\(\int (a+a \cos (c+d x))^{3/2} \sec ^2(c+d x) \, dx\) [109]

Optimal result

Integrand size = 23, antiderivative size = 65 \[ \int (a+a \cos (c+d x))^{3/2} \sec ^2(c+d x) \, dx=\frac {3 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}+\frac {a^2 \tan (c+d x)}{d \sqrt {a+a \cos (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.25 \[ \int (a+a \cos (c+d x))^{3/2} \sec ^2(c+d x) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (3 \sqrt {2} \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos (c+d x)+2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d} \] Input:

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

Output:

(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]*(3*Sqrt[2]*Arc 
Tanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x] + 2*Sin[(c + d*x)/2]))/(2*d)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3241, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3241

\(\displaystyle \frac {a^2 \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-a \int -\frac {3}{2} \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3}{2} a \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx+\frac {a^2 \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3252

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(3*a^(3/2)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + ( 
a^2*Tan[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 3042

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

rule 3241

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(b*c - a*d)*Cos[e + f*x]*(a + b 
*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a* 
d))), x] + Simp[b^2/(d*(n + 1)*(b*c + a*d))   Int[(a + b*Sin[e + f*x])^(m - 
 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b* 
c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], 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] 
 && 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. \(384\) vs. \(2(57)=114\).

Time = 1.12 (sec) , antiderivative size = 385, normalized size of antiderivative = 5.92

method	result	size
default	\(\frac {\sqrt {a}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-6 a \left (\ln \left (-\frac {4 \left (a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right )+\ln \left (\frac {4 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right )\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}+3 \ln \left (-\frac {4 \left (a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) a +3 \ln \left (\frac {4 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a \right )}{\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\right ) \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(385\)

Input:

int((a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^2,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)*(-6*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))+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)))*sin(1/2*d*x+1/2*c)^2+2*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^( 
1/2)+3*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+3*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)/(2*cos(1/2*d*x+1/2*c)-2^(1/2))/(2*cos(1/2*d 
*x+1/2*c)+2^(1/2))/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. 146 vs. \(2 (57) = 114\).

Time = 0.09 (sec) , antiderivative size = 146, normalized size of antiderivative = 2.25 \[ \int (a+a \cos (c+d x))^{3/2} \sec ^2(c+d x) \, dx=\frac {3 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, \sqrt {a \cos \left (d x + c\right ) + a} a \sin \left (d x + c\right )}{4 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}} \] Input:

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

Output:

1/4*(3*(a*cos(d*x + c)^2 + a*cos(d*x + c))*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)^2 + d*cos(d*x + c))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1314 vs. \(2 (57) = 114\).

Time = 0.19 (sec) , antiderivative size = 1314, normalized size of antiderivative = 20.22 \[ \int (a+a \cos (c+d x))^{3/2} \sec ^2(c+d x) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/4*(2*sqrt(2)*a*cos(7/2*d*x + 7/2*c)*sin(2*d*x + 2*c) + 6*sqrt(2)*a*cos( 
5/2*d*x + 5/2*c)*sin(2*d*x + 2*c) + (2*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 6* 
sqrt(2)*a*sin(1/2*d*x + 1/2*c) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin( 
1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d* 
x + 1/2*c) + 2) + 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c 
)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) 
 - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2) 
*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 3*a*log(2*co 
s(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 
1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*cos(2*d*x + 2*c)^2 + (2*sqrt 
(2)*a*sin(3/2*d*x + 3/2*c) + 6*sqrt(2)*a*sin(1/2*d*x + 1/2*c) - 3*a*log(2* 
cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x 
+ 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 3*a*log(2*cos(1/2*d*x + 1 
/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sq 
rt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin 
(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d 
*x + 1/2*c) + 2) + 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2* 
c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2 
))*sin(2*d*x + 2*c)^2 - 4*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 4*sqrt(2)*a*sin 
(1/2*d*x + 1/2*c) - 2*(sqrt(2)*a*sin(3/2*d*x + 3/2*c) - 5*sqrt(2)*a*sin...
                                                                                    
                                                                                    
 

Giac [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.65 \[ \int (a+a \cos (c+d x))^{3/2} \sec ^2(c+d x) \, dx=-\frac {\sqrt {2} {\left (3 \, \sqrt {2} a \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + \frac {4 \, a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}\right )} \sqrt {a}}{4 \, d} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(a)*a*(int(sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec(c + d*x)**2,x) + in 
t(sqrt(cos(c + d*x) + 1)*sec(c + d*x)**2,x))