\(\int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \, dx\) [349]
Optimal result
Integrand size = 25, antiderivative size = 96 \[
\int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {2 a^{3/2} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}+\frac {2 a^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{d \sqrt {a+a \cos (c+d x)}}
\]
[Out]
2*a^(3/2)*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+2*a^2*sin(d*x+
c)*sec(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(1/2)
Rubi [A] (verified)
Time = 0.23 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of
steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4307, 2841, 21, 2853, 222}
\[
\int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {2 a^{3/2} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {2 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}
\]
[In]
Int[(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(3/2),x]
[Out]
(2*a^(3/2)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/d +
(2*a^2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]])
Rule 21
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(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 222
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]
Rule 2841
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-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] + Dist[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 2853
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, a/b]
Rule 4307
Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sin[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u,
x]
Rubi steps \begin{align*}
\text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 a^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{d \sqrt {a+a \cos (c+d x)}}-\left (2 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {a}{2}-\frac {1}{2} a \cos (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx \\ & = \frac {2 a^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{d \sqrt {a+a \cos (c+d x)}}+\left (a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {2 a^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{d \sqrt {a+a \cos (c+d x)}}-\frac {\left (2 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{a}}} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d} \\ & = \frac {2 a^{3/2} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}+\frac {2 a^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{d \sqrt {a+a \cos (c+d x)}} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.13 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89
\[
\int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (\sqrt {2} \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\cos (c+d x)}+2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}
\]
[In]
Integrate[(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(3/2),x]
[Out]
(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*(Sqrt[2]*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Sq
rt[Cos[c + d*x]] + 2*Sin[(c + d*x)/2]))/d
Maple [A] (verified)
Time = 6.61 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.56
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method | result | size |
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default |
\(\frac {2 \left (\sec ^{\frac {3}{2}}\left (d x +c \right )\right ) \left (\cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+\sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, a}{d \left (1+\cos \left (d x +c \right )\right )}\) |
\(150\) |
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[In]
int((a+cos(d*x+c)*a)^(3/2)*sec(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
[Out]
2/d*sec(d*x+c)^(3/2)*(cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)
))^(1/2))+(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+sin(d*x+c))*c
os(d*x+c)*(a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))*a
Fricas [A] (verification not implemented)
none
Time = 0.29 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.95
\[
\int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \, dx=-\frac {2 \, {\left ({\left (a \cos \left (d x + c\right ) + a\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {\sqrt {a \cos \left (d x + c\right ) + a} a \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{d \cos \left (d x + c\right ) + d}
\]
[In]
integrate((a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^(3/2),x, algorithm="fricas")
[Out]
-2*((a*cos(d*x + c) + a)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) -
sqrt(a*cos(d*x + c) + a)*a*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d)
Sympy [F(-1)]
Timed out. \[
\int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out}
\]
[In]
integrate((a+a*cos(d*x+c))**(3/2)*sec(d*x+c)**(3/2),x)
[Out]
Timed out
Maxima [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 997 vs. \(2 (82) = 164\).
Time = 0.43 (sec) , antiderivative size = 997, normalized size of antiderivative = 10.39
\[
\int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Too large to display}
\]
[In]
integrate((a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^(3/2),x, algorithm="maxima")
[Out]
1/2*((a*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(
2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2(si
n(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))), (cos(2*d*x + 2*c
)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) +
1))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)
+ 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))) + 1) - a*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x
+ 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/2*arctan2
(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) +
1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x
+ 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d
*x + 2*c)))) - 1) - a*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2
*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*
c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) + a*arctan2((cos(2*d*x + 2*c)^2 +
sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (
cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(
2*d*x + 2*c) + 1)) - 1))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) + 4*
(a*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)
)) - (a*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - a)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x
+ 2*c) + 1)))*sqrt(a))/((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*d)
Giac [F]
\[
\int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \, dx=\int { {\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x }
\]
[In]
integrate((a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^(3/2),x, algorithm="giac")
[Out]
integrate((a*cos(d*x + c) + a)^(3/2)*sec(d*x + c)^(3/2), x)
Mupad [F(-1)]
Timed out. \[
\int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \, dx=\int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2} \,d x
\]
[In]
int((1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))^(3/2),x)
[Out]
int((1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))^(3/2), x)