\(\int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec (c+d x) \, dx\) [86]

Optimal result

Integrand size = 31, antiderivative size = 105 \[ \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {2 a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}+\frac {2 a^2 (3 A+4 B) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a B \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 d} \]

[Out]

2*a^(3/2)*A*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d+2/3*a^2*(3*A+4*B)*sin(d*x+c)/d/(a+a*cos(d*x+c
))^(1/2)+2/3*a*B*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3055, 3060, 2852, 212} \[ \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {2 a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {2 a^2 (3 A+4 B) \sin (c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a B \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d} \]

[In]

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

[Out]

(2*a^(3/2)*A*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (2*a^2*(3*A + 4*B)*Sin[c + d*x])/(3
*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*B*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d)

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 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 3055

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

Rule 3060

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt
[a + b*Sin[e + f*x]])), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, 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[n, -1]

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.81 \[ \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (3 \sqrt {2} A \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 (3 A+5 B+B \cos (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{3 d} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(268\) vs. \(2(91)=182\).

Time = 4.25 (sec) , antiderivative size = 269, normalized size of antiderivative = 2.56

method	result	size
parts	\(\frac {A \sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (2 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+\ln \left (\frac {4 \sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a +\ln \left (-\frac {4 \left (\sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) a \right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}+\frac {4 B \,a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+2\right ) \sqrt {2}}{3 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(269\)
default	\(\frac {\sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-8 B \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 A \sqrt {2}\, \ln \left (-\frac {2 \left (\sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) a +3 A \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+4 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a +12 A \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+24 B \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right ) \sqrt {2}}{6 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(274\)

[In]

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

[Out]

A*a^(1/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+
ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/
2)+2*a))*a+ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)
^(1/2)*a^(1/2)-2*a))*a)/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d+4/3*B*a^2*cos(1/2*d*x+1/2*c)*sin(1
/2*d*x+1/2*c)*(cos(1/2*d*x+1/2*c)^2+2)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.42 \[ \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {3 \, {\left (A a \cos \left (d x + c\right ) + A a\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (B a \cos \left (d x + c\right ) + {\left (3 \, A + 5 \, B\right )} a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

1/6*(3*(A*a*cos(d*x + c) + A*a)*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*(B*a*cos(d*x + c) + (3
*A + 5*B)*a)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c) + d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.37 \[ \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {{\left (\sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 9 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a}}{3 \, d} \]

[In]

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

[Out]

1/3*(sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 9*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*B*sqrt(a)/d

Giac [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.33 \[ \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec (c+d x) \, dx=-\frac {\sqrt {2} {\left (8 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, \sqrt {2} A a \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 12 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{6 \, d} \]

[In]

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

[Out]

-1/6*sqrt(2)*(8*B*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 + 3*sqrt(2)*A*a*log(abs(-2*sqrt(2) + 4*si
n(1/2*d*x + 1/2*c))/abs(2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c)))*sgn(cos(1/2*d*x + 1/2*c)) - 12*A*a*sgn(cos(1/2*d*
x + 1/2*c))*sin(1/2*d*x + 1/2*c) - 24*B*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c))*sqrt(a)/d

Mupad [F(-1)]

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

[In]

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

[Out]

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