 5/2*c)*sin(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) - 5*cos(5/2*d*x + 5/2*c)
*sin(2/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 6*sin(5/2*d*x + 5/2*c) + 5*sin(3/5*arctan2(sin
(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 30*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))*C
*sqrt(a))/d

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Fricas [A]  time = 0.491079, size = 328, normalized size = 1.54 \begin{align*} \frac{2 \,{\left (35 \, A \cos \left (d x + c\right )^{5} + 40 \, A \cos \left (d x + c\right )^{4} + 3 \,{\left (16 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \,{\left (16 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \,{\left (16 \, A + 21 \, C\right )} \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 )}{315 \,{\left (d \cos \left (d x + c\right ) + d\right )} \sqrt{\cos \left (d x + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

2/315*(35*A*cos(d*x + c)^5 + 40*A*cos(d*x + c)^4 + 3*(16*A + 21*C)*cos(d*x + c)^3 + 4*(16*A + 21*C)*cos(d*x +
c)^2 + 8*(16*A + 21*C)*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)]  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((A+C*sec(d*x+c)**2)*(a+a*sec(d*x+c))**(1/2)/sec(d*x+c)**(9/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt{a \sec \left (d x + c\right ) + a}}{\sec \left (d x + c\right )^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*sec(d*x + c)^2 + A)*sqrt(a*sec(d*x + c) + a)/sec(d*x + c)^(9/2), x)