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

Optimal. Leaf size=133 \[ \frac {\sqrt {2} (A+C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {C \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d} \]

[Out]

-C*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d/a^(1/2)+(A+C)*arctanh(1/2*sin(d*x+c)*a^(1/2)*sec(d*x+c
)^(1/2)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)+C*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/
2)

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Rubi [A]  time = 0.37, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {4089, 4023, 3808, 206, 3801, 215} \[ \frac {\sqrt {2} (A+C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {C \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((C*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(Sqrt[a]*d)) + (Sqrt[2]*(A + C)*ArcTanh[(Sqrt[a
]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d) + (C*Sec[c + d*x]^(3/2)*Si
n[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])

Rule 206

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

Rule 215

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

Rule 3801

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

Rule 3808

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

Rule 4023

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

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]

Rubi steps

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

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Mathematica [B]  time = 6.77, size = 717, normalized size = 5.39 \[ \frac {(2 A+C) \sin (c+d x) \cos ^4(c+d x) (\sec (c+d x)+1)^{3/2} \sqrt {\sec ^2(c+d x)-1} \left (\log \left (-3 \sec ^2(c+d x)-2 \sqrt {2} \sqrt {\sec (c+d x)+1} \sqrt {\sec ^2(c+d x)-1} \sqrt {\sec (c+d x)}-2 \sec (c+d x)+1\right )-\log \left (-3 \sec ^2(c+d x)+2 \sqrt {2} \sqrt {\sec (c+d x)+1} \sqrt {\sec ^2(c+d x)-1} \sqrt {\sec (c+d x)}-2 \sec (c+d x)+1\right )\right ) \left (A+C \sec ^2(c+d x)\right )}{2 d (\cos (c+d x)+1) \sqrt {2-2 \cos ^2(c+d x)} \sqrt {1-\cos ^2(c+d x)} \sqrt {a (\sec (c+d x)+1)} (A \cos (2 c+2 d x)+A+2 C)}-\frac {C \sin (c+d x) \cos ^4(c+d x) (\sec (c+d x)+1)^{3/2} \sqrt {\sec ^2(c+d x)-1} \left (\sqrt {2} \left (\log \left (-3 \sec ^2(c+d x)+2 \sqrt {2} \sqrt {\sec (c+d x)+1} \sqrt {\sec ^2(c+d x)-1} \sqrt {\sec (c+d x)}-2 \sec (c+d x)+1\right )-\log \left (-3 \sec ^2(c+d x)-2 \sqrt {2} \sqrt {\sec (c+d x)+1} \sqrt {\sec ^2(c+d x)-1} \sqrt {\sec (c+d x)}-2 \sec (c+d x)+1\right )\right )+8 \log \left (\sec ^{\frac {3}{2}}(c+d x)+\sqrt {\sec (c+d x)+1} \sqrt {\sec ^2(c+d x)-1}+\sqrt {\sec (c+d x)}\right )-8 \log (\sec (c+d x)+1)\right ) \left (A+C \sec ^2(c+d x)\right )}{4 d (\cos (c+d x)+1) \left (1-\cos ^2(c+d x)\right ) \sqrt {a (\sec (c+d x)+1)} (A \cos (2 c+2 d x)+A+2 C)}+\frac {\sqrt {\sec (c+d x)+1} \sqrt {(\cos (c+d x)+1) \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \left (\frac {2 C \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}+\frac {2 C \tan \left (\frac {c}{2}\right )}{d}\right )}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a (\sec (c+d x)+1)} (A \cos (2 c+2 d x)+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((2*A + C)*Cos[c + d*x]^4*(Log[1 - 2*Sec[c + d*x] - 3*Sec[c + d*x]^2 - 2*Sqrt[2]*Sqrt[Sec[c + d*x]]*Sqrt[1 + S
ec[c + d*x]]*Sqrt[-1 + Sec[c + d*x]^2]] - Log[1 - 2*Sec[c + d*x] - 3*Sec[c + d*x]^2 + 2*Sqrt[2]*Sqrt[Sec[c + d
*x]]*Sqrt[1 + Sec[c + d*x]]*Sqrt[-1 + Sec[c + d*x]^2]])*(1 + Sec[c + d*x])^(3/2)*Sqrt[-1 + Sec[c + d*x]^2]*(A
+ C*Sec[c + d*x]^2)*Sin[c + d*x])/(2*d*(1 + Cos[c + d*x])*Sqrt[2 - 2*Cos[c + d*x]^2]*Sqrt[1 - Cos[c + d*x]^2]*
(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[a*(1 + Sec[c + d*x])]) - (C*Cos[c + d*x]^4*(-8*Log[1 + Sec[c + d*x]] + 8*L
og[Sqrt[Sec[c + d*x]] + Sec[c + d*x]^(3/2) + Sqrt[1 + Sec[c + d*x]]*Sqrt[-1 + Sec[c + d*x]^2]] + Sqrt[2]*(-Log
[1 - 2*Sec[c + d*x] - 3*Sec[c + d*x]^2 - 2*Sqrt[2]*Sqrt[Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]*Sqrt[-1 + Sec[c +
d*x]^2]] + Log[1 - 2*Sec[c + d*x] - 3*Sec[c + d*x]^2 + 2*Sqrt[2]*Sqrt[Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]*Sq
rt[-1 + Sec[c + d*x]^2]]))*(1 + Sec[c + d*x])^(3/2)*Sqrt[-1 + Sec[c + d*x]^2]*(A + C*Sec[c + d*x]^2)*Sin[c + d
*x])/(4*d*(1 + Cos[c + d*x])*(1 - Cos[c + d*x]^2)*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[a*(1 + Sec[c + d*x])]) +
(Sqrt[(1 + Cos[c + d*x])*Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]*(A + C*Sec[c + d*x]^2)*((2*C*Sec[c/2]*Sec[c/2 +
(d*x)/2]*Sin[(d*x)/2])/d + (2*C*Tan[c/2])/d))/((A + 2*C + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(3/2)*Sqrt[a*(1 +
Sec[c + d*x])])

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fricas [A]  time = 0.49, size = 501, normalized size = 3.77 \[ \left [\frac {{\left (C \cos \left (d x + c\right ) + C\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {2 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right ) + {\left (A + C\right )} a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - \frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}} + \frac {4 \, C \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}, -\frac {2 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right ) + {\left (A + C\right )} a\right )} \sqrt {-\frac {1}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) + {\left (C \cos \left (d x + c\right ) + C\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right ) - \frac {2 \, C \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}\right ] \]

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))^(1/2),x, algorithm="fricas")

[Out]

[1/4*((C*cos(d*x + c) + C)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 2*cos(d*x
+ c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 +
cos(d*x + c)^2)) + 2*sqrt(2)*((A + C)*a*cos(d*x + c) + (A + C)*a)*log(-(cos(d*x + c)^2 - 2*sqrt(2)*sqrt((a*co
s(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2 +
2*cos(d*x + c) + 1))/sqrt(a) + 4*C*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a
*d*cos(d*x + c) + a*d), -1/2*(2*sqrt(2)*((A + C)*a*cos(d*x + c) + (A + C)*a)*sqrt(-1/a)*arctan(sqrt(2)*sqrt((a
*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*sqrt(cos(d*x + c))/sin(d*x + c)) + (C*cos(d*x + c) + C)*sqrt(-a)*a
rctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a
*cos(d*x + c) - 2*a)) - 2*C*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(
d*x + c) + a*d)]

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

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))^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 2.53, size = 252, normalized size = 1.89 \[ \frac {\sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (C \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \sqrt {2}\, \cos \left (d x +c \right )-C \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \sqrt {2}\, \cos \left (d x +c \right )+4 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \cos \left (d x +c \right )+2 C \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+4 C \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \cos \left (d x +c \right )\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )-1\right )}{4 d \sin \left (d x +c \right )^{2} a} \]

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))^(1/2),x)

[Out]

1/4/d*(1/cos(d*x+c))^(1/2)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(C*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*
x+c)+1-sin(d*x+c))*2^(1/2))*2^(1/2)*cos(d*x+c)-C*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)
)*2^(1/2))*2^(1/2)*cos(d*x+c)+4*A*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)+2*C*(-2/(1+cos(d
*x+c)))^(1/2)*sin(d*x+c)+4*C*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*cos(d*x+c))*(-2/(1+cos(d*x+c)))^
(1/2)/sin(d*x+c)^2*(cos(d*x+c)^2-1)/a

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maxima [B]  time = 0.87, size = 968, normalized size = 7.28 \[ \text {result too large to display} \]

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))^(1/2),x, algorithm="maxima")

[Out]

1/4*(2*(sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - sqrt(2)*lo
g(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*A/sqrt(a) - (4*sqrt(2)*cos(3/
2*arctan2(sin(d*x + c), cos(d*x + c)))*sin(2*d*x + 2*c) - 4*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)
))*sin(2*d*x + 2*c) + (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2
(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1/2*arctan2
(sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) - (cos(2*d*x + 2*c
)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*si
n(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) - 2*sqrt
(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) + (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x
+ 2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x +
c)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(
d*x + c))) + 2) - (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(sin
(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin
(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) - 2*(sqrt(2)*cos(2*d*x
+ 2*c)^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 2*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*log(cos(1/2*arctan2(sin(d*x + c
), cos(d*x + c)))^2 + sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x
+ c))) + 1) + 2*(sqrt(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 2*sqrt(2)*cos(2*d*x + 2*c) + sqrt(
2))*log(cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 - 2*si
n(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 1) - 4*(sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*sin(3/2*arctan2(sin(d
*x + c), cos(d*x + c))) + 4*(sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))))
*C/((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sqrt(a)))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))**(1/2),x)

[Out]

Timed out

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