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

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.81 \[ \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (3 \sqrt {2} A \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 (3 A+5 B+B \cos (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{3 d} \] Input:

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

Output:

(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)
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {3042, 3455, 27, 3042, 3460, 3042, 3252, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 3455

\(\displaystyle \frac {2}{3} \int \frac {1}{2} \sqrt {\cos (c+d x) a+a} (3 a A+a (3 A+4 B) \cos (c+d x)) \sec (c+d x)dx+\frac {2 a B \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int \sqrt {\cos (c+d x) a+a} (3 a A+a (3 A+4 B) \cos (c+d x)) \sec (c+d x)dx+\frac {2 a B \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 a A+a (3 A+4 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a B \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3460

\(\displaystyle \frac {1}{3} \left (3 a A \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx+\frac {2 a^2 (3 A+4 B) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a B \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (3 a A \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a^2 (3 A+4 B) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a B \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {1}{3} \left (\frac {2 a^2 (3 A+4 B) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {6 a^2 A \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {2 a B \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{3} \left (\frac {6 a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {2 a^2 (3 A+4 B) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a B \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\)

Input:

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

Output:

(2*a*B*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d) + ((6*a^(3/2)*A*ArcTan 
h[(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])/(d*Sqrt[a + a*Cos[c + d*x]]))/3
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3252

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)]), x_Symbol] :> Simp[-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 3455

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(-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] + Simp[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 3460

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] + Simp[(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]
 

Maple [B] (verified)

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

Time = 3.88 (sec) , antiderivative size = 269, normalized size of antiderivative = 2.56

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

Input:

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

Output:

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))*( 
a*2^(1/2)*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))*(a*2^(1/2)*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)

Time = 0.09 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.42 \[ \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {3 \, {\left (A a \cos \left (d x + c\right ) + A a\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (B a \cos \left (d x + c\right ) + {\left (3 \, A + 5 \, B\right )} a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:

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

Output:

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.37 \[ \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {{\left (\sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 9 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a}}{3 \, d} \] Input:

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

Output:

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)

Time = 0.45 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.33 \[ \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec (c+d x) \, dx=-\frac {\sqrt {2} {\left (8 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, \sqrt {2} A a \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 12 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{6 \, d} \] Input:

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

Output:

-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*s 
qrt(2)*A*a*log(abs(-2*sqrt(2) + 4*sin(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. \[ \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec (c+d x) \, dx=\int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}}{\cos \left (c+d\,x\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec (c+d x) \, dx=\sqrt {a}\, a \left (\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )d x \right ) b +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )d x \right ) b +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )d x \right ) a \right ) \] Input:

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

Output:

sqrt(a)*a*(int(sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec(c + d*x),x)*a + int 
(sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec(c + d*x),x)*b + int(sqrt(cos(c + 
d*x) + 1)*cos(c + d*x)**2*sec(c + d*x),x)*b + int(sqrt(cos(c + d*x) + 1)*s 
ec(c + d*x),x)*a)