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

Optimal. Leaf size=236 \[ \frac {3 e^{3/2} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \tan ^{-1}\left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{4 d (\sin (c+d x)+\cos (c+d x)+1)}-\frac {3 e^{3/2} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \sinh ^{-1}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right )}{4 d (\sin (c+d x)+\cos (c+d x)+1)}-\frac {a (e \cos (c+d x))^{5/2}}{2 d e \sqrt {a \sin (c+d x)+a}}+\frac {3 e \sqrt {a \sin (c+d x)+a} \sqrt {e \cos (c+d x)}}{4 d} \]

[Out]

-1/2*a*(e*cos(d*x+c))^(5/2)/d/e/(a+a*sin(d*x+c))^(1/2)+3/4*e*(e*cos(d*x+c))^(1/2)*(a+a*sin(d*x+c))^(1/2)/d-3/4
*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))+3/4*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 = 0.36, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ \frac {3 e^{3/2} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \tan ^{-1}\left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{4 d (\sin (c+d x)+\cos (c+d x)+1)}-\frac {3 e^{3/2} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \sinh ^{-1}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right )}{4 d (\sin (c+d x)+\cos (c+d x)+1)}-\frac {a (e \cos (c+d x))^{5/2}}{2 d e \sqrt {a \sin (c+d x)+a}}+\frac {3 e \sqrt {a \sin (c+d x)+a} \sqrt {e \cos (c+d x)}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(e*Cos[c + d*x])^(5/2))/(2*d*e*Sqrt[a + a*Sin[c + d*x]]) + (3*e*Sqrt[e*Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*
x]])/(4*d) - (3*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]])
/(4*d*(1 + Cos[c + d*x] + Sin[c + d*x])) + (3*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]])/(4*d*(1 + Cos[c + d*x] + Sin[c + d*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 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} \sqrt {a+a \sin (c+d x)} \, dx &=-\frac {a (e \cos (c+d x))^{5/2}}{2 d e \sqrt {a+a \sin (c+d x)}}+\frac {1}{4} (3 a) \int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=-\frac {a (e \cos (c+d x))^{5/2}}{2 d e \sqrt {a+a \sin (c+d x)}}+\frac {3 e \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {1}{8} \left (3 e^2\right ) \int \frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {e \cos (c+d x)}} \, dx\\ &=-\frac {a (e \cos (c+d x))^{5/2}}{2 d e \sqrt {a+a \sin (c+d x)}}+\frac {3 e \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {\left (3 a 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}{8 (a+a \cos (c+d x)+a \sin (c+d x))}+\frac {\left (3 a 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}{8 (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=-\frac {a (e \cos (c+d x))^{5/2}}{2 d e \sqrt {a+a \sin (c+d x)}}+\frac {3 e \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{4 d}-\frac {\left (3 a 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 )}{8 d (a+a \cos (c+d x)+a \sin (c+d x))}-\frac {\left (3 a 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 )}{4 d (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=-\frac {a (e \cos (c+d x))^{5/2}}{2 d e \sqrt {a+a \sin (c+d x)}}+\frac {3 e \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {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)}}{4 d (1+\cos (c+d x)+\sin (c+d x))}-\frac {\left (3 a 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 )}{4 d (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=-\frac {a (e \cos (c+d x))^{5/2}}{2 d e \sqrt {a+a \sin (c+d x)}}+\frac {3 e \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{4 d}-\frac {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)}}{4 d (1+\cos (c+d x)+\sin (c+d x))}+\frac {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)}}{4 d (1+\cos (c+d x)+\sin (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 0.95, size = 269, normalized size = 1.14 \[ -\frac {i e e^{-i (c+d x)} \sqrt {a (\sin (c+d x)+1)} \sqrt {e \cos (c+d x)} \left (-3 d x e^{2 i (c+d x)}-2 e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}}+2 i e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}}+e^{3 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}}-i \sqrt {1+e^{2 i (c+d x)}}-3 i e^{2 i (c+d x)} \log \left (1+\sqrt {1+e^{2 i (c+d x)}}\right )+3 e^{2 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{4 d \left (e^{i (c+d x)}+i\right ) \sqrt {1+e^{2 i (c+d x)}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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))^(1/2),x, algorithm="fricas")

[Out]

Timed out

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

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))^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.35, size = 241, normalized size = 1.02 \[ -\frac {\left (3 \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sin \left (d x +c \right )-3 \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sin \left (d x +c \right )-4 \left (\cos ^{3}\left (d x +c \right )\right )-4 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-2 \left (\cos ^{2}\left (d x +c \right )\right )+6 \cos \left (d x +c \right ) \sin \left (d x +c \right )+6 \cos \left (d x +c \right )\right ) \left (e \cos \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {a \left (1+\sin \left (d x +c \right )\right )}}{8 d \left (-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )\right ) \cos \left (d x +c \right )^{2}} \]

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))^(1/2),x)

[Out]

-1/8/d*(3*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)-3*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)*s
in(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)-4*cos(d*x+c)^3-4*cos(d*x+c)^2*sin(d*x+c)-2*cos(d*x+c)^2+6*cos(d*x+c)*
sin(d*x+c)+6*cos(d*x+c))*(e*cos(d*x+c))^(3/2)*(a*(1+sin(d*x+c)))^(1/2)/(-1+cos(d*x+c)-sin(d*x+c))/cos(d*x+c)^2

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

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

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))^(1/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \left (e \cos {\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx \]

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]

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

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