\(\int (a+a \cos (c+d x))^{3/2} \sec (c+d x) \, dx\) [108]

Optimal result

Integrand size = 21, antiderivative size = 66 \[ \int (a+a \cos (c+d x))^{3/2} \sec (c+d x) \, dx=\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}+\frac {2 a^2 \sin (c+d x)}{d \sqrt {a+a \cos (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.98 \[ \int (a+a \cos (c+d x))^{3/2} \sec (c+d x) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (\sqrt {2} \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{d} \] Input:

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

Output:

(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(Sqrt[2]*ArcTanh[Sqrt[2]*Si 
n[(c + d*x)/2]] + 2*Sin[(c + d*x)/2]))/d
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3242, 27, 2011, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3242

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2011

\(\displaystyle a \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx+\frac {2 a^2 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3252

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

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 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + b*x])
 

rule 3042

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

rule 3242

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x 
])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m 
+ n))   Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c* 
(m - 2) + b^2*d*(n + 1) + a^2*d*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 
 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c 
 - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] &&  !LtQ[ 
n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[ 
c, 0]))
 

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(208\) vs. \(2(58)=116\).

Time = 1.04 (sec) , antiderivative size = 209, normalized size of antiderivative = 3.17

method	result	size
default	\(\frac {\sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}+\ln \left (-\frac {4 \left (a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) a +\ln \left (\frac {4 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a \right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(209\)

Input:

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

Output:

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*si 
n(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
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 127 vs. \(2 (58) = 116\).

Time = 0.09 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.92 \[ \int (a+a \cos (c+d x))^{3/2} \sec (c+d x) \, dx=\frac {{\left (a \cos \left (d x + c\right ) + a\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, \sqrt {a \cos \left (d x + c\right ) + a} a \sin \left (d x + c\right )}{2 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:

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

Output:

1/2*((a*cos(d*x + c) + 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*sqrt(a*cos(d*x + c) + a)*a*sin 
(d*x + c))/(d*cos(d*x + c) + d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.35 \[ \int (a+a \cos (c+d x))^{3/2} \sec (c+d x) \, dx=-\frac {\sqrt {2} {\left (\sqrt {2} a \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 4 \, a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{2 \, d} \] Input:

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

Output:

-1/2*sqrt(2)*(sqrt(2)*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)) - 4*a*sgn(co 
s(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c))*sqrt(a)/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((a+a*cos(d*x+c))^(3/2)*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) + int(s 
qrt(cos(c + d*x) + 1)*sec(c + d*x),x))