3.2.14 \(\int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx\) [114]

Optimal. Leaf size=221 \[ \frac {(19 A-12 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 a^{3/2} d}-\frac {(13 A-9 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(7 A-6 B) \tan (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(2 A-B) \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}} \]

[Out]

1/4*(19*A-12*B)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(3/2)/d-1/4*(13*A-9*B)*arctanh(1/2*sin(d*
x+c)*a^(1/2)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)-1/2*(A-B)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+
c))^(3/2)-1/4*(7*A-6*B)*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)+1/2*(2*A-B)*sec(d*x+c)*tan(d*x+c)/a/d/(a+a*cos(d
*x+c))^(1/2)

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Rubi [A]
time = 0.46, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3057, 3063, 3064, 2728, 212, 2852} \begin {gather*} \frac {(19 A-12 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{4 a^{3/2} d}-\frac {(13 A-9 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(7 A-6 B) \tan (c+d x)}{4 a d \sqrt {a \cos (c+d x)+a}}+\frac {(2 A-B) \tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a \cos (c+d x)+a}}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((19*A - 12*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(4*a^(3/2)*d) - ((13*A - 9*B)*ArcTanh
[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((7*A - 6*B)*Tan[c + d*x]
)/(4*a*d*Sqrt[a + a*Cos[c + d*x]]) - ((A - B)*Sec[c + d*x]*Tan[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((
2*A - B)*Sec[c + d*x]*Tan[c + d*x])/(2*a*d*Sqrt[a + a*Cos[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 2728

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, b*(C
os[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2852

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[-2*(
b/f), Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*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 3057

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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*
x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 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 3064

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[(
B*c - A*d)/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(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]

Rubi steps

\begin {align*} \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx &=-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\left (2 a (2 A-B)-\frac {5}{2} a (A-B) \cos (c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(2 A-B) \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\left (-a^2 (7 A-6 B)+3 a^2 (2 A-B) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{4 a^3}\\ &=-\frac {(7 A-6 B) \tan (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(2 A-B) \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\left (\frac {1}{2} a^3 (19 A-12 B)-\frac {1}{2} a^3 (7 A-6 B) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{4 a^4}\\ &=-\frac {(7 A-6 B) \tan (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(2 A-B) \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}+\frac {(19 A-12 B) \int \sqrt {a+a \cos (c+d x)} \sec (c+d x) \, dx}{8 a^2}-\frac {(13 A-9 B) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{4 a}\\ &=-\frac {(7 A-6 B) \tan (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(2 A-B) \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}-\frac {(19 A-12 B) \text {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 a d}+\frac {(13 A-9 B) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{2 a d}\\ &=\frac {(19 A-12 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 a^{3/2} d}-\frac {(13 A-9 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(7 A-6 B) \tan (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(2 A-B) \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 8.59, size = 1402, normalized size = 6.34 \begin {gather*} \frac {(13 A-9 B) \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \log \left (\cos \left (\frac {c}{4}+\frac {d x}{4}\right )-\sin \left (\frac {c}{4}+\frac {d x}{4}\right )\right )}{d (a (1+\cos (c+d x)))^{3/2}}+\frac {(-13 A+9 B) \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \log \left (\cos \left (\frac {c}{4}+\frac {d x}{4}\right )+\sin \left (\frac {c}{4}+\frac {d x}{4}\right )\right )}{d (a (1+\cos (c+d x)))^{3/2}}+\frac {(19 A-12 B) \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \log \left (\sqrt {2}+2 \sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{2 \sqrt {2} d (a (1+\cos (c+d x)))^{3/2}}+\frac {i \text {ArcTan}\left (\frac {\cos \left (\frac {c}{4}+\frac {d x}{4}\right )-\sin \left (\frac {c}{4}+\frac {d x}{4}\right )-\sqrt {2} \sin \left (\frac {c}{4}+\frac {d x}{4}\right )}{-\cos \left (\frac {c}{4}+\frac {d x}{4}\right )+\sqrt {2} \cos \left (\frac {c}{4}+\frac {d x}{4}\right )-\sin \left (\frac {c}{4}+\frac {d x}{4}\right )}\right ) \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (19 \sqrt {2} A-12 \sqrt {2} B-38 A \sin \left (\frac {c}{2}\right )+24 B \sin \left (\frac {c}{2}\right )\right )}{4 d (a (1+\cos (c+d x)))^{3/2} \left (-1+\sqrt {2} \sin \left (\frac {c}{2}\right )\right )}+\frac {i \text {ArcTan}\left (\frac {\cos \left (\frac {c}{4}+\frac {d x}{4}\right )+\sin \left (\frac {c}{4}+\frac {d x}{4}\right )-\sqrt {2} \sin \left (\frac {c}{4}+\frac {d x}{4}\right )}{\cos \left (\frac {c}{4}+\frac {d x}{4}\right )+\sqrt {2} \cos \left (\frac {c}{4}+\frac {d x}{4}\right )-\sin \left (\frac {c}{4}+\frac {d x}{4}\right )}\right ) \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (19 \sqrt {2} A-12 \sqrt {2} B-38 A \sin \left (\frac {c}{2}\right )+24 B \sin \left (\frac {c}{2}\right )\right )}{4 d (a (1+\cos (c+d x)))^{3/2} \left (-1+\sqrt {2} \sin \left (\frac {c}{2}\right )\right )}+\frac {\cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \log \left (2-\sqrt {2} \cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sqrt {2} \sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \left (19 \sqrt {2} A-12 \sqrt {2} B-38 A \sin \left (\frac {c}{2}\right )+24 B \sin \left (\frac {c}{2}\right )\right )}{8 d (a (1+\cos (c+d x)))^{3/2} \left (-1+\sqrt {2} \sin \left (\frac {c}{2}\right )\right )}+\frac {\cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \log \left (2+\sqrt {2} \cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sqrt {2} \sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \left (19 \sqrt {2} A-12 \sqrt {2} B-38 A \sin \left (\frac {c}{2}\right )+24 B \sin \left (\frac {c}{2}\right )\right )}{8 d (a (1+\cos (c+d x)))^{3/2} \left (-1+\sqrt {2} \sin \left (\frac {c}{2}\right )\right )}+\frac {(-A+B) \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right )}{2 d (a (1+\cos (c+d x)))^{3/2} \left (\cos \left (\frac {c}{4}+\frac {d x}{4}\right )-\sin \left (\frac {c}{4}+\frac {d x}{4}\right )\right )^2}+\frac {(A-B) \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right )}{2 d (a (1+\cos (c+d x)))^{3/2} \left (\cos \left (\frac {c}{4}+\frac {d x}{4}\right )+\sin \left (\frac {c}{4}+\frac {d x}{4}\right )\right )^2}+\frac {A \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \sin \left (\frac {d x}{2}\right )}{d (a (1+\cos (c+d x)))^{3/2} \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {\cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-5 A \cos \left (\frac {c}{2}\right )+4 B \cos \left (\frac {c}{2}\right )+7 A \sin \left (\frac {c}{2}\right )-4 B \sin \left (\frac {c}{2}\right )\right )}{2 d (a (1+\cos (c+d x)))^{3/2} \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {A \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \sin \left (\frac {d x}{2}\right )}{d (a (1+\cos (c+d x)))^{3/2} \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {\cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (5 A \cos \left (\frac {c}{2}\right )-4 B \cos \left (\frac {c}{2}\right )+7 A \sin \left (\frac {c}{2}\right )-4 B \sin \left (\frac {c}{2}\right )\right )}{2 d (a (1+\cos (c+d x)))^{3/2} \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((13*A - 9*B)*Cos[c/2 + (d*x)/2]^3*Log[Cos[c/4 + (d*x)/4] - Sin[c/4 + (d*x)/4]])/(d*(a*(1 + Cos[c + d*x]))^(3/
2)) + ((-13*A + 9*B)*Cos[c/2 + (d*x)/2]^3*Log[Cos[c/4 + (d*x)/4] + Sin[c/4 + (d*x)/4]])/(d*(a*(1 + Cos[c + d*x
]))^(3/2)) + ((19*A - 12*B)*Cos[c/2 + (d*x)/2]^3*Log[Sqrt[2] + 2*Sin[c/2 + (d*x)/2]])/(2*Sqrt[2]*d*(a*(1 + Cos
[c + d*x]))^(3/2)) + ((I/4)*ArcTan[(Cos[c/4 + (d*x)/4] - Sin[c/4 + (d*x)/4] - Sqrt[2]*Sin[c/4 + (d*x)/4])/(-Co
s[c/4 + (d*x)/4] + Sqrt[2]*Cos[c/4 + (d*x)/4] - Sin[c/4 + (d*x)/4])]*Cos[c/2 + (d*x)/2]^3*(19*Sqrt[2]*A - 12*S
qrt[2]*B - 38*A*Sin[c/2] + 24*B*Sin[c/2]))/(d*(a*(1 + Cos[c + d*x]))^(3/2)*(-1 + Sqrt[2]*Sin[c/2])) + ((I/4)*A
rcTan[(Cos[c/4 + (d*x)/4] + Sin[c/4 + (d*x)/4] - Sqrt[2]*Sin[c/4 + (d*x)/4])/(Cos[c/4 + (d*x)/4] + Sqrt[2]*Cos
[c/4 + (d*x)/4] - Sin[c/4 + (d*x)/4])]*Cos[c/2 + (d*x)/2]^3*(19*Sqrt[2]*A - 12*Sqrt[2]*B - 38*A*Sin[c/2] + 24*
B*Sin[c/2]))/(d*(a*(1 + Cos[c + d*x]))^(3/2)*(-1 + Sqrt[2]*Sin[c/2])) + (Cos[c/2 + (d*x)/2]^3*Log[2 - Sqrt[2]*
Cos[c/2 + (d*x)/2] - Sqrt[2]*Sin[c/2 + (d*x)/2]]*(19*Sqrt[2]*A - 12*Sqrt[2]*B - 38*A*Sin[c/2] + 24*B*Sin[c/2])
)/(8*d*(a*(1 + Cos[c + d*x]))^(3/2)*(-1 + Sqrt[2]*Sin[c/2])) + (Cos[c/2 + (d*x)/2]^3*Log[2 + Sqrt[2]*Cos[c/2 +
(d*x)/2] - Sqrt[2]*Sin[c/2 + (d*x)/2]]*(19*Sqrt[2]*A - 12*Sqrt[2]*B - 38*A*Sin[c/2] + 24*B*Sin[c/2]))/(8*d*(a
*(1 + Cos[c + d*x]))^(3/2)*(-1 + Sqrt[2]*Sin[c/2])) + ((-A + B)*Cos[c/2 + (d*x)/2]^3)/(2*d*(a*(1 + Cos[c + d*x
]))^(3/2)*(Cos[c/4 + (d*x)/4] - Sin[c/4 + (d*x)/4])^2) + ((A - B)*Cos[c/2 + (d*x)/2]^3)/(2*d*(a*(1 + Cos[c + d
*x]))^(3/2)*(Cos[c/4 + (d*x)/4] + Sin[c/4 + (d*x)/4])^2) + (A*Cos[c/2 + (d*x)/2]^3*Sin[(d*x)/2])/(d*(a*(1 + Co
s[c + d*x]))^(3/2)*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2) + (Cos[c/2 + (d*x)/2]^3*
(-5*A*Cos[c/2] + 4*B*Cos[c/2] + 7*A*Sin[c/2] - 4*B*Sin[c/2]))/(2*d*(a*(1 + Cos[c + d*x]))^(3/2)*(Cos[c/2] - Si
n[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (A*Cos[c/2 + (d*x)/2]^3*Sin[(d*x)/2])/(d*(a*(1 + Cos[c +
d*x]))^(3/2)*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2) + (Cos[c/2 + (d*x)/2]^3*(5*A*C
os[c/2] - 4*B*Cos[c/2] + 7*A*Sin[c/2] - 4*B*Sin[c/2]))/(2*d*(a*(1 + Cos[c + d*x]))^(3/2)*(Cos[c/2] + Sin[c/2])
*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1539\) vs. \(2(190)=380\).
time = 0.53, size = 1540, normalized size = 6.97

method	result	size
default	\(\text {Expression too large to display}\)	\(1540\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(104*A*ln(2*(2*a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a)/cos(1/2*d*x+1/2
*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^6*a-72*B*ln(2*(2*a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a)/cos(1/2*d*x+1/2*c
))*2^(1/2)*cos(1/2*d*x+1/2*c)^6*a-76*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/
2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^6*a-76*A*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2
))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^6*a+4
8*B*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)
^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^6*a+48*B*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^
(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^6*a-104*A*2^(1/2)*ln(2*(2*a^(1/2)*(sin(1
/2*d*x+1/2*c)^2*a)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*cos(1/2*d*x+1/2*c)^4*a+72*B*2^(1/2)*ln(2*(2*a^(1/2)*(sin(1/2
*d*x+1/2*c)^2*a)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*cos(1/2*d*x+1/2*c)^4*a+28*A*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(
1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^4+76*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/
2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^4*a+76*A*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2
))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^4*a-2
4*B*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^4-48*B*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2
))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^4*a-4
8*B*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a
)^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^4*a+26*A*ln(2*(2*a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a)/cos(1/2*d*x+1/2*
c))*2^(1/2)*cos(1/2*d*x+1/2*c)^2*a-18*B*ln(2*(2*a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a)/cos(1/2*d*x+1/2*c)
)*2^(1/2)*cos(1/2*d*x+1/2*c)^2*a-22*A*a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*cos(1/2*d*x+1/2*c)^2-19*A
*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1
/2)+2*a))*cos(1/2*d*x+1/2*c)^2*a-19*A*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/
2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^2*a+16*B*a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c
)^2*a)^(1/2)*cos(1/2*d*x+1/2*c)^2+12*B*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/
2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^2*a+12*B*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2
))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^2*a+2
*A*a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*B*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2))/a^(5/2)/
cos(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)-2^(1/2))^2/(2*cos(1/2*d*x+1/2*c)+2^(1/2))^2/sin(1/2*d*x+1/2*c)/(a*cos
(1/2*d*x+1/2*c)^2)^(1/2)/d

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Maxima [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))*sec(d*x+c)^3/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]
time = 0.42, size = 361, normalized size = 1.63 \begin {gather*} -\frac {2 \, \sqrt {2} {\left ({\left (13 \, A - 9 \, B\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (13 \, A - 9 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (13 \, A - 9 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \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 ) + {\left ({\left (19 \, A - 12 \, B\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (19 \, A - 12 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (19 \, A - 12 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left ({\left (7 \, A - 6 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, A - 4 \, B\right )} \cos \left (d x + c\right ) - 2 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{16 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(2*sqrt(2)*((13*A - 9*B)*cos(d*x + c)^4 + 2*(13*A - 9*B)*cos(d*x + c)^3 + (13*A - 9*B)*cos(d*x + c)^2)*s
qrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sin(d*x + c) - 2*a*cos(d*x + c) - 3
*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + ((19*A - 12*B)*cos(d*x + c)^4 + 2*(19*A - 12*B)*cos(d*x + c)^3 +
(19*A - 12*B)*cos(d*x + c)^2)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 + 4*sqrt(a*cos(d*x + c) + a)*
sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*((7*A - 6*B)*cos(d*x + c
)^2 + (3*A - 4*B)*cos(d*x + c) - 2*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*c
os(d*x + c)^3 + a^2*d*cos(d*x + c)^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.61, size = 366, normalized size = 1.66 \begin {gather*} -\frac {\frac {\sqrt {2} {\left (13 \, A \sqrt {a} - 9 \, B \sqrt {a}\right )} \log \left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\sqrt {2} {\left (13 \, A \sqrt {a} - 9 \, B \sqrt {a}\right )} \log \left (-\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {{\left (19 \, A \sqrt {a} - 12 \, B \sqrt {a}\right )} \log \left ({\left | \frac {1}{2} \, \sqrt {2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {{\left (19 \, A \sqrt {a} - 12 \, B \sqrt {a}\right )} \log \left ({\left | -\frac {1}{2} \, \sqrt {2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, \sqrt {2} {\left (A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, {\left (10 \, \sqrt {2} A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, \sqrt {2} B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, \sqrt {2} A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, \sqrt {2} B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(sqrt(2)*(13*A*sqrt(a) - 9*B*sqrt(a))*log(sin(1/2*d*x + 1/2*c) + 1)/(a^2*sgn(cos(1/2*d*x + 1/2*c))) - sqr
t(2)*(13*A*sqrt(a) - 9*B*sqrt(a))*log(-sin(1/2*d*x + 1/2*c) + 1)/(a^2*sgn(cos(1/2*d*x + 1/2*c))) - (19*A*sqrt(
a) - 12*B*sqrt(a))*log(abs(1/2*sqrt(2) + sin(1/2*d*x + 1/2*c)))/(a^2*sgn(cos(1/2*d*x + 1/2*c))) + (19*A*sqrt(a
) - 12*B*sqrt(a))*log(abs(-1/2*sqrt(2) + sin(1/2*d*x + 1/2*c)))/(a^2*sgn(cos(1/2*d*x + 1/2*c))) - 2*sqrt(2)*(A
*sqrt(a)*sin(1/2*d*x + 1/2*c) - B*sqrt(a)*sin(1/2*d*x + 1/2*c))/((sin(1/2*d*x + 1/2*c)^2 - 1)*a^2*sgn(cos(1/2*
d*x + 1/2*c))) - 2*(10*sqrt(2)*A*sqrt(a)*sin(1/2*d*x + 1/2*c)^3 - 8*sqrt(2)*B*sqrt(a)*sin(1/2*d*x + 1/2*c)^3 -
3*sqrt(2)*A*sqrt(a)*sin(1/2*d*x + 1/2*c) + 4*sqrt(2)*B*sqrt(a)*sin(1/2*d*x + 1/2*c))/((2*sin(1/2*d*x + 1/2*c)
^2 - 1)^2*a^2*sgn(cos(1/2*d*x + 1/2*c))))/d

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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 )}{{\cos \left (c+d\,x\right )}^3\,{\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))/(cos(c + d*x)^3*(a + a*cos(c + d*x))^(3/2)),x)

[Out]

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

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