\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx\) [1284]

Optimal result

Integrand size = 45, antiderivative size = 242 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\frac {(2 B-3 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{a^{3/2} d}+\frac {(A-5 B+9 C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(A-B+3 C) \sin (c+d x)}{2 a d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \]

[Out]

-1/2*(A-B+C)*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2)+(2*B-3*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*se
c(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^(3/2)/d+1/4*(A-5*B+9*C)*arctanh(1/2*sin(d*x+c)*a^(1/2)*se
c(d*x+c)^(1/2)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^(3/2)/d*2^(1/2)+1/2*(A-B+3*
C)*sin(d*x+c)/a/d/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.178, Rules used = {4350, 4169, 4106, 4108, 3893, 212, 3886, 221} \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\frac {(A-5 B+9 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(2 B-3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{3/2} d}+\frac {(A-B+3 C) \sin (c+d x)}{2 a d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}} \]

[In]

Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2)),x]

[Out]

((2*B - 3*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]])/(
a^(3/2)*d) + ((A - 5*B + 9*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*
x]])]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(2*Sqrt[2]*a^(3/2)*d) - ((A - B + C)*Sin[c + d*x])/(2*d*Cos[c + d
*x]^(5/2)*(a + a*Sec[c + d*x])^(3/2)) + ((A - B + 3*C)*Sin[c + d*x])/(2*a*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Sec[
c + d*x]])

Rule 212

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

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 3893

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 4106

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

Rule 4108

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 4169

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[(-(a*A - b*B + a*C))*Cot[e + f*x]*(a + b*C
sc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*(2*m + 1))), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m
+ 1)*(d*Csc[e + f*x])^n*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m -
n)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

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 \frac {\sec ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx \\ & = -\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (\frac {1}{2} a (A+3 B-3 C)+a (A-B+3 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{2 a^2} \\ & = -\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(A-B+3 C) \sin (c+d x)}{2 a d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)} \left (\frac {1}{2} a^2 (A-B+3 C)+a^2 (2 B-3 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{2 a^3} \\ & = -\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(A-B+3 C) \sin (c+d x)}{2 a d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {\left ((2 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}{2 a^2}+\frac {\left ((A-5 B+9 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a} \\ & = -\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(A-B+3 C) \sin (c+d x)}{2 a d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {\left ((2 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 )}{a^2 d}-\frac {\left ((A-5 B+9 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \text {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 )}{2 a d} \\ & = \frac {(2 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)}}{a^{3/2} d}+\frac {(A-5 B+9 C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(A-B+3 C) \sin (c+d x)}{2 a d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.44 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.82 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\frac {\cos ^3\left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (2 (A-5 B+9 C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-\frac {2 \left (2 \sqrt {2} (2 B-3 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )+(A-B+3 C+2 C \sec (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{-1+\sin ^2\left (\frac {1}{2} (c+d x)\right )}\right )}{d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (a (1+\sec (c+d x)))^{3/2}} \]

[In]

Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2)),x]

[Out]

(Cos[(c + d*x)/2]^3*Sqrt[Cos[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(2*(A - 5*B + 9*C)*ArcTanh[Sin[
(c + d*x)/2]] - (2*(2*Sqrt[2]*(2*B - 3*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^2 + (A - B + 3*C
+ 2*C*Sec[c + d*x])*Sin[(c + d*x)/2]))/(-1 + Sin[(c + d*x)/2]^2)))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c
+ d*x)])*(a*(1 + Sec[c + d*x]))^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(873\) vs. \(2(203)=406\).

Time = 0.97 (sec) , antiderivative size = 874, normalized size of antiderivative = 3.61

method	result	size
default	\(\text {Expression too large to display}\)	\(874\)

[In]

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/4/a^2/d*(-A*2^(1/2)*cos(d*x+c)^2*arctan(1/2*sin(d*x+c)*2^(1/2)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))+5*B
*2^(1/2)*cos(d*x+c)^2*arctan(1/2*sin(d*x+c)*2^(1/2)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))-9*C*2^(1/2)*cos(
d*x+c)^2*arctan(1/2*sin(d*x+c)*2^(1/2)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))+2*A*cos(d*x+c)*sin(d*x+c)*(-1
/(1+cos(d*x+c)))^(1/2)-A*cos(d*x+c)*2^(1/2)*arctan(1/2*sin(d*x+c)*2^(1/2)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(
1/2))-2*B*cos(d*x+c)*sin(d*x+c)*(-1/(1+cos(d*x+c)))^(1/2)-4*B*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))*cos(d*x+c)^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))+5*B*2^(1/2)*cos(d*x+c)*arctan(1/2*sin(d*x+c)*2^(1/2)/(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)+6*C*cos(d*x+c)^2*arctan(1/2*(-cos(d*x+c)+s
in(d*x+c)-1)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))+6*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))-9*C*cos(d*x+c)*2^(1/2)*arctan(1/2*sin(d*x+c)*2^(1/2)/(1+cos(d*x+c))/(
-1/(1+cos(d*x+c)))^(1/2))-4*B*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))*
cos(d*x+c)-4*B*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))*cos(d*x+c)+4*C*s
in(d*x+c)*(-1/(1+cos(d*x+c)))^(1/2)+6*C*cos(d*x+c)*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)*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)))*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))^2/(-1/(1+cos(d*x+c)))^(1/2)/cos(d*x+c)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 728, normalized size of antiderivative = 3.01 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\left [\frac {\sqrt {2} {\left ({\left (A - 5 \, B + 9 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (A - 5 \, B + 9 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (A - 5 \, B + 9 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {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 ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (A - 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 ) - 2 \, {\left ({\left (2 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (2 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, B - 3 \, C\right )} \cos \left (d x + c\right )\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 )}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}}, -\frac {\sqrt {2} {\left ({\left (A - 5 \, B + 9 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (A - 5 \, B + 9 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (A - 5 \, B + 9 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {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 )}}{a \sin \left (d x + c\right )}\right ) - 2 \, {\left ({\left (A - 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 ) - 2 \, {\left ({\left (2 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (2 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, B - 3 \, C\right )} \cos \left (d x + c\right )\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 )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}}\right ] \]

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/8*(sqrt(2)*((A - 5*B + 9*C)*cos(d*x + c)^3 + 2*(A - 5*B + 9*C)*cos(d*x + c)^2 + (A - 5*B + 9*C)*cos(d*x + c
))*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c
))*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*((A - 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) - 2*((2*B - 3*C)*cos(d*x + c
)^3 + 2*(2*B - 3*C)*cos(d*x + c)^2 + (2*B - 3*C)*cos(d*x + c))*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)))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c)),
-1/4*(sqrt(2)*((A - 5*B + 9*C)*cos(d*x + c)^3 + 2*(A - 5*B + 9*C)*cos(d*x + c)^2 + (A - 5*B + 9*C)*cos(d*x + c
))*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))/(a*sin(d*x + c)
)) - 2*((A - 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) - 2*((2*B - 3*C)*cos(d*x + c)^3 + 2*(2*B - 3*C)*cos(d*x + c)^2 + (2*B - 3*C)*cos(d*x + c))*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)))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))]

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/cos(d*x+c)**(3/2)/(a+a*sec(d*x+c))**(3/2),x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^(3/2)*cos(d*x + c)^(3/2)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

[In]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^(3/2)*(a + a/cos(c + d*x))^(3/2)),x)

[Out]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^(3/2)*(a + a/cos(c + d*x))^(3/2)), x)