3.4.7 \(\int \frac {(e \cos (c+d x))^{3/2}}{(a+a \sin (c+d x))^{3/2}} \, dx\) [307]

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

[Out]

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

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Rubi [A]
time = 0.24, 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 = {2760, 2764, 2756, 2854, 209, 2912, 65, 221} \begin {gather*} -\frac {2 e^{3/2} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \text {ArcTan}\left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{a^2 d (\sin (c+d x)+\cos (c+d x)+1)}+\frac {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 )}{a^2 d (\sin (c+d x)+\cos (c+d x)+1)}-\frac {2 e \sqrt {a \sin (c+d x)+a} \sqrt {e \cos (c+d x)}}{a^2 d}-\frac {2 (e \cos (c+d x))^{5/2}}{d e (a \sin (c+d x)+a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

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 209

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

Rule 221

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 2756

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 2760

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

Rule 2764

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(3/2)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[g*Sqrt
[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 2854

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 2912

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/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rubi steps

\begin {align*} \int \frac {(e \cos (c+d x))^{3/2}}{(a+a \sin (c+d x))^{3/2}} \, dx &=-\frac {2 (e \cos (c+d x))^{5/2}}{d e (a+a \sin (c+d x))^{3/2}}-\frac {2 \int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+a \sin (c+d x)}} \, dx}{a}\\ &=-\frac {2 (e \cos (c+d x))^{5/2}}{d e (a+a \sin (c+d x))^{3/2}}-\frac {2 e \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{a^2 d}-\frac {e^2 \int \frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {e \cos (c+d x)}} \, dx}{a^2}\\ &=-\frac {2 (e \cos (c+d x))^{5/2}}{d e (a+a \sin (c+d x))^{3/2}}-\frac {2 e \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{a^2 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}{a (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}{a (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=-\frac {2 (e \cos (c+d x))^{5/2}}{d e (a+a \sin (c+d x))^{3/2}}-\frac {2 e \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{a^2 d}+\frac {\left (e^2 \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {e x} \sqrt {1+x}} \, dx,x,\cos (c+d x)\right )}{a d (a+a \cos (c+d x)+a \sin (c+d x))}+\frac {\left (2 e^2 \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}\right ) \text {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 {2 (e \cos (c+d x))^{5/2}}{d e (a+a \sin (c+d x))^{3/2}}-\frac {2 e \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{a^2 d}-\frac {2 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 \left (a^2+a^2 \cos (c+d x)+a^2 \sin (c+d x)\right )}+\frac {\left (2 e \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}\right ) \text {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 {2 (e \cos (c+d x))^{5/2}}{d e (a+a \sin (c+d x))^{3/2}}-\frac {2 e \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{a^2 d}+\frac {2 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 \left (a^2+a^2 \cos (c+d x)+a^2 \sin (c+d x)\right )}-\frac {2 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 \left (a^2+a^2 \cos (c+d x)+a^2 \sin (c+d x)\right )}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.13, size = 80, normalized size = 0.34 \begin {gather*} -\frac {2^{3/4} (e \cos (c+d x))^{5/2} \, _2F_1\left (\frac {5}{4},\frac {5}{4};\frac {9}{4};\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt {a (1+\sin (c+d x))}}{5 a^2 d e (1+\sin (c+d x))^{7/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(2^(3/4)*(e*Cos[c + d*x])^(5/2)*Hypergeometric2F1[5/4, 5/4, 9/4, (1 - Sin[c + d*x])/2]*Sqrt[a*(1 + Sin[c
+ d*x])])/(a^2*d*e*(1 + Sin[c + d*x])^(7/4))

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Maple [A]
time = 0.16, size = 321, normalized size = 1.36

method	result	size
default	\(\frac {2 \left (e \cos \left (d x +c \right )\right )^{\frac {3}{2}} \left (-1+\cos \left (d x +c \right )\right ) \left (-\sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sin \left (d x +c \right )+\sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sin \left (d x +c \right )+2 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )-\sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right )+\sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right )+2 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+2 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )}{d \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (a \left (1+\sin \left (d x +c \right )\right )\right )^{\frac {3}{2}} \left (\cos \left (d x +c \right )-1+\sin \left (d x +c \right )\right )}\)	\(321\)

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))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 3519 vs. \(2 (189) = 378\).
time = 188.74, size = 3519, normalized size = 14.91 \begin {gather*} \text {Too large to display} \end {gather*}

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))^(3/2),x, algorithm="fricas")

[Out]

-1/4*(4*(sqrt(2)*a^2*d*cos(d*x + c)^2 - 2*sqrt(2)*a^2*d*sin(d*x + c) - 2*sqrt(2)*a^2*d)*(1/(a^6*d^4))^(1/4)*ar
ctan(-1/4*(sqrt(2)*((sqrt(2)*a^5*d^3*cos(d*x + c)^6 + 5*sqrt(2)*a^5*d^3*cos(d*x + c)^5 - 8*sqrt(2)*a^5*d^3*cos
(d*x + c)^4 - 20*sqrt(2)*a^5*d^3*cos(d*x + c)^3 + 8*sqrt(2)*a^5*d^3*cos(d*x + c)^2 + 16*sqrt(2)*a^5*d^3*cos(d*
x + c) + (sqrt(2)*a^5*d^3*cos(d*x + c)^5 - 4*sqrt(2)*a^5*d^3*cos(d*x + c)^4 - 12*sqrt(2)*a^5*d^3*cos(d*x + c)^
3 + 8*sqrt(2)*a^5*d^3*cos(d*x + c)^2 + 16*sqrt(2)*a^5*d^3*cos(d*x + c))*sin(d*x + c))*(1/(a^6*d^4))^(3/4)*e^(9
/2) + (sqrt(2)*a^2*d*cos(d*x + c)^6*e^3 - 3*sqrt(2)*a^2*d*cos(d*x + c)^5*e^3 - 8*sqrt(2)*a^2*d*cos(d*x + c)^4*
e^3 + 4*sqrt(2)*a^2*d*cos(d*x + c)^3*e^3 + 8*sqrt(2)*a^2*d*cos(d*x + c)^2*e^3 - (sqrt(2)*a^2*d*cos(d*x + c)^5*
e^3 + 4*sqrt(2)*a^2*d*cos(d*x + c)^4*e^3 - 4*sqrt(2)*a^2*d*cos(d*x + c)^3*e^3 - 8*sqrt(2)*a^2*d*cos(d*x + c)^2
*e^3)*sin(d*x + c))*(1/(a^6*d^4))^(1/4)*e^(3/2) - (cos(d*x + c)^4*e^(9/2) - 3*cos(d*x + c)^3*e^(9/2) - 8*cos(d
*x + c)^2*e^(9/2) + (2*a^3*d^2*cos(d*x + c)^5*e^(3/2) - 5*a^3*d^2*cos(d*x + c)^4*e^(3/2) - 19*a^3*d^2*cos(d*x
+ c)^3*e^(3/2) + 20*a^3*d^2*cos(d*x + c)*e^(3/2) + 8*a^3*d^2*e^(3/2) - (2*a^3*d^2*cos(d*x + c)^4*e^(3/2) + 9*a
^3*d^2*cos(d*x + c)^3*e^(3/2) - 4*a^3*d^2*cos(d*x + c)^2*e^(3/2) - 20*a^3*d^2*cos(d*x + c)*e^(3/2) - 8*a^3*d^2
*e^(3/2))*sin(d*x + c))*sqrt(1/(a^6*d^4))*e^3 + 4*cos(d*x + c)*e^(9/2) - (cos(d*x + c)^3*e^(9/2) + 4*cos(d*x +
c)^2*e^(9/2) - 4*cos(d*x + c)*e^(9/2) - 8*e^(9/2))*sin(d*x + c) + 8*e^(9/2))*sqrt(a*sin(d*x + c) + a)*sqrt(co
s(d*x + c)))*sqrt((2*a*cos(d*x + c)*e^9*sin(d*x + c) + 2*a*cos(d*x + c)*e^9 + (a^4*d^2*e^6*sin(d*x + c) + a^4*
d^2*e^6)*sqrt(1/(a^6*d^4))*e^3 + (sqrt(2)*a^5*d^3*(1/(a^6*d^4))^(3/4)*cos(d*x + c)*e^9 + (sqrt(2)*a^2*d*e^(15/
2)*sin(d*x + c) + sqrt(2)*a^2*d*e^(15/2))*(1/(a^6*d^4))^(1/4)*e^(3/2))*sqrt(a*sin(d*x + c) + a)*sqrt(cos(d*x +
c)))/(sin(d*x + c) + 1)) - ((7*sqrt(2)*a^5*d^3*cos(d*x + c)^4*e^(9/2) + 3*sqrt(2)*a^5*d^3*cos(d*x + c)^3*e^(9
/2) - 16*sqrt(2)*a^5*d^3*cos(d*x + c)^2*e^(9/2) - 4*sqrt(2)*a^5*d^3*cos(d*x + c)*e^(9/2) + 8*sqrt(2)*a^5*d^3*e
^(9/2) + (2*sqrt(2)*a^5*d^3*cos(d*x + c)^4*e^(9/2) + sqrt(2)*a^5*d^3*cos(d*x + c)^3*e^(9/2) - 12*sqrt(2)*a^5*d
^3*cos(d*x + c)^2*e^(9/2) - 4*sqrt(2)*a^5*d^3*cos(d*x + c)*e^(9/2) + 8*sqrt(2)*a^5*d^3*e^(9/2))*sin(d*x + c))*
(1/(a^6*d^4))^(3/4)*e^(9/2) + (2*sqrt(2)*a^2*d*cos(d*x + c)^5*e^(15/2) + sqrt(2)*a^2*d*cos(d*x + c)^4*e^(15/2)
- 13*sqrt(2)*a^2*d*cos(d*x + c)^3*e^(15/2) - 8*sqrt(2)*a^2*d*cos(d*x + c)^2*e^(15/2) + 12*sqrt(2)*a^2*d*cos(d
*x + c)*e^(15/2) + 8*sqrt(2)*a^2*d*e^(15/2) - (7*sqrt(2)*a^2*d*cos(d*x + c)^3*e^(15/2) + 4*sqrt(2)*a^2*d*cos(d
*x + c)^2*e^(15/2) - 12*sqrt(2)*a^2*d*cos(d*x + c)*e^(15/2) - 8*sqrt(2)*a^2*d*e^(15/2))*sin(d*x + c))*(1/(a^6*
d^4))^(1/4)*e^(3/2))*sqrt(a*sin(d*x + c) + a)*sqrt(cos(d*x + c)))/(a*cos(d*x + c)^6*e^9 + a*cos(d*x + c)^5*e^9
- 8*a*cos(d*x + c)^4*e^9 - 8*a*cos(d*x + c)^3*e^9 + 8*a*cos(d*x + c)^2*e^9 + 8*a*cos(d*x + c)*e^9 - 4*(a*cos(
d*x + c)^4*e^9 + a*cos(d*x + c)^3*e^9 - 2*a*cos(d*x + c)^2*e^9 - 2*a*cos(d*x + c)*e^9)*sin(d*x + c)))*e^(3/2)
- 4*(sqrt(2)*a^2*d*cos(d*x + c)^2 - 2*sqrt(2)*a^2*d*sin(d*x + c) - 2*sqrt(2)*a^2*d)*(1/(a^6*d^4))^(1/4)*arctan
(1/4*(sqrt(2)*((sqrt(2)*a^5*d^3*cos(d*x + c)^6 + 5*sqrt(2)*a^5*d^3*cos(d*x + c)^5 - 8*sqrt(2)*a^5*d^3*cos(d*x
+ c)^4 - 20*sqrt(2)*a^5*d^3*cos(d*x + c)^3 + 8*sqrt(2)*a^5*d^3*cos(d*x + c)^2 + 16*sqrt(2)*a^5*d^3*cos(d*x + c
) + (sqrt(2)*a^5*d^3*cos(d*x + c)^5 - 4*sqrt(2)*a^5*d^3*cos(d*x + c)^4 - 12*sqrt(2)*a^5*d^3*cos(d*x + c)^3 + 8
*sqrt(2)*a^5*d^3*cos(d*x + c)^2 + 16*sqrt(2)*a^5*d^3*cos(d*x + c))*sin(d*x + c))*(1/(a^6*d^4))^(3/4)*e^(9/2) +
(sqrt(2)*a^2*d*cos(d*x + c)^6*e^3 - 3*sqrt(2)*a^2*d*cos(d*x + c)^5*e^3 - 8*sqrt(2)*a^2*d*cos(d*x + c)^4*e^3 +
4*sqrt(2)*a^2*d*cos(d*x + c)^3*e^3 + 8*sqrt(2)*a^2*d*cos(d*x + c)^2*e^3 - (sqrt(2)*a^2*d*cos(d*x + c)^5*e^3 +
4*sqrt(2)*a^2*d*cos(d*x + c)^4*e^3 - 4*sqrt(2)*a^2*d*cos(d*x + c)^3*e^3 - 8*sqrt(2)*a^2*d*cos(d*x + c)^2*e^3)
*sin(d*x + c))*(1/(a^6*d^4))^(1/4)*e^(3/2) + (cos(d*x + c)^4*e^(9/2) - 3*cos(d*x + c)^3*e^(9/2) - 8*cos(d*x +
c)^2*e^(9/2) + (2*a^3*d^2*cos(d*x + c)^5*e^(3/2) - 5*a^3*d^2*cos(d*x + c)^4*e^(3/2) - 19*a^3*d^2*cos(d*x + c)^
3*e^(3/2) + 20*a^3*d^2*cos(d*x + c)*e^(3/2) + 8*a^3*d^2*e^(3/2) - (2*a^3*d^2*cos(d*x + c)^4*e^(3/2) + 9*a^3*d^
2*cos(d*x + c)^3*e^(3/2) - 4*a^3*d^2*cos(d*x + c)^2*e^(3/2) - 20*a^3*d^2*cos(d*x + c)*e^(3/2) - 8*a^3*d^2*e^(3
/2))*sin(d*x + c))*sqrt(1/(a^6*d^4))*e^3 + 4*cos(d*x + c)*e^(9/2) - (cos(d*x + c)^3*e^(9/2) + 4*cos(d*x + c)^2
*e^(9/2) - 4*cos(d*x + c)*e^(9/2) - 8*e^(9/2))*sin(d*x + c) + 8*e^(9/2))*sqrt(a*sin(d*x + c) + a)*sqrt(cos(d*x
+ c)))*sqrt((2*a*cos(d*x + c)*e^9*sin(d*x + c) + 2*a*cos(d*x + c)*e^9 + (a^4*d^2*e^6*sin(d*x + c) + a^4*d^2*e
^6)*sqrt(1/(a^6*d^4))*e^3 - (sqrt(2)*a^5*d^3*(1/(a^6*d^4))^(3/4)*cos(d*x + c)*e^9 + (sqrt(2)*a^2*d*e^(15/2)*si
n(d*x + c) + sqrt(2)*a^2*d*e^(15/2))*(1/(a^6*d^4))^(1/4)*e^(3/2))*sqrt(a*sin(d*x + c) + a)*sqrt(cos(d*x + c)))
/(sin(d*x + c) + 1)) - ((7*sqrt(2)*a^5*d^3*cos(...

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

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))**(3/2),x)

[Out]

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

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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))^(3/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \end {gather*}

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))^(3/2),x)

[Out]

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

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