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

Optimal. Leaf size=152 \[ \frac{(A-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{a d}-\frac{(A+C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}+\frac{(A+3 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{a d}-\frac{(A+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d} \]

[Out]

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

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Rubi [A]  time = 0.189359, antiderivative size = 152, 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, 3787, 3771, 2641, 3768, 2639} \[ -\frac{(A+C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}+\frac{(A+3 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{a d}+\frac{(A-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}-\frac{(A+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 3768

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

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

Mathematica [C]  time = 6.63623, size = 776, normalized size = 5.11 \[ \frac{\sqrt{2} A \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) 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)}} \cos (c+d x) \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \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 ) \left (A+C \sec ^2(c+d x)\right )}{3 d (a \sec (c+d x)+a) (A \cos (2 c+2 d x)+A+2 C)}+\frac{\sqrt{2} C \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) 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)}} \cos (c+d x) \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \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 ) \left (A+C \sec ^2(c+d x)\right )}{d (a \sec (c+d x)+a) (A \cos (2 c+2 d x)+A+2 C)}+\frac{2 A \sin (c) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \sqrt{\cos (c+d x)} \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (A+C \sec ^2(c+d x)\right )}{d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a) (A \cos (2 c+2 d x)+A+2 C)}-\frac{2 C \sin (c) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \sqrt{\cos (c+d x)} \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (A+C \sec ^2(c+d x)\right )}{d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a) (A \cos (2 c+2 d x)+A+2 C)}+\frac{\cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (A+C \sec ^2(c+d x)\right ) \left (-\frac{4 \sec \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right ) \left (A \sin \left (\frac{d x}{2}\right )+C \sin \left (\frac{d x}{2}\right )\right )}{d}+\frac{2 (A+3 C) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \cos (d x)}{d}-\frac{4 (A+C) \tan \left (\frac{c}{2}\right )}{d}\right )}{\sqrt{\sec (c+d x)} (a \sec (c+d x)+a) (A \cos (2 c+2 d x)+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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]^2*
Cos[c + d*x]*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])) + (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]^2*Cos[c + d*x]*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))/(d*E^(I*d*
x)*(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*Sqrt[Cos[c + d*x]]*Csc[c/2
]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*Sin[c])/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[Sec
[c + d*x]]*(a + a*Sec[c + d*x])) - (2*C*Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2
, 2]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*Sin[c])/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[Sec[c + d*x]]*(a + a*Sec[c
+ d*x])) + (Cos[c/2 + (d*x)/2]^2*(A + C*Sec[c + d*x]^2)*((2*(A + 3*C)*Cos[d*x]*Csc[c/2]*Sec[c/2])/d - (4*Sec[
c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/d - (4*(A + C)*Tan[c/2])/d))/((A + 2*C + A*Cos[2*c
+ 2*d*x])*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x]))

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Maple [A]  time = 4.051, size = 316, normalized size = 2.1 \begin{align*} -{\frac{1}{ad}\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 ( -\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \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}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( A{\it EllipticF} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ) +A{\it EllipticE} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ) -C{\it EllipticF} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ) +3\,C{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \right ) -2\,\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \left ( A+3\,C \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+\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 ( A+5\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3} \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) ^{-1} \left ( \cos \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)*sec(d*x+c)^(1/2)/(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)*(-cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+s
in(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(A*EllipticF(cos(1/2*
d*x+1/2*c),2^(1/2))+A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*C*Ellipt
icE(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)*(A+3*C)*sin(1/2*d*x+1/
2*c)^4+(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(A+5*C)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^3
/(2*sin(1/2*d*x+1/2*c)^2-1)/cos(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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt{\sec \left (d x + c\right )}}{a \sec \left (d x + c\right ) + a}\,{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)^(1/2)/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

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

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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 \sec \left (d x + c\right ) + a}, 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)^(1/2)/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^2 + A)*sqrt(sec(d*x + c))/(a*sec(d*x + c) + a), 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)**(1/2)/(a+a*sec(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt{\sec \left (d x + c\right )}}{a \sec \left (d x + c\right ) + a}\,{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)^(1/2)/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

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