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

Optimal. Leaf size=254 \[ \frac{2 \left (5 a (a B+2 A b)+7 b^2 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{21 d}+\frac{2 \left (7 a^2 A+18 a b B+9 A b^2\right ) \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (7 a^2 A+18 a b B+9 A b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 a^2 A \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 \left (5 a (a B+2 A b)+7 b^2 B\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 a (a B+2 A b) \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

[Out]

(2*(7*a^2*A + 9*A*b^2 + 18*a*b*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (2
*(7*b^2*B + 5*a*(2*A*b + a*B))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*a^
2*A*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2)) + (2*a*(2*A*b + a*B)*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (2*(7
*a^2*A + 9*A*b^2 + 18*a*b*B)*Sin[c + d*x])/(45*d*Sec[c + d*x]^(3/2)) + (2*(7*b^2*B + 5*a*(2*A*b + a*B))*Sin[c
+ d*x])/(21*d*Sqrt[Sec[c + d*x]])

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Rubi [A]  time = 0.337205, antiderivative size = 254, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {4024, 4047, 3769, 3771, 2639, 4045, 2641} \[ \frac{2 \left (7 a^2 A+18 a b B+9 A b^2\right ) \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (7 a^2 A+18 a b B+9 A b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 a^2 A \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 \left (5 a (a B+2 A b)+7 b^2 B\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 \left (5 a (a B+2 A b)+7 b^2 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 a (a B+2 A b) \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(7*a^2*A + 9*A*b^2 + 18*a*b*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (2
*(7*b^2*B + 5*a*(2*A*b + a*B))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*a^
2*A*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2)) + (2*a*(2*A*b + a*B)*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (2*(7
*a^2*A + 9*A*b^2 + 18*a*b*B)*Sin[c + d*x])/(45*d*Sec[c + d*x]^(3/2)) + (2*(7*b^2*B + 5*a*(2*A*b + a*B))*Sin[c
+ d*x])/(21*d*Sqrt[Sec[c + d*x]])

Rule 4024

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

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{(a+b \sec (c+d x))^2 (A+B \sec (c+d x))}{\sec ^{\frac{9}{2}}(c+d x)} \, dx &=\frac{2 a^2 A \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}-\frac{2}{9} \int \frac{-\frac{9}{2} a (2 A b+a B)+\left (A \left (-\frac{7 a^2}{2}-\frac{9 b^2}{2}\right )-9 a b B\right ) \sec (c+d x)-\frac{9}{2} b^2 B \sec ^2(c+d x)}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 A \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}-\frac{2}{9} \int \frac{-\frac{9}{2} a (2 A b+a B)-\frac{9}{2} b^2 B \sec ^2(c+d x)}{\sec ^{\frac{7}{2}}(c+d x)} \, dx-\frac{1}{9} \left (-7 a^2 A-9 A b^2-18 a b B\right ) \int \frac{1}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 A \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 a (2 A b+a B) \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (7 a^2 A+9 A b^2+18 a b B\right ) \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{1}{15} \left (-7 a^2 A-9 A b^2-18 a b B\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx-\frac{1}{7} \left (-7 b^2 B-5 a (2 A b+a B)\right ) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 A \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 a (2 A b+a B) \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (7 a^2 A+9 A b^2+18 a b B\right ) \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (7 b^2 B+5 a (2 A b+a B)\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}-\frac{1}{21} \left (-7 b^2 B-5 a (2 A b+a B)\right ) \int \sqrt{\sec (c+d x)} \, dx-\frac{1}{15} \left (\left (-7 a^2 A-9 A b^2-18 a b B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{2 \left (7 a^2 A+9 A b^2+18 a b B\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{15 d}+\frac{2 a^2 A \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 a (2 A b+a B) \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (7 a^2 A+9 A b^2+18 a b B\right ) \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (7 b^2 B+5 a (2 A b+a B)\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}-\frac{1}{21} \left (\left (-7 b^2 B-5 a (2 A b+a B)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 \left (7 a^2 A+9 A b^2+18 a b B\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{15 d}+\frac{2 \left (7 b^2 B+5 a (2 A b+a B)\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{21 d}+\frac{2 a^2 A \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 a (2 A b+a B) \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (7 a^2 A+9 A b^2+18 a b B\right ) \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (7 b^2 B+5 a (2 A b+a B)\right ) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.94013, size = 189, normalized size = 0.74 \[ \frac{\sqrt{\sec (c+d x)} \left (120 \left (5 a^2 B+10 a A b+7 b^2 B\right ) \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+\sin (2 (c+d x)) \left (7 \left (43 a^2 A+72 a b B+36 A b^2\right ) \cos (c+d x)+5 \left (7 a^2 A \cos (3 (c+d x))+78 a^2 B+18 a (a B+2 A b) \cos (2 (c+d x))+156 a A b+84 b^2 B\right )\right )+168 \left (7 a^2 A+18 a b B+9 A b^2\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{1260 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sec[c + d*x]]*(168*(7*a^2*A + 9*A*b^2 + 18*a*b*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 120*(10
*a*A*b + 5*a^2*B + 7*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (7*(43*a^2*A + 36*A*b^2 + 72*a*b*B)
*Cos[c + d*x] + 5*(156*a*A*b + 78*a^2*B + 84*b^2*B + 18*a*(2*A*b + a*B)*Cos[2*(c + d*x)] + 7*a^2*A*Cos[3*(c +
d*x)]))*Sin[2*(c + d*x)]))/(1260*d)

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Maple [B]  time = 1.985, size = 610, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-1120*a^2*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2
*c)^10+(2240*A*a^2+1440*A*a*b+720*B*a^2)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-2072*A*a^2-2160*A*a*b-504*A
*b^2-1080*B*a^2-1008*B*a*b)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(952*A*a^2+1680*A*a*b+504*A*b^2+840*B*a^2+
1008*B*a*b+420*B*b^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-168*A*a^2-480*A*a*b-126*A*b^2-240*B*a^2-252*B*
a*b-210*B*b^2)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-147*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c
)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2-189*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c
)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^2+150*A*a*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1
/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-378*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c
)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b+75*B*a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/
2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+105*B*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/
2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/si
n(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [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+b*sec(d*x+c))^2*(A+B*sec(d*x+c))/sec(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{B b^{2} \sec \left (d x + c\right )^{3} + A a^{2} +{\left (2 \, B a b + A b^{2}\right )} \sec \left (d x + c\right )^{2} +{\left (B a^{2} + 2 \, A a b\right )} \sec \left (d x + c\right )}{\sec \left (d x + c\right )^{\frac{9}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*b^2*sec(d*x + c)^3 + A*a^2 + (2*B*a*b + A*b^2)*sec(d*x + c)^2 + (B*a^2 + 2*A*a*b)*sec(d*x + c))/se
c(d*x + c)^(9/2), x)

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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+b*sec(d*x+c))**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 (b \sec \left (d x + c\right ) + a\right )}^{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+b*sec(d*x+c))^2*(A+B*sec(d*x+c))/sec(d*x+c)^(9/2),x, algorithm="giac")

[Out]

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