\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx\) [1242]

Optimal result

Integrand size = 37, antiderivative size = 172 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {(7 A-C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(5 A+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}} \]

[Out]

-1/2*(A+C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(3/2)-1/4*(7*A-C)*arctan(1/2*sin(d*x+c)*a^(1/2)*2^(1
/2)/cos(d*x+c)^(1/2)/(a+a*cos(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*(5*A+C)*s
in(d*x+c)*sec(d*x+c)^(1/2)/a/d/(a+a*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {4306, 3121, 3063, 12, 2861, 211} \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {(7 A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(5 A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \sqrt {a \cos (c+d x)+a}}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}} \]

[In]

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

[Out]

-1/2*((7*A - C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Cos[
c + d*x]]*Sqrt[Sec[c + d*x]])/(Sqrt[2]*a^(3/2)*d) - ((A + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(2*d*(a + a*Cos[
c + d*x])^(3/2)) + ((5*A + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(2*a*d*Sqrt[a + a*Cos[c + d*x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

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

Rule 2861

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[-2*(a/f), Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c +
d*Sin[e + f*x]]))], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 -
d^2, 0]

Rule 3063

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

Rule 3121

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[a*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x
])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x] + Dist[1/(b*(b*c - a*d)*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)
*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2
) + C*(b*c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]

Rule 4306

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx \\ & = -\frac {(A+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} a (5 A+C)-a (A-C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2} \\ & = -\frac {(A+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(5 A+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int -\frac {a^2 (7 A-C)}{4 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{a^3} \\ & = -\frac {(A+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(5 A+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}-\frac {\left ((7 A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{4 a} \\ & = -\frac {(A+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(5 A+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}+\frac {\left ((7 A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{2 a^2+a x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{2 d} \\ & = -\frac {(7 A-C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(5 A+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 3.31 (sec) , antiderivative size = 460, normalized size of antiderivative = 2.67 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {2 \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {3}{2} (A+C) \arctan \left (\frac {1-2 \sin \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\cos (c+d x)}}\right )-\frac {3}{2} (A+C) \arctan \left (\frac {1+2 \sin \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\cos (c+d x)}}\right )-\frac {(A+C) \sqrt {\cos (c+d x)}}{-1+\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {4 C \sin \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\cos (c+d x)}}-\frac {(A+C) \sqrt {\cos (c+d x)}}{1+\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {(A+C) \left (-1+2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{4 \sqrt {\cos (c+d x)} \left (\cos \left (\frac {1}{4} (c+d x)\right )+\sin \left (\frac {1}{4} (c+d x)\right )\right )^2}-\frac {(A+C) \left (1+2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{4 \sqrt {\cos (c+d x)} \left (-1+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {(A-7 C) \csc ^3\left (\frac {1}{2} (c+d x)\right ) \left (5 (1+4 \cos (c+d x)+\cos (2 (c+d x))) \left (1-\cos (c+d x)+\text {arctanh}\left (\sqrt {-\sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )}\right ) \cos (c+d x) \sqrt {2-2 \sec (c+d x)}\right )-2 \operatorname {Hypergeometric2F1}\left (2,\frac {5}{2},\frac {7}{2},-\sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sin ^4\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x) \tan (c+d x)\right )}{40 \cos ^{\frac {3}{2}}(c+d x)}\right )}{d (a (1+\cos (c+d x)))^{3/2}} \]

[In]

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

[Out]

(2*Cos[(c + d*x)/2]^3*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((3*(A + C)*ArcTan[(1 - 2*Sin[(c + d*x)/2])/Sqrt[C
os[c + d*x]]])/2 - (3*(A + C)*ArcTan[(1 + 2*Sin[(c + d*x)/2])/Sqrt[Cos[c + d*x]]])/2 - ((A + C)*Sqrt[Cos[c + d
*x]])/(-1 + Sin[(c + d*x)/2]) + (4*C*Sin[(c + d*x)/2])/Sqrt[Cos[c + d*x]] - ((A + C)*Sqrt[Cos[c + d*x]])/(1 +
Sin[(c + d*x)/2]) + ((A + C)*(-1 + 2*Sin[(c + d*x)/2]))/(4*Sqrt[Cos[c + d*x]]*(Cos[(c + d*x)/4] + Sin[(c + d*x
)/4])^2) - ((A + C)*(1 + 2*Sin[(c + d*x)/2]))/(4*Sqrt[Cos[c + d*x]]*(-1 + Sin[(c + d*x)/2])) + ((A - 7*C)*Csc[
(c + d*x)/2]^3*(5*(1 + 4*Cos[c + d*x] + Cos[2*(c + d*x)])*(1 - Cos[c + d*x] + ArcTanh[Sqrt[-(Sec[c + d*x]*Sin[
(c + d*x)/2]^2)]]*Cos[c + d*x]*Sqrt[2 - 2*Sec[c + d*x]]) - 2*Hypergeometric2F1[2, 5/2, 7/2, -(Sec[c + d*x]*Sin
[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^4*Sin[c + d*x]*Tan[c + d*x]))/(40*Cos[c + d*x]^(3/2))))/(d*(a*(1 + Cos[c +
d*x]))^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(351\) vs. \(2(143)=286\).

Time = 1.31 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.05

method	result	size
default	\(\frac {\sqrt {2}\, \left (7 A \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-C \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+5 A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}+14 A \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+C \sqrt {2}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )-2 C \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+4 A \sqrt {2}\, \sin \left (d x +c \right )+7 A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-C \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \left (\sec ^{\frac {3}{2}}\left (d x +c \right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \cos \left (d x +c \right )}{4 a^{2} d \left (1+\cos \left (d x +c \right )\right )^{2}}\)	\(352\)
parts	\(\frac {A \sqrt {2}\, \left (7 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )+14 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )+5 \sqrt {2}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )+7 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+4 \sqrt {2}\, \sin \left (d x +c \right )\right ) \left (\sec ^{\frac {3}{2}}\left (d x +c \right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \cos \left (d x +c \right )}{4 d \,a^{2} \left (1+\cos \left (d x +c \right )\right )^{2}}-\frac {C \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (\sec ^{\frac {3}{2}}\left (d x +c \right )\right ) \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )+2 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )-\sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )\right )}{4 d \,a^{2} \left (1+\cos \left (d x +c \right )\right )^{2}}\)	\(397\)

[In]

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

[Out]

1/4/a^2/d*2^(1/2)*(7*A*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))-C*cos(d*x+
c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))+5*A*cos(d*x+c)*sin(d*x+c)*2^(1/2)+14*A*co
s(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))+C*2^(1/2)*cos(d*x+c)*sin(d*x+c)-2*C*c
os(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))+4*A*2^(1/2)*sin(d*x+c)+7*A*arcsin(co
t(d*x+c)-csc(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-C*arcsin(cot(d*x+c)-csc(d*x+c))*(cos(d*x+c)/(1+cos(d*x+
c)))^(1/2))*sec(d*x+c)^(3/2)*((1+cos(d*x+c))*a)^(1/2)*cos(d*x+c)/(1+cos(d*x+c))^2

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.94 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {2} {\left ({\left (7 \, A - C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (7 \, A - C\right )} \cos \left (d x + c\right ) + 7 \, A - C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left ({\left (5 \, A + C\right )} \cos \left (d x + c\right ) + 4 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

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

[Out]

1/4*(sqrt(2)*((7*A - C)*cos(d*x + c)^2 + 2*(7*A - C)*cos(d*x + c) + 7*A - C)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos
(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*((5*A + C)*cos(d*x + c) + 4*A)*sqrt(a*cos(d*x +
c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

[In]

int(((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(3/2))/(a + a*cos(c + d*x))^(3/2),x)

[Out]

int(((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(3/2))/(a + a*cos(c + d*x))^(3/2), x)