3.351 \(\int \frac{(a+a \cos (c+d x))^{3/2}}{\sqrt{\sec (c+d x)}} \, dx\)
Optimal. Leaf size=140 \[ \frac{a^2 \sin (c+d x)}{2 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{7 a^{3/2} \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{4 d}+\frac{7 a^2 \sin (c+d x)}{4 d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}} \]
[Out]
(7*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]])/(4*d
) + (a^2*Sin[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(3/2)) + (7*a^2*Sin[c + d*x])/(4*d*Sqrt[a +
a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]])
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Rubi [A] time = 0.238039, antiderivative size = 140, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used =
{4222, 2763, 21, 2770, 2774, 216} \[ \frac{a^2 \sin (c+d x)}{2 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{7 a^{3/2} \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{4 d}+\frac{7 a^2 \sin (c+d x)}{4 d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Cos[c + d*x])^(3/2)/Sqrt[Sec[c + d*x]],x]
[Out]
(7*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]])/(4*d
) + (a^2*Sin[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(3/2)) + (7*a^2*Sin[c + d*x])/(4*d*Sqrt[a +
a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]])
Rule 4222
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]
Rule 2763
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(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] + Dist[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 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 2770
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(2*n*(b*c + a*d)
)/(b*(2*n + 1)), 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] && GtQ[n, 0] && IntegerQ[2*n]
Rule 2774
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 216
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]
Rubi steps
\begin{align*} \int \frac{(a+a \cos (c+d x))^{3/2}}{\sqrt{\sec (c+d x)}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2} \, dx\\ &=\frac{a^2 \sin (c+d x)}{2 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)}+\frac{1}{2} \left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\cos (c+d x)} \left (\frac{7 a^2}{2}+\frac{7}{2} a^2 \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx\\ &=\frac{a^2 \sin (c+d x)}{2 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)}+\frac{1}{4} \left (7 a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{a^2 \sin (c+d x)}{2 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)}+\frac{7 a^2 \sin (c+d x)}{4 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}+\frac{1}{8} \left (7 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{a^2 \sin (c+d x)}{2 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)}+\frac{7 a^2 \sin (c+d x)}{4 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}-\frac{\left (7 a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{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 )}{4 d}\\ &=\frac{7 a^{3/2} \sin ^{-1}\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)}}{4 d}+\frac{a^2 \sin (c+d x)}{2 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)}+\frac{7 a^2 \sin (c+d x)}{4 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.253478, size = 111, normalized size = 0.79 \[ \frac{a \sec \left (\frac{1}{2} (c+d x)\right ) \sqrt{\sec (c+d x)} \sqrt{a (\cos (c+d x)+1)} \left (-5 \sin \left (\frac{1}{2} (c+d x)\right )+6 \sin \left (\frac{3}{2} (c+d x)\right )+\sin \left (\frac{5}{2} (c+d x)\right )+7 \sqrt{2} \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right ) \sqrt{\cos (c+d x)}\right )}{8 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Cos[c + d*x])^(3/2)/Sqrt[Sec[c + d*x]],x]
[Out]
(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*(7*Sqrt[2]*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*
Sqrt[Cos[c + d*x]] - 5*Sin[(c + d*x)/2] + 6*Sin[(3*(c + d*x))/2] + Sin[(5*(c + d*x))/2]))/(8*d)
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Maple [A] time = 0.463, size = 170, normalized size = 1.2 \begin{align*}{\frac{a \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2}\cos \left ( dx+c \right ) }{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}} \left ( 2\,\sin \left ( dx+c \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\cos \left ( dx+c \right ) +7\,\sin \left ( dx+c \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+7\,\arctan \left ({\frac{\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) \right ) \sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{-1}}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+cos(d*x+c)*a)^(3/2)/sec(d*x+c)^(1/2),x)
[Out]
1/4/d*a*(-1+cos(d*x+c))^2*(2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+7*sin(d*x+c)*(cos(d*x+c)/
(1+cos(d*x+c)))^(1/2)+7*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c)))*cos(d*x+c)*(a*(1+cos(
d*x+c)))^(1/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)/(1/cos(d*x+c))^(1/2)/sin(d*x+c)^4
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Maxima [B] time = 2.03693, size = 1458, normalized size = 10.41 \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))^(3/2)/sec(d*x+c)^(1/2),x, algorithm="maxima")
[Out]
1/16*(2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*((a*cos(1/2*arctan2(sin(2*d*x
+ 2*c), cos(2*d*x + 2*c)))*sin(2*d*x + 2*c) + a*sin(2*d*x + 2*c) - (a*cos(2*d*x + 2*c) - 6*a)*sin(1/2*arctan2
(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + (a*sin(2*d*x
+ 2*c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - a*cos(2*d*x + 2*c) + (a*cos(2*d*x + 2*c) - 6*a)
*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 6*a)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c
) + 1)))*sqrt(a) + 7*(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)) - co
s(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)))),
(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), co
s(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*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)*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))*sqrt(a))/d
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Fricas [A] time = 1.73359, size = 316, normalized size = 2.26 \begin{align*} -\frac{7 \,{\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{{\left (2 \, a \cos \left (d x + c\right )^{2} + 7 \, a \cos \left (d x + c\right )\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{4 \,{\left (d \cos \left (d x + c\right ) + d\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))^(3/2)/sec(d*x+c)^(1/2),x, algorithm="fricas")
[Out]
-1/4*(7*(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))
) - (2*a*cos(d*x + c)^2 + 7*a*cos(d*x + c))*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d
*x + c) + d)
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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))**(3/2)/sec(d*x+c)**(1/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}{\sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))^(3/2)/sec(d*x+c)^(1/2),x, algorithm="giac")
[Out]
integrate((a*cos(d*x + c) + a)^(3/2)/sqrt(sec(d*x + c)), x)