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

Optimal. Leaf size=204 \[ -\frac{2 a^2 \sqrt{\cos (c+d x)+1} \sqrt{a \sin (c+d x)+a} \tan ^{-1}\left (\frac{\sqrt{e} \sin (c+d x)}{\sqrt{\cos (c+d x)+1} \sqrt{e \cos (c+d x)}}\right )}{d e^{5/2} (\sin (c+d x)+\cos (c+d x)+1)}+\frac{2 a^2 \sqrt{\cos (c+d x)+1} \sqrt{a \sin (c+d x)+a} \sinh ^{-1}\left (\frac{\sqrt{e \cos (c+d x)}}{\sqrt{e}}\right )}{d e^{5/2} (\sin (c+d x)+\cos (c+d x)+1)}+\frac{4 a (a \sin (c+d x)+a)^{3/2}}{3 d e (e \cos (c+d x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.29602, antiderivative size = 204, 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 = {2676, 2677, 2775, 203, 2833, 63, 215} \[ -\frac{2 a^2 \sqrt{\cos (c+d x)+1} \sqrt{a \sin (c+d x)+a} \tan ^{-1}\left (\frac{\sqrt{e} \sin (c+d x)}{\sqrt{\cos (c+d x)+1} \sqrt{e \cos (c+d x)}}\right )}{d e^{5/2} (\sin (c+d x)+\cos (c+d x)+1)}+\frac{2 a^2 \sqrt{\cos (c+d x)+1} \sqrt{a \sin (c+d x)+a} \sinh ^{-1}\left (\frac{\sqrt{e \cos (c+d x)}}{\sqrt{e}}\right )}{d e^{5/2} (\sin (c+d x)+\cos (c+d x)+1)}+\frac{4 a (a \sin (c+d x)+a)^{3/2}}{3 d e (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2676

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

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{(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{5/2}} \, dx &=\frac{4 a (a+a \sin (c+d x))^{3/2}}{3 d e (e \cos (c+d x))^{3/2}}-\frac{a^2 \int \frac{\sqrt{a+a \sin (c+d x)}}{\sqrt{e \cos (c+d x)}} \, dx}{e^2}\\ &=\frac{4 a (a+a \sin (c+d x))^{3/2}}{3 d e (e \cos (c+d x))^{3/2}}-\frac{\left (a^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}{e^2 (a+a \cos (c+d x)+a \sin (c+d x))}-\frac{\left (a^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}{e^2 (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=\frac{4 a (a+a \sin (c+d x))^{3/2}}{3 d e (e \cos (c+d x))^{3/2}}+\frac{\left (a^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 )}{d e^2 (a+a \cos (c+d x)+a \sin (c+d x))}+\frac{\left (2 a^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 )}{d e^2 (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=\frac{4 a (a+a \sin (c+d x))^{3/2}}{3 d e (e \cos (c+d x))^{3/2}}-\frac{2 a^3 \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 e^{5/2} (a+a \cos (c+d x)+a \sin (c+d x))}+\frac{\left (2 a^3 \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 e^3 (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=\frac{4 a (a+a \sin (c+d x))^{3/2}}{3 d e (e \cos (c+d x))^{3/2}}+\frac{2 a^3 \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 e^{5/2} (a+a \cos (c+d x)+a \sin (c+d x))}-\frac{2 a^3 \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 e^{5/2} (a+a \cos (c+d x)+a \sin (c+d x))}\\ \end{align*}

Mathematica [C]  time = 0.188837, size = 77, normalized size = 0.38 \[ \frac{4\ 2^{3/4} (a (\sin (c+d x)+1))^{5/2} \, _2F_1\left (-\frac{3}{4},-\frac{3}{4};\frac{1}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{3 d e (\sin (c+d x)+1)^{7/4} (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*2^(3/4)*Hypergeometric2F1[-3/4, -3/4, 1/4, (1 - Sin[c + d*x])/2]*(a*(1 + Sin[c + d*x]))^(5/2))/(3*d*e*(e*Co
s[c + d*x])^(3/2)*(1 + Sin[c + d*x])^(7/4))

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Maple [B]  time = 0.137, size = 545, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/d*(3*2^(1/2)*arctan(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)-3*2^(1/2)*arc
tanh(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))*sin(d*x+c)*cos(d*x+c)-3*cos(d*x+c
)^2*2^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+3*cos(d*x+c)^2*2^(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))-4*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c
)*cos(d*x+c)-6*2^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*sin(d*x+c)+6*2^(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))*sin(d*x+c)-3*cos(d*x+c)*2^(1/2)*arctan(
1/2*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+3*cos(d*x+c)*2^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(1+c
os(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c))+6*2^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-6*
2^(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)))*(a*(1+sin(d*x+c)))^(5
/2)/(1+sin(d*x+c))/sin(d*x+c)/(e*cos(d*x+c))^(5/2)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}{\left (e \cos \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out