3.306 \(\int \frac{(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=215 \[ \frac{3 e^{5/2} \sqrt{\cos (c+d x)+1} \sqrt{a \sin (c+d x)+a} \tan ^{-1}\left (\frac{\sqrt{e} \sin (c+d x)}{\sqrt{\cos (c+d x)+1} \sqrt{e \cos (c+d x)}}\right )}{d \left (a^2 \sin (c+d x)+a^2 \cos (c+d x)+a^2\right )}+\frac{3 e^{5/2} \sqrt{\cos (c+d x)+1} \sqrt{a \sin (c+d x)+a} \sinh ^{-1}\left (\frac{\sqrt{e \cos (c+d x)}}{\sqrt{e}}\right )}{d \left (a^2 \sin (c+d x)+a^2 \cos (c+d x)+a^2\right )}+\frac{e (e \cos (c+d x))^{3/2}}{a d \sqrt{a \sin (c+d x)+a}} \]

[Out]

(e*(e*Cos[c + d*x])^(3/2))/(a*d*Sqrt[a + a*Sin[c + d*x]]) + (3*e^(5/2)*ArcSinh[Sqrt[e*Cos[c + d*x]]/Sqrt[e]]*S
qrt[1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(d*(a^2 + a^2*Cos[c + d*x] + a^2*Sin[c + d*x])) + (3*e^(5/2)*A
rcTan[(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]])/(d*(a^2 + a^2*Cos[c + d*x] + a^2*Sin[c + d*x]))

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Rubi [A]  time = 0.286914, antiderivative size = 215, 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 = {2679, 2684, 2775, 203, 2833, 63, 215} \[ \frac{3 e^{5/2} \sqrt{\cos (c+d x)+1} \sqrt{a \sin (c+d x)+a} \tan ^{-1}\left (\frac{\sqrt{e} \sin (c+d x)}{\sqrt{\cos (c+d x)+1} \sqrt{e \cos (c+d x)}}\right )}{d \left (a^2 \sin (c+d x)+a^2 \cos (c+d x)+a^2\right )}+\frac{3 e^{5/2} \sqrt{\cos (c+d x)+1} \sqrt{a \sin (c+d x)+a} \sinh ^{-1}\left (\frac{\sqrt{e \cos (c+d x)}}{\sqrt{e}}\right )}{d \left (a^2 \sin (c+d x)+a^2 \cos (c+d x)+a^2\right )}+\frac{e (e \cos (c+d x))^{3/2}}{a d \sqrt{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(5/2)/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(e*(e*Cos[c + d*x])^(3/2))/(a*d*Sqrt[a + a*Sin[c + d*x]]) + (3*e^(5/2)*ArcSinh[Sqrt[e*Cos[c + d*x]]/Sqrt[e]]*S
qrt[1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(d*(a^2 + a^2*Cos[c + d*x] + a^2*Sin[c + d*x])) + (3*e^(5/2)*A
rcTan[(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]])/(d*(a^2 + a^2*Cos[c + d*x] + a^2*Sin[c + d*x]))

Rule 2679

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + p)), x] + Dist[(g^2*(p - 1))/(a*(m + p)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p,
0] && IntegersQ[2*m, 2*p]

Rule 2684

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(g*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[(g*Sqrt[1 + Cos[e + f*x]]*Sqrt[a + b*Sin[e + f*x]])/(b + b*Cos[e + f*x] +
a*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))^{5/2}}{(a+a \sin (c+d x))^{3/2}} \, dx &=\frac{e (e \cos (c+d x))^{3/2}}{a d \sqrt{a+a \sin (c+d x)}}+\frac{\left (3 e^2\right ) \int \frac{\sqrt{e \cos (c+d x)}}{\sqrt{a+a \sin (c+d x)}} \, dx}{2 a}\\ &=\frac{e (e \cos (c+d x))^{3/2}}{a d \sqrt{a+a \sin (c+d x)}}+\frac{\left (3 e^3 \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+a \cos (c+d x)+a \sin (c+d x))}-\frac{\left (3 e^3 \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+a \cos (c+d x)+a \sin (c+d x))}\\ &=\frac{e (e \cos (c+d x))^{3/2}}{a d \sqrt{a+a \sin (c+d x)}}+\frac{\left (3 e^3 \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 a d (a+a \cos (c+d x)+a \sin (c+d x))}-\frac{\left (3 e^3 \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 )}{a d (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=\frac{e (e \cos (c+d x))^{3/2}}{a d \sqrt{a+a \sin (c+d x)}}+\frac{3 e^{5/2} \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 \left (a^2+a^2 \cos (c+d x)+a^2 \sin (c+d x)\right )}+\frac{\left (3 e^2 \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 )}{a d (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=\frac{e (e \cos (c+d x))^{3/2}}{a d \sqrt{a+a \sin (c+d x)}}+\frac{3 e^{5/2} \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 \left (a^2+a^2 \cos (c+d x)+a^2 \sin (c+d x)\right )}+\frac{3 e^{5/2} \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 \left (a^2+a^2 \cos (c+d x)+a^2 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 0.183856, size = 80, normalized size = 0.37 \[ -\frac{2 \sqrt [4]{2} \sqrt{a (\sin (c+d x)+1)} (e \cos (c+d x))^{7/2} \, _2F_1\left (\frac{3}{4},\frac{7}{4};\frac{11}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{7 a^2 d e (\sin (c+d x)+1)^{9/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(5/2)/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-2*2^(1/4)*(e*Cos[c + d*x])^(7/2)*Hypergeometric2F1[3/4, 7/4, 11/4, (1 - Sin[c + d*x])/2]*Sqrt[a*(1 + Sin[c +
d*x])])/(7*a^2*d*e*(1 + Sin[c + d*x])^(9/4))

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Maple [A]  time = 0.113, size = 232, normalized size = 1.1 \begin{align*} -{\frac{1}{2\,d \left ( \cos \left ( dx+c \right ) \sin \left ( dx+c \right ) - \left ( \cos \left ( dx+c \right ) \right ) ^{2}-2\,\sin \left ( dx+c \right ) -\cos \left ( dx+c \right ) +2 \right ) } \left ( e\cos \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}} \left ( -3\,\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\arctan \left ( 1/2\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}} \right ) \sqrt{2}\sin \left ( dx+c \right ) -3\,\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\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 ( a \left ( 1+\sin \left ( dx+c \right ) \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^(3/2),x)

[Out]

-1/2/d*(e*cos(d*x+c))^(5/2)*(-3*(-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)*sin(d*x+c)-3*(-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))/(cos(d*x+c)*sin(d*x+c)-cos(d*x+c)^2-2*sin(d*x+c)-cos(d*x+c)+2)/(a*(1+sin(d*x+c)))^(3/2)

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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{5}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(5/2)/(a*sin(d*x + c) + a)^(3/2), 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))^(5/2)/(a+a*sin(d*x+c))^(3/2),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))**(5/2)/(a+a*sin(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [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))^(5/2)/(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out