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

Optimal. Leaf size=124 \[ -\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}-\frac {(A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {(3 A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d} \]

[Out]

-(A+C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))+(3*A+C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*
EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a/d-(A-C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)
/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a/d

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Rubi [A]  time = 0.17, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4085, 3787, 3771, 2639, 2641} \[ -\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}-\frac {(A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {(3 A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*A + 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 + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(a + a*Se
c[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]

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 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 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 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)]

Rubi steps

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

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Mathematica [C]  time = 6.46, size = 795, normalized size = 6.41 \[ -\frac {\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)}} \cos (c+d x) \csc \left (\frac {c}{2}\right ) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \, _2F_1\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 ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{d (\cos (2 c+2 d x) A+A+2 C) (\sec (c+d x) a+a)}-\frac {\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)}} \cos (c+d x) \csc \left (\frac {c}{2}\right ) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \, _2F_1\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 ^2\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)}-\frac {2 A \sqrt {\cos (c+d x)} \csc \left (\frac {c}{2}\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sec \left (\frac {c}{2}\right ) \left (C \sec ^2(c+d x)+A\right ) \sin (c) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{d (\cos (2 c+2 d x) A+A+2 C) \sqrt {\sec (c+d x)} (\sec (c+d x) a+a)}+\frac {2 C \sqrt {\cos (c+d x)} \csc \left (\frac {c}{2}\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sec \left (\frac {c}{2}\right ) \left (C \sec ^2(c+d x)+A\right ) \sin (c) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{d (\cos (2 c+2 d x) A+A+2 C) \sqrt {\sec (c+d x)} (\sec (c+d x) a+a)}+\frac {\left (C \sec ^2(c+d x)+A\right ) \left (-\frac {2 (\cos (2 c) A+2 A+C) \cos (d x) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right )}{d}+\frac {4 \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 ) \sec \left (\frac {c}{2}\right )}{d}+\frac {8 A \cos (c) \sin (d x)}{d}+\frac {4 (A+C) \tan \left (\frac {c}{2}\right )}{d}\right ) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{(\cos (2 c+2 d x) A+A+2 C) \sqrt {\sec (c+d x)} (\sec (c+d x) a+a)} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Sec[c + d*x]^2)/(Sqrt[Sec[c + d*x]]*(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))/(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))/(3*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*Se
c[c + d*x])) + (Cos[c/2 + (d*x)/2]^2*(A + C*Sec[c + d*x]^2)*((-2*(2*A + C + A*Cos[2*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 + (8*A*Cos[c]*Sin[d*x])/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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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt {\sec \left (d x + c\right )}}{a \sec \left (d x + c\right )^{2} + a \sec \left (d x + c\right )}, x\right ) \]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \sec \left (d x + c\right )^{2} + A}{{\left (a \sec \left (d x + c\right ) + a\right )} \sqrt {\sec \left (d x + c\right )}}\,{d x} \]

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

[Out]

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

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maple [A]  time = 5.28, size = 245, normalized size = 1.98 \[ \frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (A \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 A \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-C \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+C \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )+\left (2 A +2 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-A -C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

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))/sec(d*x+c)^(1/2),x)

[Out]

((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)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si
n(1/2*d*x+1/2*c)^2-1)^(1/2)*(A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*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))+C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+(2*A+2*C)*sin(1/2*d*x+1/2*c)
^4+(-A-C)*sin(1/2*d*x+1/2*c)^2)/a/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)/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]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \sec \left (d x + c\right )^{2} + A}{{\left (a \sec \left (d x + c\right ) + a\right )} \sqrt {\sec \left (d x + c\right )}}\,{d x} \]

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))/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C/cos(c + d*x)^2)/((a + a/cos(c + d*x))*(1/cos(c + d*x))^(1/2)),x)

[Out]

int((A + C/cos(c + d*x)^2)/((a + a/cos(c + d*x))*(1/cos(c + d*x))^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A}{\sec ^{\frac {3}{2}}{\left (c + d x \right )} + \sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\sec ^{\frac {3}{2}}{\left (c + d x \right )} + \sqrt {\sec {\left (c + d x \right )}}}\, dx}{a} \]

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))/sec(d*x+c)**(1/2),x)

[Out]

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

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