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

Optimal. Leaf size=230 \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) (5 a B+5 A b+7 b C)}{21 d}+\frac{2 \sin (c+d x) (7 a A+9 a C+9 b B)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \sin (c+d x) (5 a B+5 A b+7 b C)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (7 a A+9 a C+9 b B)}{15 d}+\frac{2 (a B+A b) \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 a A \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.280518, antiderivative size = 230, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4074, 4047, 3769, 3771, 2639, 4045, 2641} \[ \frac{2 \sin (c+d x) (7 a A+9 a C+9 b B)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \sin (c+d x) (5 a B+5 A b+7 b C)}{21 d \sqrt{\sec (c+d x)}}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (5 a B+5 A b+7 b C)}{21 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (7 a A+9 a C+9 b B)}{15 d}+\frac{2 (a B+A b) \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 a A \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4074

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_)), x_Symbol] :> Simp[(A*a*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*n), x]
+ Dist[1/(d*n), Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && LtQ[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)) \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 \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}-\frac{2}{9} \int \frac{-\frac{9}{2} (A b+a B)-\frac{1}{2} (7 a A+9 b B+9 a C) \sec (c+d x)-\frac{9}{2} b C \sec ^2(c+d x)}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 a A \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}-\frac{2}{9} \int \frac{-\frac{9}{2} (A b+a B)-\frac{9}{2} b C \sec ^2(c+d x)}{\sec ^{\frac{7}{2}}(c+d x)} \, dx-\frac{1}{9} (-7 a A-9 b B-9 a C) \int \frac{1}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a A \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 (A b+a B) \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (7 a A+9 b B+9 a C) \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{1}{15} (-7 a A-9 b B-9 a C) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx-\frac{1}{7} (-5 A b-5 a B-7 b C) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a A \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 (A b+a B) \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (7 a A+9 b B+9 a C) \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (5 A b+5 a B+7 b C) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}-\frac{1}{21} (-5 A b-5 a B-7 b C) \int \sqrt{\sec (c+d x)} \, dx-\frac{1}{15} \left ((-7 a A-9 b B-9 a C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{2 (7 a A+9 b B+9 a C) \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 A \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 (A b+a B) \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (7 a A+9 b B+9 a C) \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (5 A b+5 a B+7 b C) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}-\frac{1}{21} \left ((-5 A b-5 a B-7 b C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 (7 a A+9 b B+9 a C) \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 (5 A b+5 a B+7 b C) \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 A \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 (A b+a B) \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 (7 a A+9 b B+9 a C) \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (5 A b+5 a B+7 b C) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 4.68094, size = 249, normalized size = 1.08 \[ \frac{e^{-i d x} \sqrt{\sec (c+d x)} (\cos (d x)+i \sin (d x)) \left (-56 i e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right ) (7 a A+9 a C+9 b B)+120 \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) (5 a B+5 A b+7 b C)+\cos (c+d x) (30 \sin (c+d x) (23 a B+23 A b+28 b C)+14 \sin (2 (c+d x)) (19 a A+18 a C+18 b B)+35 a A \sin (4 (c+d x))+1176 i a A+90 a B \sin (3 (c+d x))+1512 i a C+90 A b \sin (3 (c+d x))+1512 i b B)\right )}{1260 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])*(120*(5*A*b + 5*a*B + 7*b*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x
)/2, 2] - (56*I)*(7*a*A + 9*b*B + 9*a*C)*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2,
3/4, 7/4, -E^((2*I)*(c + d*x))] + Cos[c + d*x]*((1176*I)*a*A + (1512*I)*b*B + (1512*I)*a*C + 30*(23*A*b + 23*a
*B + 28*b*C)*Sin[c + d*x] + 14*(19*a*A + 18*b*B + 18*a*C)*Sin[2*(c + d*x)] + 90*A*b*Sin[3*(c + d*x)] + 90*a*B*
Sin[3*(c + d*x)] + 35*a*A*Sin[4*(c + d*x)])))/(1260*d*E^(I*d*x))

________________________________________________________________________________________

Maple [B]  time = 2.036, size = 565, normalized size = 2.5 \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))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/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*a*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^10+(2240*A*a+720*A*b+720*B*a)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-2072*A*a-1080*A*b-1080*B*a-504*B*b-5
04*C*a)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(952*A*a+840*A*b+840*B*a+504*B*b+504*C*a+420*C*b)*sin(1/2*d*x+
1/2*c)^4*cos(1/2*d*x+1/2*c)+(-168*A*a-240*A*b-240*B*a-126*B*b-126*C*a-210*C*b)*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+75*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))-189*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))*
b+75*B*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1
89*C*(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+105
*C*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)))/(-2*
sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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))*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(9/2),x)

[Out]

Timed out

________________________________________________________________________________________

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