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

Optimal. Leaf size=228 \[ \frac{2 a^3 (124 A+135 B) \sin (c+d x)}{315 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (4 A+3 B) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{21 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{4 a^3 (292 A+345 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{315 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^3 (292 A+345 B) \sin (c+d x)}{315 d \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac{7}{2}}(c+d x)} \]

[Out]

(2*a^3*(124*A + 135*B)*Sin[c + d*x])/(315*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(292*A + 345
*B)*Sin[c + d*x])/(315*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (4*a^3*(292*A + 345*B)*Sqrt[Sec[c + d*
x]]*Sin[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(4*A + 3*B)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])
/(21*d*Sec[c + d*x]^(5/2)) + (2*a*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2))

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Rubi [A]  time = 0.631508, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {4017, 4015, 3805, 3804} \[ \frac{2 a^3 (124 A+135 B) \sin (c+d x)}{315 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (4 A+3 B) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{21 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{4 a^3 (292 A+345 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{315 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^3 (292 A+345 B) \sin (c+d x)}{315 d \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*(124*A + 135*B)*Sin[c + d*x])/(315*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(292*A + 345
*B)*Sin[c + d*x])/(315*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (4*a^3*(292*A + 345*B)*Sqrt[Sec[c + d*
x]]*Sin[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(4*A + 3*B)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])
/(21*d*Sec[c + d*x]^(5/2)) + (2*a*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2))

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 \frac{(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac{9}{2}}(c+d x)} \, dx &=\frac{2 a A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2}{9} \int \frac{(a+a \sec (c+d x))^{3/2} \left (\frac{3}{2} a (4 A+3 B)+\frac{1}{2} a (4 A+9 B) \sec (c+d x)\right )}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 (4 A+3 B) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{21 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 a A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{4}{63} \int \frac{\sqrt{a+a \sec (c+d x)} \left (\frac{1}{4} a^2 (124 A+135 B)+\frac{1}{4} a^2 (76 A+99 B) \sec (c+d x)\right )}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a^3 (124 A+135 B) \sin (c+d x)}{315 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (4 A+3 B) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{21 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 a A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{1}{105} \left (a^2 (292 A+345 B)\right ) \int \frac{\sqrt{a+a \sec (c+d x)}}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a^3 (124 A+135 B) \sin (c+d x)}{315 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{2 a^3 (292 A+345 B) \sin (c+d x)}{315 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (4 A+3 B) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{21 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 a A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{1}{315} \left (2 a^2 (292 A+345 B)\right ) \int \frac{\sqrt{a+a \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{2 a^3 (124 A+135 B) \sin (c+d x)}{315 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{2 a^3 (292 A+345 B) \sin (c+d x)}{315 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{4 a^3 (292 A+345 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{315 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (4 A+3 B) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{21 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 a A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 0.763781, size = 108, normalized size = 0.47 \[ \frac{2 a^3 \sin (c+d x) \left ((584 A+690 B) \sec ^4(c+d x)+(292 A+345 B) \sec ^3(c+d x)+3 (73 A+60 B) \sec ^2(c+d x)+5 (26 A+9 B) \sec (c+d x)+35 A\right )}{315 d \sec ^{\frac{7}{2}}(c+d x) \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*(35*A + 5*(26*A + 9*B)*Sec[c + d*x] + 3*(73*A + 60*B)*Sec[c + d*x]^2 + (292*A + 345*B)*Sec[c + d*x]^3 +
(584*A + 690*B)*Sec[c + d*x]^4)*Sin[c + d*x])/(315*d*Sec[c + d*x]^(7/2)*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [A]  time = 0.306, size = 143, 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}+45\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+219\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+180\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+292\,A\cos \left ( dx+c \right ) +345\,B\cos \left ( dx+c \right ) +584\,A+690\,B \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+B*sec(d*x+c))/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+45*B*cos(d*x+c)^3+219*A*cos(d*x+c)^2+180*B*
cos(d*x+c)^2+292*A*cos(d*x+c)+345*B*cos(d*x+c)+584*A+690*B)*(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 = 2.20182, size = 1007, normalized size = 4.42 \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+B*sec(d*x+c))/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))/d

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Fricas [A]  time = 0.478615, size = 371, normalized size = 1.63 \begin{align*} \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\right )} a^{2} \cos \left (d x + c\right )^{3} +{\left (292 \, A + 345 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \,{\left (292 \, A + 345 \, B\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+B*sec(d*x+c))/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)*a^2*cos(d*x + c)^3 + (292
*A + 345*B)*a^2*cos(d*x + c)^2 + 2*(292*A + 345*B)*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*s
in(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+B*sec(d*x+c))/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 (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} \end{align*}

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))/sec(d*x+c)^(9/2),x, algorithm="giac")

[Out]

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