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

Optimal. Leaf size=109 \[ \frac{\sqrt{2} (A+C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 C \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{3 a d}-\frac{4 C \tan (c+d x)}{3 d \sqrt{a \sec (c+d x)+a}} \]

[Out]

(Sqrt[2]*(A + C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d) - (4*C*Tan[c +
d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*C*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*a*d)

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Rubi [A]  time = 0.205821, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {4083, 4001, 3795, 203} \[ \frac{\sqrt{2} (A+C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 C \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{3 a d}-\frac{4 C \tan (c+d x)}{3 d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*(A + C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d) - (4*C*Tan[c +
d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*C*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*a*d)

Rule 4083

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(
m_), x_Symbol] :> -Simp[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)),
Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) - a*C*Csc[e + f*x], x], x], x] /; FreeQ
[{a, b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

Rule 4001

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*B*m + A*b*(m + 1))/(b*(
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, x] && NeQ[A*b - a*B,
0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] &&  !LtQ[m, -2^(-1)]

Rule 3795

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx &=\frac{2 C \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3 a d}+\frac{2 \int \frac{\sec (c+d x) \left (\frac{1}{2} a (3 A+C)-a C \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{3 a}\\ &=-\frac{4 C \tan (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3 a d}+(A+C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=-\frac{4 C \tan (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3 a d}-\frac{(2 (A+C)) \operatorname{Subst}\left (\int \frac{1}{2 a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}\\ &=\frac{\sqrt{2} (A+C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{\sqrt{a} d}-\frac{4 C \tan (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3 a d}\\ \end{align*}

Mathematica [A]  time = 1.5188, size = 125, normalized size = 1.15 \[ -\frac{2 \cos \left (\frac{c}{2}\right ) \cos (c) \sin (c+d x) \left (3 \sqrt{2} (A+C) \sqrt{\sec (c+d x)-1} \tan ^{-1}\left (\frac{\sqrt{\sec (c+d x)-1}}{\sqrt{2}}\right )+8 C \sin ^4\left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x)\right )}{3 d \left (\cos \left (\frac{c}{2}\right )+\cos \left (\frac{3 c}{2}\right )\right ) (\cos (c+d x)-1) \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[c/2]*Cos[c]*(3*Sqrt[2]*(A + C)*ArcTan[Sqrt[-1 + Sec[c + d*x]]/Sqrt[2]]*Sqrt[-1 + Sec[c + d*x]] + 8*C*S
ec[c + d*x]^2*Sin[(c + d*x)/2]^4)*Sin[c + d*x])/(3*d*(Cos[c/2] + Cos[(3*c)/2])*(-1 + Cos[c + d*x])*Sqrt[a*(1 +
Sec[c + d*x])])

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Maple [B]  time = 0.326, size = 385, normalized size = 3.5 \begin{align*} -{\frac{1}{6\,ad\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) } \left ( 3\,A\sin \left ( dx+c \right ) \ln \left ( -{\frac{1}{\sin \left ( dx+c \right ) } \left ( -\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) -1 \right ) } \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\cos \left ( dx+c \right ) +3\,C\sin \left ( dx+c \right ) \ln \left ( -{\frac{1}{\sin \left ( dx+c \right ) } \left ( -\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) -1 \right ) } \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\cos \left ( dx+c \right ) +3\,A\ln \left ( -{\frac{1}{\sin \left ( dx+c \right ) } \left ( -\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) -1 \right ) } \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\sin \left ( dx+c \right ) +3\,C\ln \left ( -{\frac{1}{\sin \left ( dx+c \right ) } \left ( -\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) -1 \right ) } \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\sin \left ( dx+c \right ) -4\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+8\,C\cos \left ( dx+c \right ) -4\,C \right ) \sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/d/a*(3*A*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*c
os(d*x+c)/(cos(d*x+c)+1))^(3/2)*cos(d*x+c)+3*C*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c
)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*cos(d*x+c)+3*A*ln(-(-(-2*cos(d*x+c)/(cos(d*x+
c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*sin(d*x+c)+3*C*ln(-(-(-
2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*s
in(d*x+c)-4*C*cos(d*x+c)^2+8*C*cos(d*x+c)-4*C)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)/cos(d*x+c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.600128, size = 895, normalized size = 8.21 \begin{align*} \left [\frac{3 \, \sqrt{2}{\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{2} +{\left (A + C\right )} a \cos \left (d x + c\right )\right )} \sqrt{-\frac{1}{a}} \log \left (-\frac{2 \, \sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{-\frac{1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \,{\left (C \cos \left (d x + c\right ) - C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{6 \,{\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}}, -\frac{2 \,{\left (C \cos \left (d x + c\right ) - C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) + \frac{3 \, \sqrt{2}{\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{2} +{\left (A + C\right )} a \cos \left (d x + c\right )\right )} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right )}{\sqrt{a}}}{3 \,{\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*sqrt(2)*((A + C)*a*cos(d*x + c)^2 + (A + C)*a*cos(d*x + c))*sqrt(-1/a)*log(-(2*sqrt(2)*sqrt((a*cos(d*x
+ c) + a)/cos(d*x + c))*sqrt(-1/a)*cos(d*x + c)*sin(d*x + c) - 3*cos(d*x + c)^2 - 2*cos(d*x + c) + 1)/(cos(d*
x + c)^2 + 2*cos(d*x + c) + 1)) - 4*(C*cos(d*x + c) - C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))
/(a*d*cos(d*x + c)^2 + a*d*cos(d*x + c)), -1/3*(2*(C*cos(d*x + c) - C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))
*sin(d*x + c) + 3*sqrt(2)*((A + C)*a*cos(d*x + c)^2 + (A + C)*a*cos(d*x + c))*arctan(sqrt(2)*sqrt((a*cos(d*x +
c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d*x + c)^2 + a*d*cos(d*x + c))]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 9.10977, size = 193, normalized size = 1.77 \begin{align*} \frac{\frac{4 \, \sqrt{2} C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} + \frac{3 \, \sqrt{2}{\left (A + C\right )} \log \left ({\left | -\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt{-a} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(4*sqrt(2)*C*a*tan(1/2*d*x + 1/2*c)^3/((a*tan(1/2*d*x + 1/2*c)^2 - a)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*
sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) + 3*sqrt(2)*(A + C)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2*
d*x + 1/2*c)^2 + a)))/(sqrt(-a)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d