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

Optimal. Leaf size=231 \[ \frac {64 a^3 (13 A+15 B+21 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{315 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 (13 A+15 B+21 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{315 d \sqrt {\sec (c+d x)}}+\frac {2 a (13 A+15 B+21 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (5 A+9 B) \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)} \]

[Out]

2/105*a*(13*A+15*B+21*C)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/9*A*(a+a*sec(d*x+c))^(5/2)*sin
(d*x+c)/d/sec(d*x+c)^(7/2)+2/63*(5*A+9*B)*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d/sec(d*x+c)^(5/2)+64/315*a^3*(13*
A+15*B+21*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+16/315*a^2*(13*A+15*B+21*C)*sin(d*x+c)*(a+a*
sec(d*x+c))^(1/2)/d/sec(d*x+c)^(1/2)

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Rubi [A]  time = 0.56, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {4086, 4013, 3809, 3804} \[ \frac {64 a^3 (13 A+15 B+21 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{315 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 (13 A+15 B+21 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{315 d \sqrt {\sec (c+d x)}}+\frac {2 a (13 A+15 B+21 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{105 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (5 A+9 B) \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*a^3*(13*A + 15*B + 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
+ 15*B + 21*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(315*d*Sqrt[Sec[c + d*x]]) + (2*a*(13*A + 15*B + 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)) + (2*(5*A + 9*B)*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(63*d*Sec[c + d*x]^(5
/2))

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]

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 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 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])

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 {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 {1}{2} a (5 A+9 B)+\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 {2 (5 A+9 B) (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+15 B+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+15 B+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 {2 (5 A+9 B) (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+15 B+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+15 B+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+15 B+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 {2 (5 A+9 B) (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+15 B+21 C)\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {64 a^3 (13 A+15 B+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+15 B+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+15 B+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 {2 (5 A+9 B) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 1.91, size = 124, normalized size = 0.54 \[ \frac {a^2 \tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sec (c+d x)+1)} ((3116 A+3030 B+2352 C) \cos (c+d x)+4 (254 A+180 B+63 C) \cos (2 (c+d x))+260 A \cos (3 (c+d x))+35 A \cos (4 (c+d x))+5653 A+90 B \cos (3 (c+d x))+6240 B+7476 C)}{1260 d \sqrt {\sec (c+d x)}} \]

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]^(9/2),x]

[Out]

(a^2*(5653*A + 6240*B + 7476*C + (3116*A + 3030*B + 2352*C)*Cos[c + d*x] + 4*(254*A + 180*B + 63*C)*Cos[2*(c +
d*x)] + 260*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])]*Ta
n[(c + d*x)/2])/(1260*d*Sqrt[Sec[c + d*x]])

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fricas [A]  time = 0.44, size = 149, normalized size = 0.65 \[ \frac {2 \, {\left (35 \, A a^{2} \cos \left (d x + c\right )^{5} + 5 \, {\left (26 \, A + 9 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 3 \, {\left (73 \, A + 60 \, B + 21 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (292 \, A + 345 \, B + 294 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (584 \, A + 690 \, B + 903 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \]

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)^(9/2),x, algorithm="fricas")

[Out]

2/315*(35*A*a^2*cos(d*x + c)^5 + 5*(26*A + 9*B)*a^2*cos(d*x + c)^4 + 3*(73*A + 60*B + 21*C)*a^2*cos(d*x + c)^3
+ (292*A + 345*B + 294*C)*a^2*cos(d*x + c)^2 + (584*A + 690*B + 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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\sec \left (d x + c\right )^{\frac {9}{2}}}\,{d x} \]

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)^(9/2),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)^(9/2), x)

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maple [A]  time = 2.58, size = 166, normalized size = 0.72 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (35 A \left (\cos ^{4}\left (d x +c \right )\right )+130 A \left (\cos ^{3}\left (d x +c \right )\right )+45 B \left (\cos ^{3}\left (d x +c \right )\right )+219 A \left (\cos ^{2}\left (d x +c \right )\right )+180 B \left (\cos ^{2}\left (d x +c \right )\right )+63 C \left (\cos ^{2}\left (d x +c \right )\right )+292 A \cos \left (d x +c \right )+345 B \cos \left (d x +c \right )+294 C \cos \left (d x +c \right )+584 A +690 B +903 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\cos ^{5}\left (d x +c \right )\right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {9}{2}} a^{2}}{315 d \sin \left (d x +c \right )} \]

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)^(9/2),x)

[Out]

-2/315/d*(-1+cos(d*x+c))*(35*A*cos(d*x+c)^4+130*A*cos(d*x+c)^3+45*B*cos(d*x+c)^3+219*A*cos(d*x+c)^2+180*B*cos(
d*x+c)^2+63*C*cos(d*x+c)^2+292*A*cos(d*x+c)+345*B*cos(d*x+c)+294*C*cos(d*x+c)+584*A+690*B+903*C)*(a*(1+cos(d*x
+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^5*(1/cos(d*x+c))^(9/2)/sin(d*x+c)*a^2

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maxima [B]  time = 0.80, size = 804, normalized size = 3.48 \[ \text {result too large to display} \]

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)^(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) + 30*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*arctan2(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*ar
ctan2(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))))*B*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*s
qrt(2)*a^2*sin(1/2*d*x + 1/2*c))*C*sqrt(a))/d

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mupad [B]  time = 8.02, size = 190, normalized size = 0.82 \[ \frac {a^2\,\cos \left (c+d\,x\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}\,\left (10290\,A\,\sin \left (c+d\,x\right )+11760\,B\,\sin \left (c+d\,x\right )+14700\,C\,\sin \left (c+d\,x\right )+2856\,A\,\sin \left (2\,c+2\,d\,x\right )+981\,A\,\sin \left (3\,c+3\,d\,x\right )+260\,A\,\sin \left (4\,c+4\,d\,x\right )+35\,A\,\sin \left (5\,c+5\,d\,x\right )+2940\,B\,\sin \left (2\,c+2\,d\,x\right )+720\,B\,\sin \left (3\,c+3\,d\,x\right )+90\,B\,\sin \left (4\,c+4\,d\,x\right )+2352\,C\,\sin \left (2\,c+2\,d\,x\right )+252\,C\,\sin \left (3\,c+3\,d\,x\right )\right )}{2520\,d\,\left (\cos \left (c+d\,x\right )+1\right )} \]

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))^(9/2),x)

[Out]

(a^2*cos(c + d*x)*(1/cos(c + d*x))^(1/2)*((a*(cos(c + d*x) + 1))/cos(c + d*x))^(1/2)*(10290*A*sin(c + d*x) + 1
1760*B*sin(c + d*x) + 14700*C*sin(c + d*x) + 2856*A*sin(2*c + 2*d*x) + 981*A*sin(3*c + 3*d*x) + 260*A*sin(4*c
+ 4*d*x) + 35*A*sin(5*c + 5*d*x) + 2940*B*sin(2*c + 2*d*x) + 720*B*sin(3*c + 3*d*x) + 90*B*sin(4*c + 4*d*x) +
2352*C*sin(2*c + 2*d*x) + 252*C*sin(3*c + 3*d*x)))/(2520*d*(cos(c + d*x) + 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)**(9/2),x)

[Out]

Timed out

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