3.102 \(\int \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \, dx\)

Optimal. Leaf size=102 \[ \frac{3 a \tan (c+d x)}{4 d \sqrt{a \cos (c+d x)+a}}+\frac{3 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{4 d}+\frac{a \tan (c+d x) \sec (c+d x)}{2 d \sqrt{a \cos (c+d x)+a}} \]

[Out]

(3*Sqrt[a]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(4*d) + (3*a*Tan[c + d*x])/(4*d*Sqrt[a +
a*Cos[c + d*x]]) + (a*Sec[c + d*x]*Tan[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.162247, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2772, 2773, 206} \[ \frac{3 a \tan (c+d x)}{4 d \sqrt{a \cos (c+d x)+a}}+\frac{3 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{4 d}+\frac{a \tan (c+d x) \sec (c+d x)}{2 d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^3,x]

[Out]

(3*Sqrt[a]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(4*d) + (3*a*Tan[c + d*x])/(4*d*Sqrt[a +
a*Cos[c + d*x]]) + (a*Sec[c + d*x]*Tan[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]])

Rule 2772

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[((b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x]
+ Dist[((2*n + 3)*(b*c - a*d))/(2*b*(n + 1)*(c^2 - d^2)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), 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] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

Rule 2773

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \, dx &=\frac{a \sec (c+d x) \tan (c+d x)}{2 d \sqrt{a+a \cos (c+d x)}}+\frac{3}{4} \int \sqrt{a+a \cos (c+d x)} \sec ^2(c+d x) \, dx\\ &=\frac{3 a \tan (c+d x)}{4 d \sqrt{a+a \cos (c+d x)}}+\frac{a \sec (c+d x) \tan (c+d x)}{2 d \sqrt{a+a \cos (c+d x)}}+\frac{3}{8} \int \sqrt{a+a \cos (c+d x)} \sec (c+d x) \, dx\\ &=\frac{3 a \tan (c+d x)}{4 d \sqrt{a+a \cos (c+d x)}}+\frac{a \sec (c+d x) \tan (c+d x)}{2 d \sqrt{a+a \cos (c+d x)}}-\frac{(3 a) \operatorname{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 d}\\ &=\frac{3 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{4 d}+\frac{3 a \tan (c+d x)}{4 d \sqrt{a+a \cos (c+d x)}}+\frac{a \sec (c+d x) \tan (c+d x)}{2 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.186922, size = 94, normalized size = 0.92 \[ \frac{\sec \left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) \sqrt{a (\cos (c+d x)+1)} \left (\sin \left (\frac{1}{2} (c+d x)\right )+3 \sin \left (\frac{3}{2} (c+d x)\right )+3 \sqrt{2} \cos ^2(c+d x) \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^3,x]

[Out]

(Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^2*(3*Sqrt[2]*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c
+ d*x]^2 + Sin[(c + d*x)/2] + 3*Sin[(3*(c + d*x))/2]))/(8*d)

________________________________________________________________________________________

Maple [B]  time = 2.656, size = 545, normalized size = 5.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(12*a*(ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^
(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))+ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a
^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a)))*sin(1/2*d*x+1/2*c)^4+(-12*ln
(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*
c)+2*a))*a-12*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a*2^(1/2)*co
s(1/2*d*x+1/2*c)+2*a))*a-12*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2))*sin(1/2*d*x+1/2*c)^2+3*ln(-4/(-2*c
os(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*
a+3*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a*2^(1/2)*cos(1/2*d*x+
1/2*c)+2*a))*a+10*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2))/a^(1/2)/(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)/(cos(1/2*d*x+1/2*c)^2*a)^(1/2)/d

________________________________________________________________________________________

Maxima [B]  time = 22.5067, size = 3567, normalized size = 34.97 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(1/2)*sec(d*x+c)^3,x, algorithm="maxima")

[Out]

1/16*(3*(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) - 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) + log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqr
t(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 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))*cos(4*d*x + 4*c)^2 + 12*(l
og(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) - 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) + 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) - 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))*cos(2*d*x + 2*c)^2 + 3*(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) - 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)*si
n(1/2*d*x + 1/2*c) + 2) + 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) - 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(4*d*x + 4*c)^2 + 12*(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) - 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) + 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*s
qrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 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 - 24*sqrt(2)*cos(7/2*d*x + 7/2*c)*sin(
2*d*x + 2*c) - 8*sqrt(2)*cos(5/2*d*x + 5/2*c)*sin(2*d*x + 2*c) + 2*(6*(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) - 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) + l
og(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) - 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))*cos(2*d*x + 2*c) + 6*sqrt(2)*sin(7/2*d*x + 7/2*c) + 2*sqrt(2)*sin(5/2*d*x
+ 5/2*c) - 2*sqrt(2)*sin(3/2*d*x + 3/2*c) - 6*sqrt(2)*sin(1/2*d*x + 1/2*c) + 3*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*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*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*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))*cos(4*d*x + 4*c) - 4*(2*sqrt(2)*sin(3/2*d*x + 3/2*c) + 6*sqr
t(2)*sin(1/2*d*x + 1/2*c) - 3*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*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*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*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))
*cos(2*d*x + 2*c) + 4*(3*(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) - 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) + 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) - 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) - 3*sqrt(2)*cos(7/2*d*x + 7/2*c) - sqrt(2)*cos(5/2*d*x + 5/2*c) + sqrt(2)*cos(3/2*d*x + 3/2*c) + 3*sqr
t(2)*cos(1/2*d*x + 1/2*c))*sin(4*d*x + 4*c) + 12*(2*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*sin(7/2*d*x + 7/2*c) +
4*(2*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*sin(5/2*d*x + 5/2*c) + 8*(sqrt(2)*cos(3/2*d*x + 3/2*c) + 3*sqrt(2)*c
os(1/2*d*x + 1/2*c))*sin(2*d*x + 2*c) - 4*sqrt(2)*sin(3/2*d*x + 3/2*c) - 12*sqrt(2)*sin(1/2*d*x + 1/2*c) + 3*l
og(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*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*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*c
os(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3*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))*sqrt(a)/((2*(2*cos(2*d*x + 2*c
) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*
sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*d)

________________________________________________________________________________________

Fricas [A]  time = 1.72247, size = 414, normalized size = 4.06 \begin{align*} \frac{3 \,{\left (\cos \left (d x + c\right )^{3} + \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 \, \sqrt{a \cos \left (d x + c\right ) + a}{\left (3 \, \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{16 \,{\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(1/2)*sec(d*x+c)^3,x, algorithm="fricas")

[Out]

1/16*(3*(cos(d*x + c)^3 + 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*sqrt(a*cos(d
*x + c) + a)*(3*cos(d*x + c) + 2)*sin(d*x + c))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 3.85229, size = 443, normalized size = 4.34 \begin{align*} \frac{\sqrt{2}{\left (\frac{3 \, \sqrt{2} a^{\frac{3}{2}} \log \left (\frac{{\left | 2 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt{2}{\left | a \right |} - 6 \, a \right |}}{{\left | 2 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt{2}{\left | a \right |} - 6 \, a \right |}}\right )}{{\left | a \right |}} - \frac{8 \,{\left (5 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{6} a^{\frac{3}{2}} + 19 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{4} a^{\frac{5}{2}} - 17 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} a^{\frac{7}{2}} + a^{\frac{9}{2}}\right )}}{{\left ({\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{4} - 6 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} a + a^{2}\right )}^{2}}\right )}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(1/2)*sec(d*x+c)^3,x, algorithm="giac")

[Out]

1/16*sqrt(2)*(3*sqrt(2)*a^(3/2)*log(abs(2*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^
2 - 4*sqrt(2)*abs(a) - 6*a)/abs(2*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 4*sq
rt(2)*abs(a) - 6*a))/abs(a) - 8*(5*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*a^(3/
2) + 19*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*a^(5/2) - 17*(sqrt(a)*tan(1/2*d*
x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a^(7/2) + a^(9/2))/((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*
tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a
^2)^2)/d