3.6.35 \(\int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx\) [535]

Optimal. Leaf size=237 \[ \frac {(2 A-3 B) \text {ArcSin}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{a^{3/2} d}-\frac {(5 A-9 B) \text {ArcTan}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}+\frac {(A-B) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)}-\frac {(A-3 B) \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \]

[Out]

1/2*(A-B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)/sec(d*x+c)^(3/2)-1/2*(A-3*B)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/
2)/sec(d*x+c)^(1/2)+(2*A-3*B)*arcsin(sin(d*x+c)*a^(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-1/4*(5*A-9*B)*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)

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Rubi [A]
time = 0.47, antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {3040, 3056, 3062, 3061, 2861, 211, 2853, 222} \begin {gather*} \frac {(2 A-3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \text {ArcSin}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{3/2} d}-\frac {(5 A-9 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \text {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 {(A-B) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac {(A-3 B) \sin (c+d x)}{2 a d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*A - 3*B)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(a
^(3/2)*d) - ((5*A - 9*B)*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]])/(2*Sqrt[2]*a^(3/2)*d) + ((A - B)*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x]
)^(3/2)*Sec[c + d*x]^(3/2)) - ((A - 3*B)*Sin[c + d*x])/(2*a*d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*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 222

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

Rule 2853

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, 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 3040

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[(a + b*Sin[e + f*x])^m*((
c + d*Sin[e + f*x])^n/(g*Sin[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

Rule 3056

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

Rule 3061

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

Rule 3062

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)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*
(m + n + 1))), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*b*c
*(m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*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] && GtQ[n, 0] &&
(IntegerQ[n] || EqQ[m + 1/2, 0])

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 6.68, size = 836, normalized size = 3.53 \begin {gather*} -\frac {i A e^{-\frac {1}{2} i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right ) \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right )}{d (a (1+\cos (c+d x)))^{3/2}}+\frac {3 i B e^{-\frac {1}{2} i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right ) \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right )}{d (a (1+\cos (c+d x)))^{3/2}}+\frac {2 i \sqrt {2} A e^{-\frac {1}{2} i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \left (-\sinh ^{-1}\left (e^{i (c+d x)}\right )+\sqrt {2} \tanh ^{-1}\left (\frac {-1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )+\tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right )}{d (a (1+\cos (c+d x)))^{3/2}}-\frac {3 i \sqrt {2} B e^{-\frac {1}{2} i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \left (-\sinh ^{-1}\left (e^{i (c+d x)}\right )+\sqrt {2} \tanh ^{-1}\left (\frac {-1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )+\tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right )}{d (a (1+\cos (c+d x)))^{3/2}}+\frac {\cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\sec (c+d x)} \left (-\frac {2 A \cos \left (\frac {d x}{2}\right ) \sin \left (\frac {c}{2}\right )}{d}+\frac {\sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sin \left (\frac {c}{2}\right )-B \sin \left (\frac {c}{2}\right )\right )}{d}+\frac {2 B \cos \left (\frac {3 d x}{2}\right ) \sin \left (\frac {3 c}{2}\right )}{d}-\frac {2 A \cos \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )}{d}+\frac {\sec \left (\frac {c}{2}\right ) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sin \left (\frac {d x}{2}\right )-B \sin \left (\frac {d x}{2}\right )\right )}{d}+\frac {2 B \cos \left (\frac {3 c}{2}\right ) \sin \left (\frac {3 d x}{2}\right )}{d}\right )}{(a (1+\cos (c+d x)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[(1 - E^(I*(c + d
*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])]*Cos[c/2 + (d*x)/2]^3)/(d*E^((I/2)*(c + d*x))*(a*(1 + Cos[c + d*
x]))^(3/2)) + ((3*I)*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[(
1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])]*Cos[c/2 + (d*x)/2]^3)/(d*E^((I/2)*(c + d*x))*(a*
(1 + Cos[c + d*x]))^(3/2)) + ((2*I)*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I
)*(c + d*x))]*(-ArcSinh[E^(I*(c + d*x))] + Sqrt[2]*ArcTanh[(-1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(
c + d*x))])] + ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]])*Cos[c/2 + (d*x)/2]^3)/(d*E^((I/2)*(c + d*x))*(a*(1 + Co
s[c + d*x]))^(3/2)) - ((3*I)*Sqrt[2]*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c +
d*x))]*(-ArcSinh[E^(I*(c + d*x))] + Sqrt[2]*ArcTanh[(-1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x
))])] + ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]])*Cos[c/2 + (d*x)/2]^3)/(d*E^((I/2)*(c + d*x))*(a*(1 + Cos[c + d
*x]))^(3/2)) + (Cos[c/2 + (d*x)/2]^3*Sqrt[Sec[c + d*x]]*((-2*A*Cos[(d*x)/2]*Sin[c/2])/d + (Sec[c/2]*Sec[c/2 +
(d*x)/2]*(A*Sin[c/2] - B*Sin[c/2]))/d + (2*B*Cos[(3*d*x)/2]*Sin[(3*c)/2])/d - (2*A*Cos[c/2]*Sin[(d*x)/2])/d +
(Sec[c/2]*Sec[c/2 + (d*x)/2]^2*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2]))/d + (2*B*Cos[(3*c)/2]*Sin[(3*d*x)/2])/d))/(a
*(1 + Cos[c + d*x]))^(3/2)

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Maple [A]
time = 0.42, size = 370, normalized size = 1.56

method	result	size
default	\(-\frac {\sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \cos \left (d x +c \right ) \left (-1+\cos \left (d x +c \right )\right )^{3} \left (-2 B \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right )+A \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+4 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right ) \sqrt {2}\, \sin \left (d x +c \right )-B \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )-6 B \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right ) \sqrt {2}\, \sin \left (d x +c \right )+5 A \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )-A \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-9 B \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )+3 B \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \sqrt {2}}{4 d \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sin \left (d x +c \right )^{7} a^{2}}\)	\(370\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/d*(a*(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*(-1+cos(d*x+c))^3*(-2*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*c
os(d*x+c)^2+A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+4*A*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*
x+c)))^(1/2)/cos(d*x+c))*2^(1/2)*sin(d*x+c)-B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)-6*B*arctan(
sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*2^(1/2)*sin(d*x+c)+5*A*arcsin((-1+cos(d*x+c))/sin(d*x
+c))*sin(d*x+c)-A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-9*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)+
3*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))/(1/cos(d*x+c))^(3/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)/sin(d*x+
c)^7*2^(1/2)/a^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 4.11, size = 246, normalized size = 1.04 \begin {gather*} \frac {\sqrt {2} {\left ({\left (5 \, A - 9 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A - 9 \, B\right )} \cos \left (d x + c\right ) + 5 \, A - 9 \, B\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 ) - 4 \, {\left ({\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 2 \, A - 3 \, B\right )} \sqrt {a} \arctan \left (\frac {\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 (2 \, B \cos \left (d x + c\right )^{2} - {\left (A - 3 \, B\right )} \cos \left (d x + c\right )\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(2)*((5*A - 9*B)*cos(d*x + c)^2 + 2*(5*A - 9*B)*cos(d*x + c) + 5*A - 9*B)*sqrt(a)*arctan(sqrt(2)*sqrt
(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - 4*((2*A - 3*B)*cos(d*x + c)^2 + 2*(2*A - 3*B
)*cos(d*x + c) + 2*A - 3*B)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c)))
+ 2*(2*B*cos(d*x + c)^2 - (A - 3*B)*cos(d*x + c))*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)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3436 deep

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {A+B\,\cos \left (c+d\,x\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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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