\(\int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) [1250]
Optimal result
Integrand size = 45, antiderivative size = 151 \[
\int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {\sqrt {a} (8 A+4 B+3 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{4 d}+\frac {a (4 B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {C \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x)}
\]
[Out]
1/4*(8*A+4*B+3*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))*a^(1/2)*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)
/d+1/4*a*(4*B+C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+1/2*C*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/
d/cos(d*x+c)^(3/2)
Rubi [A] (verified)
Time = 0.54 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00, number of
steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4350, 4173, 4101, 3886, 221}
\[
\int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {\sqrt {a} (8 A+4 B+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{4 d}+\frac {a (4 B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {C \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d \cos ^{\frac {3}{2}}(c+d x)}
\]
[In]
Int[(Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[Cos[c + d*x]],x]
[Out]
(Sqrt[a]*(8*A + 4*B + 3*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Se
c[c + d*x]])/(4*d) + (a*(4*B + C)*Sin[c + d*x])/(4*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (C*Sqrt[a
+ a*Sec[c + d*x]]*Sin[c + d*x])/(2*d*Cos[c + d*x]^(3/2))
Rule 221
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 3886
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2*(a/(b
*f))*Sqrt[a*(d/b)], 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 4101
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(
B_.) + (A_)), x_Symbol] :> Simp[-2*b*B*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])
), x] + Dist[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1)), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^n, x], x]
/; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n
, 0] && !LtQ[n, 0]
Rule 4173
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + 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 + b*B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A
, B, C, m, n}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]
Rule 4350
Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]
Rubi steps \begin{align*}
\text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx \\ & = \frac {C \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \left (\frac {1}{2} a (4 A+C)+\frac {1}{2} a (4 B+C) \sec (c+d x)\right ) \, dx}{2 a} \\ & = \frac {a (4 B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {C \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {1}{8} \left ((8 A+4 B+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx \\ & = \frac {a (4 B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {C \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left ((8 A+4 B+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \text {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 )}{4 d} \\ & = \frac {\sqrt {a} (8 A+4 B+3 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{4 d}+\frac {a (4 B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {C \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x)} \\
\end{align*}
Mathematica [A] (verified)
Time = 2.61 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.15
\[
\int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {a \sqrt {\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \left ((4 B+3 C) \arcsin \left (\sqrt {1-\sec (c+d x)}\right )-8 A \arcsin \left (\sqrt {\sec (c+d x)}\right )+2 C \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+4 B \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}+3 C \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}\right ) \sin (c+d x)}{4 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}}
\]
[In]
Integrate[(Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[Cos[c + d*x]],x]
[Out]
(a*Sqrt[Cos[c + d*x]]*Sec[c + d*x]^(3/2)*((4*B + 3*C)*ArcSin[Sqrt[1 - Sec[c + d*x]]] - 8*A*ArcSin[Sqrt[Sec[c +
d*x]]] + 2*C*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(3/2) + 4*B*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])] + 3*C*
Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])])*Sin[c + d*x])/(4*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])
])
Maple [B] (verified)
Leaf count of result is larger than twice the leaf count of optimal. \(449\) vs. \(2(127)=254\).
Time = 0.91 (sec) , antiderivative size = 450, normalized size of antiderivative = 2.98
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| method | result | size |
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| default |
\(\frac {\left (8 A \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-8 A \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+4 B \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-4 B \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+8 B \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+3 C \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-3 C \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+6 C \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+4 C \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{8 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )^{\frac {3}{2}}}\) |
\(450\) |
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[In]
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/cos(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
[Out]
1/8/d*(8*A*cos(d*x+c)^2*arctan(1/2*(cos(d*x+c)-sin(d*x+c)+1)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))-8*A*cos
(d*x+c)^2*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))+4*B*cos(d*x+c)^2*arct
an(1/2*(cos(d*x+c)-sin(d*x+c)+1)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))-4*B*cos(d*x+c)^2*arctan(1/2*(cos(d*
x+c)+sin(d*x+c)+1)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))+8*B*cos(d*x+c)*sin(d*x+c)*(-1/(1+cos(d*x+c)))^(1/
2)+3*C*cos(d*x+c)^2*arctan(1/2*(cos(d*x+c)-sin(d*x+c)+1)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))-3*C*cos(d*x
+c)^2*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))+6*C*cos(d*x+c)*sin(d*x+c)
*(-1/(1+cos(d*x+c)))^(1/2)+4*C*sin(d*x+c)*(-1/(1+cos(d*x+c)))^(1/2))*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))/(
-1/(1+cos(d*x+c)))^(1/2)/cos(d*x+c)^(3/2)
Fricas [A] (verification not implemented)
none
Time = 0.43 (sec) , antiderivative size = 413, normalized size of antiderivative = 2.74
\[
\int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\left [\frac {4 \, {\left ({\left (4 \, B + 3 \, C\right )} \cos \left (d x + c\right ) + 2 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left ({\left (8 \, A + 4 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (8 \, A + 4 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 4 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{16 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, \frac {2 \, {\left ({\left (4 \, B + 3 \, C\right )} \cos \left (d x + c\right ) + 2 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left ({\left (8 \, A + 4 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (8 \, A + 4 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{8 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ]
\]
[In]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/cos(d*x+c)^(1/2),x, algorithm="fricas")
[Out]
[1/16*(4*((4*B + 3*C)*cos(d*x + c) + 2*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x +
c) + ((8*A + 4*B + 3*C)*cos(d*x + c)^3 + (8*A + 4*B + 3*C)*cos(d*x + c)^2)*sqrt(a)*log((a*cos(d*x + c)^3 - 4*
sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(cos(d*x + c) - 2)*sqrt(cos(d*x + c))*sin(d*x + c) - 7*a*cos(d
*x + c)^2 + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2), 1/8*(2*((4*B + 3*C
)*cos(d*x + c) + 2*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + ((8*A + 4*B +
3*C)*cos(d*x + c)^3 + (8*A + 4*B + 3*C)*cos(d*x + c)^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/c
os(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)))/(d*cos(d*x + c)^3 + d
*cos(d*x + c)^2)]
Sympy [F]
\[
\int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\int \frac {\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )}{\sqrt {\cos {\left (c + d x \right )}}}\, dx
\]
[In]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)*(a+a*sec(d*x+c))**(1/2)/cos(d*x+c)**(1/2),x)
[Out]
Integral(sqrt(a*(sec(c + d*x) + 1))*(A + B*sec(c + d*x) + C*sec(c + d*x)**2)/sqrt(cos(c + d*x)), x)
Maxima [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 2167 vs. \(2 (127) = 254\).
Time = 0.55 (sec) , antiderivative size = 2167, normalized size of antiderivative = 14.35
\[
\int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\text {Too large to display}
\]
[In]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/cos(d*x+c)^(1/2),x, algorithm="maxima")
[Out]
1/16*(8*A*sqrt(a)*(log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) +
2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(
1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)
^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - log(2*cos(1/2*d*x + 1/2*c)^2 + 2*s
in(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2)) - 4*(4*sqrt(2)*c
os(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*ar
ctan2(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*ar
ctan2(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) + (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*arctan
2(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*arctan
2(sin(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) - 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))))*B*sqrt(a)/(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x
+ 2*c) + 1) - (12*(sqrt(2)*sin(4*d*x + 4*c) + 2*sqrt(2)*sin(2*d*x + 2*c))*cos(7/2*arctan2(sin(d*x + c), cos(d
*x + c))) + 4*(sqrt(2)*sin(4*d*x + 4*c) + 2*sqrt(2)*sin(2*d*x + 2*c))*cos(5/2*arctan2(sin(d*x + c), cos(d*x +
c))) - 4*(sqrt(2)*sin(4*d*x + 4*c) + 2*sqrt(2)*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(d*x + c), cos(d*x + c)))
- 12*(sqrt(2)*sin(4*d*x + 4*c) + 2*sqrt(2)*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) - 3*
(2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2
+ 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(s
in(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(s
in(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) + 3*(2*(2*cos(2*d*x
+ 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x +
4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*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) - 3*(2*(2*cos(2*d*x + 2*c) + 1)*cos(
4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x +
2*c) + 4*sin(2*d*x + 2*c)^2 + 4*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) + 3*(2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + c
os(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*
d*x + 2*c)^2 + 4*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*arctan
2(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) - 12*(sqrt(2)*cos(4*d*x + 4*c) + 2*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2
))*sin(7/2*arctan2(sin(d*x + c), cos(d*x + c))) - 4*(sqrt(2)*cos(4*d*x + 4*c) + 2*sqrt(2)*cos(2*d*x + 2*c) + s
qrt(2))*sin(5/2*arctan2(sin(d*x + c), cos(d*x + c))) + 4*(sqrt(2)*cos(4*d*x + 4*c) + 2*sqrt(2)*cos(2*d*x + 2*c
) + sqrt(2))*sin(3/2*arctan2(sin(d*x + c), cos(d*x + c))) + 12*(sqrt(2)*cos(4*d*x + 4*c) + 2*sqrt(2)*cos(2*d*x
+ 2*c) + sqrt(2))*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))))*C*sqrt(a)/(2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d
*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*
c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1))/d
Giac [F]
\[
\int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {a \sec \left (d x + c\right ) + a}}{\sqrt {\cos \left (d x + c\right )}} \,d x }
\]
[In]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/cos(d*x+c)^(1/2),x, algorithm="giac")
[Out]
sage0*x
Mupad [F(-1)]
Timed out. \[
\int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\int \frac {\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\sqrt {\cos \left (c+d\,x\right )}} \,d x
\]
[In]
int(((a + a/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/cos(c + d*x)^(1/2),x)
[Out]
int(((a + a/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/cos(c + d*x)^(1/2), x)