3.1112 \(\int \frac{A+C \sec ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx\)

Optimal. Leaf size=150 \[ \frac{(3 A+5 C) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 a d}+\frac{(A+3 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}-\frac{(A+C) \sin (c+d x)}{d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)}+\frac{(3 A+5 C) \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x)}-\frac{(A+3 C) \sin (c+d x)}{a d \sqrt{\cos (c+d x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.262772, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4114, 3042, 2748, 2636, 2641, 2639} \[ \frac{(3 A+5 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a d}+\frac{(A+3 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}-\frac{(A+C) \sin (c+d x)}{d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)}+\frac{(3 A+5 C) \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x)}-\frac{(A+3 C) \sin (c+d x)}{a d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4114

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*sec[(e_.)
+ (f_.)*(x_)]^2), x_Symbol] :> Dist[d^(m + 2), Int[(b + a*Cos[e + f*x])^m*(d*Cos[e + f*x])^(n - m - 2)*(C + A
*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] &&  !IntegerQ[n] && IntegerQ[m]

Rule 3042

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x
])^(n + 1))/(f*(b*c - a*d)*(2*m + 1)), x] + Dist[1/(b*(b*c - a*d)*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)
*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2
) + C*(b*c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2636

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

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)}{\cos ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx &=\int \frac{C+A \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))} \, dx\\ &=-\frac{(A+C) \sin (c+d x)}{d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))}+\frac{\int \frac{\frac{1}{2} a (3 A+5 C)-\frac{1}{2} a (A+3 C) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{a^2}\\ &=-\frac{(A+C) \sin (c+d x)}{d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))}-\frac{(A+3 C) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{2 a}+\frac{(3 A+5 C) \int \frac{1}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{2 a}\\ &=\frac{(3 A+5 C) \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x)}-\frac{(A+3 C) \sin (c+d x)}{a d \sqrt{\cos (c+d x)}}-\frac{(A+C) \sin (c+d x)}{d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))}+\frac{(A+3 C) \int \sqrt{\cos (c+d x)} \, dx}{2 a}+\frac{(3 A+5 C) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a}\\ &=\frac{(A+3 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}+\frac{(3 A+5 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a d}+\frac{(3 A+5 C) \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x)}-\frac{(A+3 C) \sin (c+d x)}{a d \sqrt{\cos (c+d x)}}-\frac{(A+C) \sin (c+d x)}{d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))}\\ \end{align*}

Mathematica [C]  time = 6.95902, size = 1337, normalized size = 8.91 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((I/2)*A*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*((2*E^((2*I)*d*x)*Hypergeo
metric2F1[1/2, 3/4, 7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-
1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1
+ E^((2*I)*d*x))*Cos[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*
d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)
]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^
((2*I)*d*x))*Sin[c])))/((A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) + (((3*I)/2)*C*Cos[c/2 + (d*x)/2]
^2*Cos[c + d*x]*Csc[c/2]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4, 7/4, -(
E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/
E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))*Cos[c] -
3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)
]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Co
s[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[c])))/((A
+ 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) + (Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]^(3/2)*(A + C*Sec[c + d*
x]^2)*((-2*(2*C + A*Cos[c] + C*Cos[c])*Csc[c/2]*Sec[c/2]*Sec[c])/d - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*
x)/2] + C*Sin[(d*x)/2]))/d + (8*C*Sec[c]*Sec[c + d*x]^2*Sin[d*x])/(3*d) + (8*Sec[c]*Sec[c + d*x]*(C*Sin[c] - 3
*C*Sin[d*x]))/(3*d)))/((A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) - (2*A*Cos[c/2 + (d*x)/2]^2*Cos[c
+ d*x]*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + C*Sec[c + d*x]
^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - A
rcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a
+ a*Sec[c + d*x])) - (10*C*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin
[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[
Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3
*d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x]))

________________________________________________________________________________________

Maple [B]  time = 6.177, size = 486, normalized size = 3.2 \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)/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c)),x)

[Out]

-1/a*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*C*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1
/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*
d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2
)))+(A+C)*(cos(1/2*d*x+1/2*c)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2
*d*x+1/2*c),2^(1/2))-EllipticE(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)/cos(1
/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-2*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)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(
1/2))+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*
d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1))/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+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

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{\cos \left (d x + c\right )}}{a \cos \left (d x + c\right )^{2} \sec \left (d x + c\right ) + a \cos \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)/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

________________________________________________________________________________________

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 )} \cos \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)/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)/((a*sec(d*x + c) + a)*cos(d*x + c)^(3/2)), x)