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

Optimal. Leaf size=323 \[ -\frac {77 a^3 (e \cos (c+d x))^{5/2}}{96 d e \sqrt {a \sin (c+d x)+a}}+\frac {77 a^2 e^{3/2} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \tan ^{-1}\left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{64 d (\sin (c+d x)+\cos (c+d x)+1)}-\frac {77 a^2 e^{3/2} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \sinh ^{-1}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right )}{64 d (\sin (c+d x)+\cos (c+d x)+1)}-\frac {11 a^2 \sqrt {a \sin (c+d x)+a} (e \cos (c+d x))^{5/2}}{24 d e}+\frac {77 a^2 e \sqrt {a \sin (c+d x)+a} \sqrt {e \cos (c+d x)}}{64 d}-\frac {a (a \sin (c+d x)+a)^{3/2} (e \cos (c+d x))^{5/2}}{4 d e} \]

[Out]

-1/4*a*(e*cos(d*x+c))^(5/2)*(a+a*sin(d*x+c))^(3/2)/d/e-77/96*a^3*(e*cos(d*x+c))^(5/2)/d/e/(a+a*sin(d*x+c))^(1/
2)-11/24*a^2*(e*cos(d*x+c))^(5/2)*(a+a*sin(d*x+c))^(1/2)/d/e+77/64*a^2*e*(e*cos(d*x+c))^(1/2)*(a+a*sin(d*x+c))
^(1/2)/d-77/64*a^2*e^(3/2)*arcsinh((e*cos(d*x+c))^(1/2)/e^(1/2))*(1+cos(d*x+c))^(1/2)*(a+a*sin(d*x+c))^(1/2)/d
/(1+cos(d*x+c)+sin(d*x+c))+77/64*a^2*e^(3/2)*arctan(sin(d*x+c)*e^(1/2)/(e*cos(d*x+c))^(1/2)/(1+cos(d*x+c))^(1/
2))*(1+cos(d*x+c))^(1/2)*(a+a*sin(d*x+c))^(1/2)/d/(1+cos(d*x+c)+sin(d*x+c))

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Rubi [A]  time = 0.54, antiderivative size = 323, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2678, 2685, 2677, 2775, 203, 2833, 63, 215} \[ \frac {77 a^2 e^{3/2} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \tan ^{-1}\left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{64 d (\sin (c+d x)+\cos (c+d x)+1)}-\frac {77 a^2 e^{3/2} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \sinh ^{-1}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right )}{64 d (\sin (c+d x)+\cos (c+d x)+1)}-\frac {77 a^3 (e \cos (c+d x))^{5/2}}{96 d e \sqrt {a \sin (c+d x)+a}}-\frac {11 a^2 \sqrt {a \sin (c+d x)+a} (e \cos (c+d x))^{5/2}}{24 d e}+\frac {77 a^2 e \sqrt {a \sin (c+d x)+a} \sqrt {e \cos (c+d x)}}{64 d}-\frac {a (a \sin (c+d x)+a)^{3/2} (e \cos (c+d x))^{5/2}}{4 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-77*a^3*(e*Cos[c + d*x])^(5/2))/(96*d*e*Sqrt[a + a*Sin[c + d*x]]) + (77*a^2*e*Sqrt[e*Cos[c + d*x]]*Sqrt[a + a
*Sin[c + d*x]])/(64*d) - (11*a^2*(e*Cos[c + d*x])^(5/2)*Sqrt[a + a*Sin[c + d*x]])/(24*d*e) - (77*a^2*e^(3/2)*A
rcSinh[Sqrt[e*Cos[c + d*x]]/Sqrt[e]]*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(64*d*(1 + Cos[c + d*x]
+ Sin[c + d*x])) + (77*a^2*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]])/(64*d*(1 + Cos[c + d*x] + Sin[c + d*x])) - (a*(e*Cos[c + d*
x])^(5/2)*(a + a*Sin[c + d*x])^(3/2))/(4*d*e)

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 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 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]

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 2678

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

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 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 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]

Rubi steps

\begin {align*} \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^{5/2} \, dx &=-\frac {a (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}}{4 d e}+\frac {1}{8} (11 a) \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^{3/2} \, dx\\ &=-\frac {11 a^2 (e \cos (c+d x))^{5/2} \sqrt {a+a \sin (c+d x)}}{24 d e}-\frac {a (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}}{4 d e}+\frac {1}{48} \left (77 a^2\right ) \int (e \cos (c+d x))^{3/2} \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {77 a^3 (e \cos (c+d x))^{5/2}}{96 d e \sqrt {a+a \sin (c+d x)}}-\frac {11 a^2 (e \cos (c+d x))^{5/2} \sqrt {a+a \sin (c+d x)}}{24 d e}-\frac {a (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}}{4 d e}+\frac {1}{64} \left (77 a^3\right ) \int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=-\frac {77 a^3 (e \cos (c+d x))^{5/2}}{96 d e \sqrt {a+a \sin (c+d x)}}+\frac {77 a^2 e \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{64 d}-\frac {11 a^2 (e \cos (c+d x))^{5/2} \sqrt {a+a \sin (c+d x)}}{24 d e}-\frac {a (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}}{4 d e}+\frac {1}{128} \left (77 a^2 e^2\right ) \int \frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {e \cos (c+d x)}} \, dx\\ &=-\frac {77 a^3 (e \cos (c+d x))^{5/2}}{96 d e \sqrt {a+a \sin (c+d x)}}+\frac {77 a^2 e \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{64 d}-\frac {11 a^2 (e \cos (c+d x))^{5/2} \sqrt {a+a \sin (c+d x)}}{24 d e}-\frac {a (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}}{4 d e}+\frac {\left (77 a^3 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}{128 (a+a \cos (c+d x)+a \sin (c+d x))}+\frac {\left (77 a^3 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}{128 (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=-\frac {77 a^3 (e \cos (c+d x))^{5/2}}{96 d e \sqrt {a+a \sin (c+d x)}}+\frac {77 a^2 e \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{64 d}-\frac {11 a^2 (e \cos (c+d x))^{5/2} \sqrt {a+a \sin (c+d x)}}{24 d e}-\frac {a (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}}{4 d e}-\frac {\left (77 a^3 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 )}{128 d (a+a \cos (c+d x)+a \sin (c+d x))}-\frac {\left (77 a^3 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 )}{64 d (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=-\frac {77 a^3 (e \cos (c+d x))^{5/2}}{96 d e \sqrt {a+a \sin (c+d x)}}+\frac {77 a^2 e \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{64 d}-\frac {11 a^2 (e \cos (c+d x))^{5/2} \sqrt {a+a \sin (c+d x)}}{24 d e}-\frac {a (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}}{4 d e}+\frac {77 a^3 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)}}{64 d (a+a \cos (c+d x)+a \sin (c+d x))}-\frac {\left (77 a^3 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 )}{64 d (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=-\frac {77 a^3 (e \cos (c+d x))^{5/2}}{96 d e \sqrt {a+a \sin (c+d x)}}+\frac {77 a^2 e \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{64 d}-\frac {11 a^2 (e \cos (c+d x))^{5/2} \sqrt {a+a \sin (c+d x)}}{24 d e}-\frac {a (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}}{4 d e}-\frac {77 a^3 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)}}{64 d (a+a \cos (c+d x)+a \sin (c+d x))}+\frac {77 a^3 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)}}{64 d (a+a \cos (c+d x)+a \sin (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 0.30, size = 77, normalized size = 0.24 \[ -\frac {16\ 2^{3/4} (a (\sin (c+d x)+1))^{5/2} (e \cos (c+d x))^{5/2} \, _2F_1\left (-\frac {11}{4},\frac {5}{4};\frac {9}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{5 d e (\sin (c+d x)+1)^{15/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*2^(3/4)*(e*Cos[c + d*x])^(5/2)*Hypergeometric2F1[-11/4, 5/4, 9/4, (1 - Sin[c + d*x])/2]*(a*(1 + Sin[c + d
*x]))^(5/2))/(5*d*e*(1 + Sin[c + d*x])^(15/4))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}} {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]

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))^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.33, size = 344, normalized size = 1.07 \[ -\frac {\left (96 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+96 \left (\cos ^{5}\left (d x +c \right )\right )-368 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )-231 \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sin \left (d x +c \right )+231 \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sin \left (d x +c \right )+272 \left (\cos ^{4}\left (d x +c \right )\right )-308 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-676 \left (\cos ^{3}\left (d x +c \right )\right )+462 \cos \left (d x +c \right ) \sin \left (d x +c \right )-154 \left (\cos ^{2}\left (d x +c \right )\right )+462 \cos \left (d x +c \right )\right ) \left (e \cos \left (d x +c \right )\right )^{\frac {3}{2}} \left (a \left (1+\sin \left (d x +c \right )\right )\right )^{\frac {5}{2}}}{384 d \left (\left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-\left (\cos ^{3}\left (d x +c \right )\right )+2 \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 \left (\cos ^{2}\left (d x +c \right )\right )-4 \sin \left (d x +c \right )+2 \cos \left (d x +c \right )-4\right ) \cos \left (d x +c \right )^{2}} \]

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))^(5/2),x)

[Out]

-1/384/d*(96*sin(d*x+c)*cos(d*x+c)^4+96*cos(d*x+c)^5-368*sin(d*x+c)*cos(d*x+c)^3-231*2^(1/2)*(-2*cos(d*x+c)/(1
+cos(d*x+c)))^(1/2)*arctanh(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)
+231*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*sin
(d*x+c)+272*cos(d*x+c)^4-308*cos(d*x+c)^2*sin(d*x+c)-676*cos(d*x+c)^3+462*cos(d*x+c)*sin(d*x+c)-154*cos(d*x+c)
^2+462*cos(d*x+c))*(e*cos(d*x+c))^(3/2)*(a*(1+sin(d*x+c)))^(5/2)/(cos(d*x+c)^2*sin(d*x+c)-cos(d*x+c)^3+2*cos(d
*x+c)*sin(d*x+c)+3*cos(d*x+c)^2-4*sin(d*x+c)+2*cos(d*x+c)-4)/cos(d*x+c)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}} {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]

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))^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))**(5/2),x)

[Out]

Timed out

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