3.299 \(\int \frac{(e \cos (c+d x))^{3/2}}{\sqrt{a+a \sin (c+d x)}} \, dx\)

Optimal. Leaf size=200 \[ \frac{e^{3/2} \sqrt{\cos (c+d x)+1} \sqrt{a \sin (c+d x)+a} \tan ^{-1}\left (\frac{\sqrt{e} \sin (c+d x)}{\sqrt{\cos (c+d x)+1} \sqrt{e \cos (c+d x)}}\right )}{a d (\sin (c+d x)+\cos (c+d x)+1)}-\frac{e^{3/2} \sqrt{\cos (c+d x)+1} \sqrt{a \sin (c+d x)+a} \sinh ^{-1}\left (\frac{\sqrt{e \cos (c+d x)}}{\sqrt{e}}\right )}{a d (\sin (c+d x)+\cos (c+d x)+1)}+\frac{e \sqrt{a \sin (c+d x)+a} \sqrt{e \cos (c+d x)}}{a d} \]

[Out]

(e*Sqrt[e*Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(a*d) - (e^(3/2)*ArcSinh[Sqrt[e*Cos[c + d*x]]/Sqrt[e]]*Sqrt[
1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(a*d*(1 + Cos[c + d*x] + Sin[c + d*x])) + (e^(3/2)*ArcTan[(Sqrt[e]
*Sin[c + d*x])/(Sqrt[e*Cos[c + d*x]]*Sqrt[1 + Cos[c + d*x]])]*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])
/(a*d*(1 + Cos[c + d*x] + Sin[c + d*x]))

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Rubi [A]  time = 0.278785, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {2685, 2677, 2775, 203, 2833, 63, 215} \[ \frac{e^{3/2} \sqrt{\cos (c+d x)+1} \sqrt{a \sin (c+d x)+a} \tan ^{-1}\left (\frac{\sqrt{e} \sin (c+d x)}{\sqrt{\cos (c+d x)+1} \sqrt{e \cos (c+d x)}}\right )}{a d (\sin (c+d x)+\cos (c+d x)+1)}-\frac{e^{3/2} \sqrt{\cos (c+d x)+1} \sqrt{a \sin (c+d x)+a} \sinh ^{-1}\left (\frac{\sqrt{e \cos (c+d x)}}{\sqrt{e}}\right )}{a d (\sin (c+d x)+\cos (c+d x)+1)}+\frac{e \sqrt{a \sin (c+d x)+a} \sqrt{e \cos (c+d x)}}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*Sqrt[e*Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(a*d) - (e^(3/2)*ArcSinh[Sqrt[e*Cos[c + d*x]]/Sqrt[e]]*Sqrt[
1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(a*d*(1 + Cos[c + d*x] + Sin[c + d*x])) + (e^(3/2)*ArcTan[(Sqrt[e]
*Sin[c + d*x])/(Sqrt[e*Cos[c + d*x]]*Sqrt[1 + Cos[c + d*x]])]*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])
/(a*d*(1 + Cos[c + d*x] + Sin[c + d*x]))

Rule 2685

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(3/2)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[(g*Sqr
t[g*Cos[e + f*x]]*Sqrt[a + b*Sin[e + f*x]])/(b*f), x] + Dist[g^2/(2*a), Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[g*Co
s[e + f*x]], x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]

Rule 2677

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)], x_Symbol] :> Dist[(a*Sqrt[
1 + Cos[e + f*x]]*Sqrt[a + b*Sin[e + f*x]])/(a + a*Cos[e + f*x] + b*Sin[e + f*x]), Int[Sqrt[1 + Cos[e + f*x]]/
Sqrt[g*Cos[e + f*x]], x], x] + Dist[(b*Sqrt[1 + Cos[e + f*x]]*Sqrt[a + b*Sin[e + f*x]])/(a + a*Cos[e + f*x] +
b*Sin[e + f*x]), Int[Sin[e + f*x]/(Sqrt[g*Cos[e + f*x]]*Sqrt[1 + Cos[e + f*x]]), x], x] /; FreeQ[{a, b, e, f,
g}, x] && EqQ[a^2 - b^2, 0]

Rule 2775

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*b)/f, Subst[Int[1/(b + d*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*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 203

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

Rule 2833

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

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

Mathematica [C]  time = 0.116269, size = 77, normalized size = 0.38 \[ -\frac{2\ 2^{3/4} (e \cos (c+d x))^{5/2} \, _2F_1\left (\frac{1}{4},\frac{5}{4};\frac{9}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{5 d e (\sin (c+d x)+1)^{3/4} \sqrt{a (\sin (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*2^(3/4)*(e*Cos[c + d*x])^(5/2)*Hypergeometric2F1[1/4, 5/4, 9/4, (1 - Sin[c + d*x])/2])/(5*d*e*(1 + Sin[c +
d*x])^(3/4)*Sqrt[a*(1 + Sin[c + d*x])])

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Maple [A]  time = 0.125, size = 212, normalized size = 1.1 \begin{align*}{\frac{1}{2\,d \left ( -1+\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) \right ) \cos \left ( dx+c \right ) } \left ( \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\arctan \left ({\frac{\sqrt{2}}{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) \sqrt{2}\sin \left ( dx+c \right ) -\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}{\it Artanh} \left ({\frac{\sqrt{2}\sin \left ( dx+c \right ) }{2\,\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) \sqrt{2}\sin \left ( dx+c \right ) +2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+2\,\cos \left ( dx+c \right ) \right ) \left ( e\cos \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{a \left ( 1+\sin \left ( dx+c \right ) \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*2^(1/2)*s
in(d*x+c)-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*
x+c)/cos(d*x+c))*2^(1/2)*sin(d*x+c)+2*cos(d*x+c)*sin(d*x+c)-2*cos(d*x+c)^2+2*cos(d*x+c))*(e*cos(d*x+c))^(3/2)/
(-1+cos(d*x+c)+sin(d*x+c))/(a*(1+sin(d*x+c)))^(1/2)/cos(d*x+c)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos \left (d x + c\right )\right )^{\frac{3}{2}}}{\sqrt{a \sin \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*cos(d*x + c))^(3/2)/sqrt(a*sin(d*x + c) + a), x)

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Fricas [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((e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

Timed out

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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((e*cos(d*x+c))**(3/2)/(a+a*sin(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 (e \cos \left (d x + c\right )\right )^{\frac{3}{2}}}{\sqrt{a \sin \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*cos(d*x + c))^(3/2)/sqrt(a*sin(d*x + c) + a), x)