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

Optimal. Leaf size=236 \[ -\frac{5 (3 A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 a^2 d}-\frac{(3 A+C) \sin (c+d x)}{a^2 d \sec ^{\frac{3}{2}}(c+d x) (\sec (c+d x)+1)}+\frac{4 (14 A+5 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 (3 A+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}+\frac{4 (14 A+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac{(A+C) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^2} \]

[Out]

(4*(14*A + 5*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*a^2*d) - (5*(3*A + C)*Sqrt
[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) + (4*(14*A + 5*C)*Sin[c + d*x])/(15*a^2
*d*Sec[c + d*x]^(3/2)) - (5*(3*A + C)*Sin[c + d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]]) - ((3*A + C)*Sin[c + d*x])/(a
^2*d*Sec[c + d*x]^(3/2)*(1 + Sec[c + d*x])) - ((A + C)*Sin[c + d*x])/(3*d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*
x])^2)

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Rubi [A]  time = 0.376041, antiderivative size = 236, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4085, 4020, 3787, 3769, 3771, 2639, 2641} \[ -\frac{(3 A+C) \sin (c+d x)}{a^2 d \sec ^{\frac{3}{2}}(c+d x) (\sec (c+d x)+1)}+\frac{4 (14 A+5 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 (3 A+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{5 (3 A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{4 (14 A+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac{(A+C) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(14*A + 5*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*a^2*d) - (5*(3*A + C)*Sqrt
[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) + (4*(14*A + 5*C)*Sin[c + d*x])/(15*a^2
*d*Sec[c + d*x]^(3/2)) - (5*(3*A + C)*Sin[c + d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]]) - ((3*A + C)*Sin[c + d*x])/(a
^2*d*Sec[c + d*x]^(3/2)*(1 + Sec[c + d*x])) - ((A + C)*Sin[c + d*x])/(3*d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*
x])^2)

Rule 4085

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*(A + C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(a*f*(
2*m + 1)), x] + Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*C*n + A*b*(
2*m + n + 1) - (a*(A*(m + n + 1) - C*(m - n)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x]
&& EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 4020

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

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, 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 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+C \sec ^2(c+d x)}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx &=-\frac{(A+C) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac{\int \frac{-\frac{1}{2} a (11 A+5 C)+\frac{1}{2} a (7 A+C) \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx}{3 a^2}\\ &=-\frac{(3 A+C) \sin (c+d x)}{a^2 d \sec ^{\frac{3}{2}}(c+d x) (1+\sec (c+d x))}-\frac{(A+C) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac{\int \frac{-2 a^2 (14 A+5 C)+\frac{15}{2} a^2 (3 A+C) \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x)} \, dx}{3 a^4}\\ &=-\frac{(3 A+C) \sin (c+d x)}{a^2 d \sec ^{\frac{3}{2}}(c+d x) (1+\sec (c+d x))}-\frac{(A+C) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac{(5 (3 A+C)) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{2 a^2}+\frac{(2 (14 A+5 C)) \int \frac{1}{\sec ^{\frac{5}{2}}(c+d x)} \, dx}{3 a^2}\\ &=\frac{4 (14 A+5 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 (3 A+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(3 A+C) \sin (c+d x)}{a^2 d \sec ^{\frac{3}{2}}(c+d x) (1+\sec (c+d x))}-\frac{(A+C) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac{(5 (3 A+C)) \int \sqrt{\sec (c+d x)} \, dx}{6 a^2}+\frac{(2 (14 A+5 C)) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{5 a^2}\\ &=\frac{4 (14 A+5 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 (3 A+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(3 A+C) \sin (c+d x)}{a^2 d \sec ^{\frac{3}{2}}(c+d x) (1+\sec (c+d x))}-\frac{(A+C) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac{\left (5 (3 A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a^2}+\frac{\left (2 (14 A+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 a^2}\\ &=\frac{4 (14 A+5 C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{5 a^2 d}-\frac{5 (3 A+C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 a^2 d}+\frac{4 (14 A+5 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 (3 A+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(3 A+C) \sin (c+d x)}{a^2 d \sec ^{\frac{3}{2}}(c+d x) (1+\sec (c+d x))}-\frac{(A+C) \sin (c+d x)}{3 d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}\\ \end{align*}

Mathematica [C]  time = 6.34301, size = 301, normalized size = 1.28 \[ -\frac{\sin (c) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) e^{-i d x} \cos \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{5}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (8 i (14 A+5 C) e^{-\frac{1}{2} i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \left (1+e^{i (c+d x)}\right )^3 \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+400 (3 A+C) \cos ^3\left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+2 \cos (c+d x) \left (-72 i (14 A+5 C) \cos \left (\frac{1}{2} (c+d x)\right )-24 i (14 A+5 C) \cos \left (\frac{3}{2} (c+d x)\right )+2 \sin \left (\frac{1}{2} (c+d x)\right ) ((179 A+60 C) \cos (c+d x)+8 A \cos (2 (c+d x))-3 A \cos (3 (c+d x))+158 A+50 C)\right )\right )}{120 a^2 d (\sec (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[(c + d*x)/2]*Csc[c/2]*Sec[c/2]*Sec[c + d*x]^(5/2)*Sin[c]*(Cos[d*x] + I*Sin[d*x])*(400*(3*A + C)*Cos[(c +
d*x)/2]^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + ((8*I)*(14*A + 5*C)*(1 + E^(I*(c + d*x)))^3*Sqrt[1 +
E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^((I/2)*(c + d*x)) + 2*Cos[c +
d*x]*((-72*I)*(14*A + 5*C)*Cos[(c + d*x)/2] - (24*I)*(14*A + 5*C)*Cos[(3*(c + d*x))/2] + 2*(158*A + 50*C + (17
9*A + 60*C)*Cos[c + d*x] + 8*A*Cos[2*(c + d*x)] - 3*A*Cos[3*(c + d*x)])*Sin[(c + d*x)/2])))/(120*a^2*d*E^(I*d*
x)*(1 + Sec[c + d*x])^2)

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Maple [A]  time = 2.474, size = 451, normalized size = 1.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(96*A*cos(1/2*d*x+1/2*c)^10-352*A*cos(1/2*d*x+1/
2*c)^8+120*A*cos(1/2*d*x+1/2*c)^6-150*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*Ellipti
cF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^3-336*A*cos(1/2*d*x+1/2*c)^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-
2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-120*C*cos(1/2*d*x+1/2*c)^6-50*C*cos(1/2*
d*x+1/2*c)^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/
2))-120*C*cos(1/2*d*x+1/2*c)^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/
2*d*x+1/2*c),2^(1/2))+266*A*cos(1/2*d*x+1/2*c)^4+190*C*cos(1/2*d*x+1/2*c)^4-135*A*cos(1/2*d*x+1/2*c)^2-75*C*co
s(1/2*d*x+1/2*c)^2+5*A+5*C)/a^2/cos(1/2*d*x+1/2*c)^3/(-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

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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+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^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{{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt{\sec \left (d x + c\right )}}{a^{2} \sec \left (d x + c\right )^{5} + 2 \, a^{2} \sec \left (d x + c\right )^{4} + a^{2} \sec \left (d x + c\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*sec(d*x + c)^2 + A)*sqrt(sec(d*x + c))/(a^2*sec(d*x + c)^5 + 2*a^2*sec(d*x + c)^4 + a^2*sec(d*x +
c)^3), 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+C*sec(d*x+c)**2)/sec(d*x+c)**(5/2)/(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \sec \left (d x + c\right )^{2} + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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