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

Optimal. Leaf size=170 \[ -\frac{(5 A-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 a^2 d}-\frac{(5 A-C) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 a^2 d (\sec (c+d x)+1)}+\frac{4 A \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A+C) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.317311, antiderivative size = 170, 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 = {4085, 4020, 3787, 3771, 2639, 2641} \[ -\frac{(5 A-C) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 a^2 d (\sec (c+d x)+1)}-\frac{(5 A-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{4 A \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A+C) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*A*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d) - ((5*A - C)*Sqrt[Cos[c + d*x]]*
EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) - ((5*A - C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*a^2*d
*(1 + Sec[c + d*x])) - ((A + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d*(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 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)}{\sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx &=-\frac{(A+C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{\int \frac{-\frac{1}{2} a (7 A+C)+\frac{3}{2} a (A-C) \sec (c+d x)}{\sqrt{\sec (c+d x)} (a+a \sec (c+d x))} \, dx}{3 a^2}\\ &=-\frac{(5 A-C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(A+C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{\int \frac{-6 a^2 A+\frac{1}{2} a^2 (5 A-C) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{3 a^4}\\ &=-\frac{(5 A-C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(A+C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{(2 A) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{a^2}-\frac{(5 A-C) \int \sqrt{\sec (c+d x)} \, dx}{6 a^2}\\ &=-\frac{(5 A-C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(A+C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{\left (2 A \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{a^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}{6 a^2}\\ &=\frac{4 A \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{(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{(5 A-C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(A+C) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}\\ \end{align*}

Mathematica [C]  time = 3.49726, size = 298, normalized size = 1.75 \[ \frac{\cos ^4\left (\frac{1}{2} (c+d x)\right ) \left (A+C \sec ^2(c+d x)\right ) \left (-\frac{8 (5 A-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{d}-\frac{32 i A e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \sqrt{\sec (c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )}{d}+\frac{8 i e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right ) \sqrt{\sec (c+d x)} \left (A \left (16 e^{i (c+d x)}+20 e^{2 i (c+d x)}+9 e^{3 i (c+d x)}+3\right )-C e^{i (c+d x)} \left (-1+e^{i (c+d x)}\right )\right )}{d \left (1+e^{i (c+d x)}\right )^3}\right )}{3 a^2 (\sec (c+d x)+1)^2 (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^4*(((8*I)*(1 + E^((2*I)*(c + d*x)))*(-(C*E^(I*(c + d*x))*(-1 + E^(I*(c + d*x)))) + A*(3 + 16
*E^(I*(c + d*x)) + 20*E^((2*I)*(c + d*x)) + 9*E^((3*I)*(c + d*x))))*Sqrt[Sec[c + d*x]])/(d*E^(I*(c + d*x))*(1
+ E^(I*(c + d*x)))^3) - (8*(5*A - C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d - ((32
*I)*A*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))]*Sqr
t[Sec[c + d*x]])/d)*(A + C*Sec[c + d*x]^2))/(3*a^2*(A + 2*C + A*Cos[2*(c + d*x)])*(1 + Sec[c + d*x])^2)

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Maple [A]  time = 2.348, size = 352, normalized size = 2.1 \begin{align*}{\frac{1}{6\,{a}^{2}d}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 24\,A \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+10\,A\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+24\,A \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -2\,C \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -38\,A \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-2\,C \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+15\,A \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+3\,C \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-A-C \right ) \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2/sec(d*x+c)^(1/2),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)*(24*A*cos(1/2*d*x+1/2*c)^6+10*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+24*A*c
os(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))-2*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(c
os(1/2*d*x+1/2*c),2^(1/2))-38*A*cos(1/2*d*x+1/2*c)^4-2*C*cos(1/2*d*x+1/2*c)^4+15*A*cos(1/2*d*x+1/2*c)^2+3*C*co
s(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)/(a+a*sec(d*x+c))^2/sec(d*x+c)^(1/2),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 )^{3} + 2 \, a^{2} \sec \left (d x + c\right )^{2} + a^{2} \sec \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A}{\sec ^{\frac{5}{2}}{\left (c + d x \right )} + 2 \sec ^{\frac{3}{2}}{\left (c + d x \right )} + \sqrt{\sec{\left (c + d x \right )}}}\, dx + \int \frac{C \sec ^{2}{\left (c + d x \right )}}{\sec ^{\frac{5}{2}}{\left (c + d x \right )} + 2 \sec ^{\frac{3}{2}}{\left (c + d x \right )} + \sqrt{\sec{\left (c + d x \right )}}}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**2/sec(d*x+c)**(1/2),x)

[Out]

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

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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} \sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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