3.274 \(\int \frac{(a+a \sec (c+d x))^{5/2} (A+C \sec ^2(c+d x))}{\sec ^{\frac{9}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=216 \[ \frac{64 a^3 (13 A+21 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{315 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a^2 (13 A+21 C) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{315 d \sqrt{\sec (c+d x)}}+\frac{2 a (13 A+21 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{10 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{9 d \sec ^{\frac{7}{2}}(c+d x)} \]

[Out]

(64*a^3*(13*A + 21*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(13*A + 21*C
)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(315*d*Sqrt[Sec[c + d*x]]) + (2*a*(13*A + 21*C)*(a + a*Sec[c + d*x])^
(3/2)*Sin[c + d*x])/(105*d*Sec[c + d*x]^(3/2)) + (2*A*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(9*d*Sec[c + d*
x]^(7/2)) + (10*A*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(63*d*Sec[c + d*x]^(5/2))

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Rubi [A]  time = 0.500494, antiderivative size = 216, 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, 4013, 3809, 3804} \[ \frac{64 a^3 (13 A+21 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{315 d \sqrt{a \sec (c+d x)+a}}+\frac{16 a^2 (13 A+21 C) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{315 d \sqrt{\sec (c+d x)}}+\frac{2 a (13 A+21 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{10 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{9 d \sec ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(9/2),x]

[Out]

(64*a^3*(13*A + 21*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(13*A + 21*C
)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(315*d*Sqrt[Sec[c + d*x]]) + (2*a*(13*A + 21*C)*(a + a*Sec[c + d*x])^
(3/2)*Sin[c + d*x])/(105*d*Sec[c + d*x]^(3/2)) + (2*A*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(9*d*Sec[c + d*
x]^(7/2)) + (10*A*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(63*d*Sec[c + d*x]^(5/2))

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 4013

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), 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[(
a*A*m - b*B*n)/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A
, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -1]

Rule 3809

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(a*Co
t[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*m), x] + Dist[(b*(2*m - 1))/(d*m), Int[(a + b*C
sc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& EqQ[m + n, 0] && GtQ[m, 1/2] && IntegerQ[2*m]

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{(a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac{9}{2}}(c+d x)} \, dx &=\frac{2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 \int \frac{(a+a \sec (c+d x))^{5/2} \left (\frac{5 a A}{2}+\frac{1}{2} a (2 A+9 C) \sec (c+d x)\right )}{\sec ^{\frac{7}{2}}(c+d x)} \, dx}{9 a}\\ &=\frac{2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{10 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{1}{21} (13 A+21 C) \int \frac{(a+a \sec (c+d x))^{5/2}}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a (13 A+21 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{105 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{10 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{1}{105} (8 a (13 A+21 C)) \int \frac{(a+a \sec (c+d x))^{3/2}}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{16 a^2 (13 A+21 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{315 d \sqrt{\sec (c+d x)}}+\frac{2 a (13 A+21 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{105 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{10 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{1}{315} \left (32 a^2 (13 A+21 C)\right ) \int \frac{\sqrt{a+a \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{64 a^3 (13 A+21 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{315 d \sqrt{a+a \sec (c+d x)}}+\frac{16 a^2 (13 A+21 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{315 d \sqrt{\sec (c+d x)}}+\frac{2 a (13 A+21 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{105 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{10 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 1.56315, size = 105, normalized size = 0.49 \[ \frac{a^2 \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)} (4 (779 A+588 C) \cos (c+d x)+4 (254 A+63 C) \cos (2 (c+d x))+260 A \cos (3 (c+d x))+35 A \cos (4 (c+d x))+5653 A+7476 C)}{1260 d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(9/2),x]

[Out]

(a^2*(5653*A + 7476*C + 4*(779*A + 588*C)*Cos[c + d*x] + 4*(254*A + 63*C)*Cos[2*(c + d*x)] + 260*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 = 0.436, size = 132, normalized size = 0.6 \begin{align*} -{\frac{2\,{a}^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 35\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+130\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+219\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+63\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+292\,A\cos \left ( dx+c \right ) +294\,C\cos \left ( dx+c \right ) +584\,A+903\,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+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(9/2),x)

[Out]

-2/315/d*a^2*(-1+cos(d*x+c))*(35*A*cos(d*x+c)^4+130*A*cos(d*x+c)^3+219*A*cos(d*x+c)^2+63*C*cos(d*x+c)^2+292*A*
cos(d*x+c)+294*C*cos(d*x+c)+584*A+903*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 = 2.03526, size = 652, normalized size = 3.02 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

1/5040*(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*a
rctan2(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*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) - 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), co
s(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*arctan
2(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))))*A*sqrt(a) + 168*(3*sqrt(2)*a
^2*sin(5/2*d*x + 5/2*c) + 25*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 150*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c))*C*sqrt(a
))/d

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Fricas [A]  time = 0.487792, size = 359, normalized size = 1.66 \begin{align*} \frac{2 \,{\left (35 \, A a^{2} \cos \left (d x + c\right )^{5} + 130 \, A a^{2} \cos \left (d x + c\right )^{4} + 3 \,{\left (73 \, A + 21 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \,{\left (146 \, A + 147 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} +{\left (584 \, A + 903 \, 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 )}{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+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

2/315*(35*A*a^2*cos(d*x + c)^5 + 130*A*a^2*cos(d*x + c)^4 + 3*(73*A + 21*C)*a^2*cos(d*x + c)^3 + 2*(146*A + 14
7*C)*a^2*cos(d*x + c)^2 + (584*A + 903*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)]  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+a*sec(d*x+c))**(5/2)*(A+C*sec(d*x+c)**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 )}{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}{\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+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(9/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(a*sec(d*x + c) + a)^(5/2)/sec(d*x + c)^(9/2), x)