3.1133 \(\int \frac {\sqrt {a+a \sec (c+d x)} (A+C \sec ^2(c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=189 \[ \frac {a (8 A+5 C) \sin (c+d x)}{8 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {\sqrt {a} (8 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{8 d}+\frac {C \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {a C \sin (c+d x)}{12 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}} \]
[Out]
1/8*(8*A+5*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
/12*a*C*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(1/2)+1/8*a*(8*A+5*C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+
a*sec(d*x+c))^(1/2)+1/3*C*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d/cos(d*x+c)^(5/2)
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Rubi [A] time = 0.48, antiderivative size = 189, 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 =
{4265, 4089, 4016, 3803, 3801, 215} \[ \frac {a (8 A+5 C) \sin (c+d x)}{8 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {\sqrt {a} (8 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{8 d}+\frac {C \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {a C \sin (c+d x)}{12 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}} \]
Antiderivative was successfully verified.
[In]
Int[(Sqrt[a + a*Sec[c + d*x]]*(A + C*Sec[c + d*x]^2))/Cos[c + d*x]^(3/2),x]
[Out]
(Sqrt[a]*(8*A + 5*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c +
d*x]])/(8*d) + (a*C*Sin[c + d*x])/(12*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) + (a*(8*A + 5*C)*Sin[c +
d*x])/(8*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])/(3*d*Cos[c
+ d*x]^(5/2))
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 3803
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*b*d
*Cot[e + f*x]*(d*Csc[e + f*x])^(n - 1))/(f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[(2*a*d*(n - 1))/(b*(
2*n - 1)), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a
^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]
Rule 4016
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 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 4265
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*} \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx\\ &=\frac {C \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {3}{2} a (2 A+C)+\frac {1}{2} a C \sec (c+d x)\right ) \, dx}{3 a}\\ &=\frac {a C \sin (c+d x)}{12 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {C \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {1}{8} \left ((8 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a C \sin (c+d x)}{12 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a (8 A+5 C) \sin (c+d x)}{8 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)}{3 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {1}{16} \left ((8 A+5 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 C \sin (c+d x)}{12 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a (8 A+5 C) \sin (c+d x)}{8 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)}{3 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\left ((8 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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 )}{8 d}\\ &=\frac {\sqrt {a} (8 A+5 C) \sinh ^{-1}\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)}}{8 d}+\frac {a C \sin (c+d x)}{12 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a (8 A+5 C) \sin (c+d x)}{8 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)}{3 d \cos ^{\frac {5}{2}}(c+d x)}\\ \end {align*}
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Mathematica [A] time = 0.90, size = 125, normalized size = 0.66 \[ \frac {\sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sec (c+d x)+1)} \left (\sin \left (\frac {1}{2} (c+d x)\right ) (3 (8 A+5 C) \cos (2 (c+d x))+24 A+20 C \cos (c+d x)+31 C)+3 \sqrt {2} (8 A+5 C) \cos ^3(c+d x) \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{48 d \cos ^{\frac {5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Integrate[(Sqrt[a + a*Sec[c + d*x]]*(A + C*Sec[c + d*x]^2))/Cos[c + d*x]^(3/2),x]
[Out]
(Sec[(c + d*x)/2]*Sqrt[a*(1 + Sec[c + d*x])]*(3*Sqrt[2]*(8*A + 5*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c +
d*x]^3 + (24*A + 31*C + 20*C*Cos[c + d*x] + 3*(8*A + 5*C)*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(48*d*Cos[c + d
*x]^(5/2))
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fricas [A] time = 0.60, size = 427, normalized size = 2.26 \[ \left [\frac {4 \, {\left (3 \, {\left (8 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{2} + 10 \, C \cos \left (d x + c\right ) + 8 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, {\left ({\left (8 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (8 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 4 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{96 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}, \frac {2 \, {\left (3 \, {\left (8 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{2} + 10 \, C \cos \left (d x + c\right ) + 8 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, {\left ({\left (8 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (8 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{48 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}\right ] \]
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))^(1/2)/cos(d*x+c)^(3/2),x, algorithm="fricas")
[Out]
[1/96*(4*(3*(8*A + 5*C)*cos(d*x + c)^2 + 10*C*cos(d*x + c) + 8*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt
(cos(d*x + c))*sin(d*x + c) + 3*((8*A + 5*C)*cos(d*x + c)^4 + (8*A + 5*C)*cos(d*x + c)^3)*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)^4 + d*cos(d*x + c)^3), 1/48
*(2*(3*(8*A + 5*C)*cos(d*x + c)^2 + 10*C*cos(d*x + c) + 8*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(
d*x + c))*sin(d*x + c) + 3*((8*A + 5*C)*cos(d*x + c)^4 + (8*A + 5*C)*cos(d*x + c)^3)*sqrt(-a)*arctan(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)))/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)]
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt {a \sec \left (d x + c\right ) + a}}{\cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]
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))^(1/2)/cos(d*x+c)^(3/2),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + A)*sqrt(a*sec(d*x + c) + a)/cos(d*x + c)^(3/2), x)
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maple [B] time = 1.96, size = 375, normalized size = 1.98 \[ -\frac {\left (-1+\cos \left (d x +c \right )\right ) \left (24 A \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right )-24 A \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right )+15 C \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right )-15 C \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right )+48 A \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}+30 C \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+20 C \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )+16 C \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{48 d \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{\frac {5}{2}}} \]
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))^(1/2)/cos(d*x+c)^(3/2),x)
[Out]
-1/48/d*(-1+cos(d*x+c))*(24*A*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)^3-24*A*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)^3
+15*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)^3-15*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)^3+48*A*sin(d*x+c)*cos(d*x
+c)^2*(-2/(1+cos(d*x+c)))^(1/2)+30*C*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*cos(d*x+c)^2+20*C*(-2/(1+cos(d*x+c))
)^(1/2)*sin(d*x+c)*cos(d*x+c)+16*C*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c))*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/(
-2/(1+cos(d*x+c)))^(1/2)/sin(d*x+c)^2/cos(d*x+c)^(5/2)
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maxima [B] time = 0.87, size = 2740, normalized size = 14.50 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/cos(d*x+c)^(3/2),x, algorithm="maxima")
[Out]
-1/96*(24*(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*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*arcta
n2(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) - 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))))*A*sqrt(a)/(cos(2*d*x + 2*c)^2 + sin(2*d*x
+ 2*c)^2 + 2*cos(2*d*x + 2*c) + 1) + (60*(sqrt(2)*sin(6*d*x + 6*c) + 3*sqrt(2)*sin(4*d*x + 4*c) + 3*sqrt(2)*si
n(2*d*x + 2*c))*cos(11/2*arctan2(sin(d*x + c), cos(d*x + c))) + 20*(sqrt(2)*sin(6*d*x + 6*c) + 3*sqrt(2)*sin(4
*d*x + 4*c) + 3*sqrt(2)*sin(2*d*x + 2*c))*cos(9/2*arctan2(sin(d*x + c), cos(d*x + c))) + 168*(sqrt(2)*sin(6*d*
x + 6*c) + 3*sqrt(2)*sin(4*d*x + 4*c) + 3*sqrt(2)*sin(2*d*x + 2*c))*cos(7/2*arctan2(sin(d*x + c), cos(d*x + c)
)) - 168*(sqrt(2)*sin(6*d*x + 6*c) + 3*sqrt(2)*sin(4*d*x + 4*c) + 3*sqrt(2)*sin(2*d*x + 2*c))*cos(5/2*arctan2(
sin(d*x + c), cos(d*x + c))) - 20*(sqrt(2)*sin(6*d*x + 6*c) + 3*sqrt(2)*sin(4*d*x + 4*c) + 3*sqrt(2)*sin(2*d*x
+ 2*c))*cos(3/2*arctan2(sin(d*x + c), cos(d*x + c))) - 60*(sqrt(2)*sin(6*d*x + 6*c) + 3*sqrt(2)*sin(4*d*x + 4
*c) + 3*sqrt(2)*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) - 15*(2*(3*cos(4*d*x + 4*c) + 3
*cos(2*d*x + 2*c) + 1)*cos(6*d*x + 6*c) + cos(6*d*x + 6*c)^2 + 6*(3*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 9
*cos(4*d*x + 4*c)^2 + 9*cos(2*d*x + 2*c)^2 + 6*(sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + sin(6*
d*x + 6*c)^2 + 9*sin(4*d*x + 4*c)^2 + 18*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 9*sin(2*d*x + 2*c)^2 + 6*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) + 15*(2*(3*cos(4*d*x + 4*c) + 3*cos(2*d*x + 2*c) + 1)*cos(6*d*x + 6*c) + cos(6*d*x + 6*c)^2 +
6*(3*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 9*cos(4*d*x + 4*c)^2 + 9*cos(2*d*x + 2*c)^2 + 6*(sin(4*d*x + 4*
c) + sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + sin(6*d*x + 6*c)^2 + 9*sin(4*d*x + 4*c)^2 + 18*sin(4*d*x + 4*c)*sin(
2*d*x + 2*c) + 9*sin(2*d*x + 2*c)^2 + 6*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) - 15*(2*(3*cos(4*d*x + 4*c) + 3*cos(2*d*x + 2
*c) + 1)*cos(6*d*x + 6*c) + cos(6*d*x + 6*c)^2 + 6*(3*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 9*cos(4*d*x + 4
*c)^2 + 9*cos(2*d*x + 2*c)^2 + 6*(sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + sin(6*d*x + 6*c)^2 +
9*sin(4*d*x + 4*c)^2 + 18*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 9*sin(2*d*x + 2*c)^2 + 6*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*sq
rt(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) + 15*(2*(3*cos(4*d*x + 4*c) + 3*cos(2*d*x + 2*c) + 1)*cos(6*d*x + 6*c) + cos(6*d*x + 6*c)^2 + 6*(3*cos(2*d*
x + 2*c) + 1)*cos(4*d*x + 4*c) + 9*cos(4*d*x + 4*c)^2 + 9*cos(2*d*x + 2*c)^2 + 6*(sin(4*d*x + 4*c) + sin(2*d*x
+ 2*c))*sin(6*d*x + 6*c) + sin(6*d*x + 6*c)^2 + 9*sin(4*d*x + 4*c)^2 + 18*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) +
9*sin(2*d*x + 2*c)^2 + 6*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) - 60*(sqrt(2)*cos(6*d*x + 6*c) + 3*sqrt(2)*cos(4*d*x + 4*c)
+ 3*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*sin(11/2*arctan2(sin(d*x + c), cos(d*x + c))) - 20*(sqrt(2)*cos(6*d*x
+ 6*c) + 3*sqrt(2)*cos(4*d*x + 4*c) + 3*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*sin(9/2*arctan2(sin(d*x + c), cos
(d*x + c))) - 168*(sqrt(2)*cos(6*d*x + 6*c) + 3*sqrt(2)*cos(4*d*x + 4*c) + 3*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2
))*sin(7/2*arctan2(sin(d*x + c), cos(d*x + c))) + 168*(sqrt(2)*cos(6*d*x + 6*c) + 3*sqrt(2)*cos(4*d*x + 4*c) +
3*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*sin(5/2*arctan2(sin(d*x + c), cos(d*x + c))) + 20*(sqrt(2)*cos(6*d*x +
6*c) + 3*sqrt(2)*cos(4*d*x + 4*c) + 3*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*sin(3/2*arctan2(sin(d*x + c), cos(d*
x + c))) + 60*(sqrt(2)*cos(6*d*x + 6*c) + 3*sqrt(2)*cos(4*d*x + 4*c) + 3*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*s
in(1/2*arctan2(sin(d*x + c), cos(d*x + c))))*C*sqrt(a)/(2*(3*cos(4*d*x + 4*c) + 3*cos(2*d*x + 2*c) + 1)*cos(6*
d*x + 6*c) + cos(6*d*x + 6*c)^2 + 6*(3*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 9*cos(4*d*x + 4*c)^2 + 9*cos(2
*d*x + 2*c)^2 + 6*(sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + sin(6*d*x + 6*c)^2 + 9*sin(4*d*x +
4*c)^2 + 18*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 9*sin(2*d*x + 2*c)^2 + 6*cos(2*d*x + 2*c) + 1))/d
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}}{{\cos \left (c+d\,x\right )}^{3/2}} \,d x \]
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))^(1/2))/cos(c + d*x)^(3/2),x)
[Out]
int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(1/2))/cos(c + d*x)^(3/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + C \sec ^{2}{\left (c + d x \right )}\right )}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
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))**(1/2)/cos(d*x+c)**(3/2),x)
[Out]
Integral(sqrt(a*(sec(c + d*x) + 1))*(A + C*sec(c + d*x)**2)/cos(c + d*x)**(3/2), x)
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