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

Optimal. Leaf size=152 \[ -\frac{\sqrt{2} (A+C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 (15 A+14 C) \tan (c+d x)}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt{a \sec (c+d x)+a}}-\frac{2 C \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{15 a d} \]

[Out]

-((Sqrt[2]*(A + C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d)) + (2*(15*A
+ 14*C)*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*C*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*Sqrt[a + a*Sec[
c + d*x]]) - (2*C*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(15*a*d)

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Rubi [A]  time = 0.416758, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4089, 4010, 4001, 3795, 203} \[ -\frac{\sqrt{2} (A+C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 (15 A+14 C) \tan (c+d x)}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt{a \sec (c+d x)+a}}-\frac{2 C \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{15 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^2*(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)) + (2*(15*A
+ 14*C)*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*C*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*Sqrt[a + a*Sec[
c + d*x]]) - (2*C*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(15*a*d)

Rule 4089

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

Rule 4010

Int[csc[(e_.) + (f_.)*(x_)]^2*(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 + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), I
nt[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !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 ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx &=\frac{2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}+\frac{2 \int \frac{\sec ^2(c+d x) \left (\frac{1}{2} a (5 A+4 C)-\frac{1}{2} a C \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{5 a}\\ &=\frac{2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}-\frac{2 C \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 a d}+\frac{4 \int \frac{\sec (c+d x) \left (-\frac{a^2 C}{4}+\frac{1}{4} a^2 (15 A+14 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{15 a^2}\\ &=\frac{2 (15 A+14 C) \tan (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}-\frac{2 C \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 a d}+(-A-C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=\frac{2 (15 A+14 C) \tan (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}-\frac{2 C \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 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{2 (15 A+14 C) \tan (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}-\frac{2 C \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 a d}\\ \end{align*}

Mathematica [A]  time = 4.33884, size = 160, normalized size = 1.05 \[ \frac{2 \cos ^2(c+d x) \sqrt{a (\sec (c+d x)+1)} \left (A+C \sec ^2(c+d x)\right ) \left (\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) ((15 A+13 C) \cos (2 (c+d x))+15 A-2 C \cos (c+d x)+19 C)-15 \sqrt{2} (A+C) \cot (c+d x) \sqrt{\sec (c+d x)-1} \tan ^{-1}\left (\frac{\sqrt{\sec (c+d x)-1}}{\sqrt{2}}\right )\right )}{15 a d (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[c + d*x]^2*Sqrt[a*(1 + Sec[c + d*x])]*(A + C*Sec[c + d*x]^2)*(-15*Sqrt[2]*(A + C)*ArcTan[Sqrt[-1 + Sec[
c + d*x]]/Sqrt[2]]*Cot[c + d*x]*Sqrt[-1 + Sec[c + d*x]] + (15*A + 19*C - 2*C*Cos[c + d*x] + (15*A + 13*C)*Cos[
2*(c + d*x)])*Sec[c + d*x]^2*Tan[(c + d*x)/2]))/(15*a*d*(A + 2*C + A*Cos[2*(c + d*x)]))

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Maple [B]  time = 0.346, size = 586, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60/d/a*(15*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))^(5/2)*sin(d*x+c)*cos(d*x+c)^2+15*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))^(5/2)*sin(d*x+c)*cos(d*x+c)^2+30*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))^(5/2)*sin(d*x+c)*cos(
d*x+c)+30*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)/(co
s(d*x+c)+1))^(5/2)*sin(d*x+c)*cos(d*x+c)+15*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))^(5/2)*sin(d*x+c)+15*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))^(5/2)*sin(d*x+c)+120*A*cos(d*x+c)^3+104*
C*cos(d*x+c)^3-120*A*cos(d*x+c)^2-112*C*cos(d*x+c)^2+32*C*cos(d*x+c)-24*C)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)
/cos(d*x+c)^2/sin(d*x+c)

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

[Out]

Timed out

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Fricas [A]  time = 0.605205, size = 998, normalized size = 6.57 \begin{align*} \left [\frac{15 \, \sqrt{2}{\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{3} +{\left (A + C\right )} a \cos \left (d x + c\right )^{2}\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 ({\left (15 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{2} - C \cos \left (d x + c\right ) + 3 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{30 \,{\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}}, \frac{2 \,{\left ({\left (15 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{2} - C \cos \left (d x + c\right ) + 3 \, 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{15 \, \sqrt{2}{\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{3} +{\left (A + C\right )} a \cos \left (d x + c\right )^{2}\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}}}{15 \,{\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30*(15*sqrt(2)*((A + C)*a*cos(d*x + c)^3 + (A + C)*a*cos(d*x + c)^2)*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*((15*A + 13*C)*cos(d*x + c)^2 - C*cos(d*x + c) + 3*C)*sqrt((a*cos(d*x +
c) + a)/cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2), 1/15*(2*((15*A + 13*C)*cos(d*x
+ c)^2 - C*cos(d*x + c) + 3*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) + 15*sqrt(2)*((A + C)*a*c
os(d*x + c)^3 + (A + C)*a*cos(d*x + c)^2)*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)^3 + a*d*cos(d*x + c)^2)]

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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 ^{2}{\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)**2*(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)**2/sqrt(a*(sec(c + d*x) + 1)), x)

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Giac [B]  time = 8.92879, size = 394, normalized size = 2.59 \begin{align*} -\frac{\frac{15 \,{\left (\sqrt{2} A + \sqrt{2} 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 )} + \frac{2 \,{\left (15 \, \sqrt{2} A a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) + 15 \, \sqrt{2} C a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) -{\left (30 \, \sqrt{2} A a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) + 20 \, \sqrt{2} C a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) -{\left (15 \, \sqrt{2} A a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) + 17 \, \sqrt{2} C a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*(sqrt(2)*A + sqrt(2)*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)) + 2*(15*sqrt(2)*A*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 15*sqrt(
2)*C*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - (30*sqrt(2)*A*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 20*sqrt(2)*C*a^
2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - (15*sqrt(2)*A*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 17*sqrt(2)*C*a^2*sgn(t
an(1/2*d*x + 1/2*c)^2 - 1))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d
*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d