3.241 \(\int \frac{A+C \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=201 \[ \frac{2 (5 A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 a^2 d}+\frac{2 (5 A+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(7 A+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (\sec (c+d x)+1)}-\frac{(7 A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A+C) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^2} \]

[Out]

-(((7*A + C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d)) + (2*(5*A + C)*Sqrt[Cos
[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) + (2*(5*A + C)*Sin[c + d*x])/(3*a^2*d*Sqrt[
Sec[c + d*x]]) - ((7*A + C)*Sin[c + d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]]*(1 + Sec[c + d*x])) - ((A + C)*Sin[c + d
*x])/(3*d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2)

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Rubi [A]  time = 0.362494, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 8, 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, 2641, 2639} \[ \frac{2 (5 A+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(7 A+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (\sec (c+d x)+1)}+\frac{2 (5 A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{(7 A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A+C) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2),x]

[Out]

-(((7*A + C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d)) + (2*(5*A + C)*Sqrt[Cos
[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) + (2*(5*A + C)*Sin[c + d*x])/(3*a^2*d*Sqrt[
Sec[c + d*x]]) - ((7*A + C)*Sin[c + d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]]*(1 + Sec[c + d*x])) - ((A + C)*Sin[c + d
*x])/(3*d*Sqrt[Sec[c + d*x]]*(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 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]

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]

Rubi steps

\begin{align*} \int \frac{A+C \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx &=-\frac{(A+C) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}-\frac{\int \frac{-\frac{3}{2} a (3 A+C)+\frac{1}{2} a (5 A-C) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx}{3 a^2}\\ &=-\frac{(7 A+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}-\frac{(A+C) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}-\frac{\int \frac{-3 a^2 (5 A+C)+\frac{3}{2} a^2 (7 A+C) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{3 a^4}\\ &=-\frac{(7 A+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}-\frac{(A+C) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac{(5 A+C) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{a^2}-\frac{(7 A+C) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{2 a^2}\\ &=\frac{2 (5 A+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(7 A+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}-\frac{(A+C) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac{(5 A+C) \int \sqrt{\sec (c+d x)} \, dx}{3 a^2}-\frac{\left ((7 A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 a^2}\\ &=-\frac{(7 A+C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^2 d}+\frac{2 (5 A+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(7 A+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}-\frac{(A+C) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac{\left ((5 A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 a^2}\\ &=-\frac{(7 A+C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^2 d}+\frac{2 (5 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{2 (5 A+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(7 A+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}-\frac{(A+C) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}\\ \end{align*}

Mathematica [C]  time = 6.78187, size = 912, normalized size = 4.54 \[ \frac{14 \sqrt{2} A e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \csc \left (\frac{c}{2}\right ) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) \sec \left (\frac{c}{2}\right ) \left (C \sec ^2(c+d x)+A\right ) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (\cos (2 c+2 d x) A+A+2 C) (\sec (c+d x) a+a)^2}+\frac{2 \sqrt{2} C e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \csc \left (\frac{c}{2}\right ) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) \sec \left (\frac{c}{2}\right ) \left (C \sec ^2(c+d x)+A\right ) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (\cos (2 c+2 d x) A+A+2 C) (\sec (c+d x) a+a)^2}+\frac{40 A \sqrt{\cos (c+d x)} \csc \left (\frac{c}{2}\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \sec \left (\frac{c}{2}\right ) \sqrt{\sec (c+d x)} \left (C \sec ^2(c+d x)+A\right ) \sin (c) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (\cos (2 c+2 d x) A+A+2 C) (\sec (c+d x) a+a)^2}+\frac{8 C \sqrt{\cos (c+d x)} \csc \left (\frac{c}{2}\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \sec \left (\frac{c}{2}\right ) \sqrt{\sec (c+d x)} \left (C \sec ^2(c+d x)+A\right ) \sin (c) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (\cos (2 c+2 d x) A+A+2 C) (\sec (c+d x) a+a)^2}+\frac{\sqrt{\sec (c+d x)} \left (C \sec ^2(c+d x)+A\right ) \left (\frac{4 \sec \left (\frac{c}{2}\right ) \left (A \sin \left (\frac{d x}{2}\right )+C \sin \left (\frac{d x}{2}\right )\right ) \sec ^3\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}+\frac{4 (A+C) \tan \left (\frac{c}{2}\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}-\frac{16 \sec \left (\frac{c}{2}\right ) \left (5 A \sin \left (\frac{d x}{2}\right )+2 C \sin \left (\frac{d x}{2}\right )\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}+\frac{4 (2 \cos (2 c) A+5 A+C) \cos (d x) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right )}{d}+\frac{8 A \cos (2 d x) \sin (2 c)}{3 d}-\frac{32 A \cos (c) \sin (d x)}{d}+\frac{8 A \cos (2 c) \sin (2 d x)}{3 d}-\frac{16 (5 A+2 C) \tan \left (\frac{c}{2}\right )}{3 d}\right ) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{(\cos (2 c+2 d x) A+A+2 C) (\sec (c+d x) a+a)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2),x]

[Out]

(14*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]
^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7
/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*(A + C*Sec[c + d*x]^2))/(3*d*E^(I*d*x)*(A + 2*C + A*Cos[2*c + 2*d*x])*(a +
a*Sec[c + d*x])^2) + (2*Sqrt[2]*C*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x)
)]*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeo
metric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*(A + C*Sec[c + d*x]^2))/(3*d*E^(I*d*x)*(A + 2*C + A*C
os[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) + (40*A*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c
+ d*x)/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*(A + C*Sec[c + d*x]^2)*Sin[c])/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(
a + a*Sec[c + d*x])^2) + (8*C*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c
/2]*Sqrt[Sec[c + d*x]]*(A + C*Sec[c + d*x]^2)*Sin[c])/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])
^2) + (Cos[c/2 + (d*x)/2]^4*Sqrt[Sec[c + d*x]]*(A + C*Sec[c + d*x]^2)*((4*(5*A + C + 2*A*Cos[2*c])*Cos[d*x]*Cs
c[c/2]*Sec[c/2])/d + (8*A*Cos[2*d*x]*Sin[2*c])/(3*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] + C*Si
n[(d*x)/2]))/(3*d) - (16*Sec[c/2]*Sec[c/2 + (d*x)/2]*(5*A*Sin[(d*x)/2] + 2*C*Sin[(d*x)/2]))/(3*d) - (32*A*Cos[
c]*Sin[d*x])/d + (8*A*Cos[2*c]*Sin[2*d*x])/(3*d) - (16*(5*A + 2*C)*Tan[c/2])/(3*d) + (4*(A + C)*Sec[c/2 + (d*x
)/2]^2*Tan[c/2])/(3*d)))/((A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2)

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Maple [A]  time = 2.401, size = 437, normalized size = 2.2 \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)^(3/2)/(a+a*sec(d*x+c))^2,x)

[Out]

-1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(16*A*cos(1/2*d*x+1/2*c)^8+12*A*cos(1/2*d*x+1/2*c
)^6+20*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*
cos(1/2*d*x+1/2*c)^3+42*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))+12*C*cos(1/2*d*x+1/2*c)^6+4*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))+6*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))-48*A*cos(1/2*d
*x+1/2*c)^4-20*C*cos(1/2*d*x+1/2*c)^4+21*A*cos(1/2*d*x+1/2*c)^2+9*C*cos(1/2*d*x+1/2*c)^2-A-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)^(3/2)/(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 )^{4} + 2 \, a^{2} \sec \left (d x + c\right )^{3} + a^{2} \sec \left (d x + c\right )^{2}}, 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)^(3/2)/(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)^4 + 2*a^2*sec(d*x + c)^3 + a^2*sec(d*x +
c)^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+C*sec(d*x+c)**2)/sec(d*x+c)**(3/2)/(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{3}{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)^(3/2)/(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)^(3/2)), x)