3.4.15 \(\int \frac {(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^{5/2}} \, dx\) [315]

Optimal. Leaf size=218 \[ -\frac {4 e (e \cos (c+d x))^{3/2}}{3 a d (a+a \sin (c+d x))^{3/2}}-\frac {2 e^{5/2} \sinh ^{-1}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d \left (a^3+a^3 \cos (c+d x)+a^3 \sin (c+d x)\right )}-\frac {2 e^{5/2} \tan ^{-1}\left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d \left (a^3+a^3 \cos (c+d x)+a^3 \sin (c+d x)\right )} \]

[Out]

-4/3*e*(e*cos(d*x+c))^(3/2)/a/d/(a+a*sin(d*x+c))^(3/2)-2*e^(5/2)*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/(a^3+a^3*cos(d*x+c)+a^3*sin(d*x+c))-2*e^(5/2)*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/(a^3+a^3*cos(d*x+c)+
a^3*sin(d*x+c))

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Rubi [A]
time = 0.20, antiderivative size = 218, normalized size of antiderivative = 1.00, 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 = {2759, 2763, 2854, 209, 2912, 65, 221} \begin {gather*} -\frac {2 e^{5/2} \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 )}{d \left (a^3 \sin (c+d x)+a^3 \cos (c+d x)+a^3\right )}-\frac {2 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^3 \sin (c+d x)+a^3 \cos (c+d x)+a^3\right )}-\frac {4 e (e \cos (c+d x))^{3/2}}{3 a d (a \sin (c+d x)+a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*e*(e*Cos[c + d*x])^(3/2))/(3*a*d*(a + a*Sin[c + d*x])^(3/2)) - (2*e^(5/2)*ArcSinh[Sqrt[e*Cos[c + d*x]]/Sqr
t[e]]*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(d*(a^3 + a^3*Cos[c + d*x] + a^3*Sin[c + d*x])) - (2*e^
(5/2)*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*(a^3 + a^3*Cos[c + d*x] + a^3*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 2759

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

Rule 2763

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 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))^{5/2}}{(a+a \sin (c+d x))^{5/2}} \, dx &=-\frac {4 e (e \cos (c+d x))^{3/2}}{3 a d (a+a \sin (c+d x))^{3/2}}-\frac {e^2 \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx}{a^2}\\ &=-\frac {4 e (e \cos (c+d x))^{3/2}}{3 a d (a+a \sin (c+d x))^{3/2}}-\frac {\left (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}{a^2 (a+a \cos (c+d x)+a \sin (c+d x))}+\frac {\left (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}{a^2 (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=-\frac {4 e (e \cos (c+d x))^{3/2}}{3 a d (a+a \sin (c+d x))^{3/2}}-\frac {\left (e^3 \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^2 d (a+a \cos (c+d x)+a \sin (c+d x))}+\frac {\left (2 e^3 \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^2 d (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=-\frac {4 e (e \cos (c+d x))^{3/2}}{3 a d (a+a \sin (c+d x))^{3/2}}-\frac {2 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^3+a^3 \cos (c+d x)+a^3 \sin (c+d x)\right )}-\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}{\sqrt {1+\frac {x^2}{e}}} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{a^2 d (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=-\frac {4 e (e \cos (c+d x))^{3/2}}{3 a d (a+a \sin (c+d x))^{3/2}}-\frac {2 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^3+a^3 \cos (c+d x)+a^3 \sin (c+d x)\right )}-\frac {2 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^3+a^3 \cos (c+d x)+a^3 \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 = 0.14, size = 80, normalized size = 0.37 \begin {gather*} -\frac {\sqrt [4]{2} (e \cos (c+d x))^{7/2} \, _2F_1\left (\frac {7}{4},\frac {7}{4};\frac {11}{4};\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt {a (1+\sin (c+d x))}}{7 a^3 d e (1+\sin (c+d x))^{9/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(544\) vs. \(2(190)=380\).
time = 0.15, size = 545, normalized size = 2.50

method	result	size
default	\(-\frac {\left (3 \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 ) \cos \left (d x +c \right )+3 \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 ) \left (\cos ^{2}\left (d x +c \right )\right )+3 \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 ) \cos \left (d x +c \right )+3 \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 ) \left (\cos ^{2}\left (d x +c \right )\right )-6 \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 )+3 \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 ) \cos \left (d x +c \right )-6 \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 )+3 \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 ) \cos \left (d x +c \right )-4 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )-6 \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 )-6 \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 )\right ) \left (e \cos \left (d x +c \right )\right )^{\frac {5}{2}}}{3 d \left (\sin \left (d x +c \right )-1\right ) \left (a \left (1+\sin \left (d x +c \right )\right )\right )^{\frac {5}{2}} \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )}\)	\(545\)

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))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3/d*(3*2^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*sin(d*x+c)*cos(d*x+c)+3*2^(1/2)*arc
tan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*cos(d*x+c)^2+3*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)*cos(d*x+c)+3*2^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+c
os(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*cos(d*x+c)^2-6*2^(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)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*cos(d*x+c)-6*
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)+3*2^(1/2)*a
rctanh(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))*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*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))-6*
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)))*(e*cos(d*x+c))^(5/2)/
(sin(d*x+c)-1)/(a*(1+sin(d*x+c)))^(5/2)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/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))^(5/2)/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 3631 vs. \(2 (176) = 352\).
time = 187.76, size = 3631, normalized size = 16.66 \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))^(5/2)/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/12*(12*(3*sqrt(2)*a^3*d*cos(d*x + c)^2 - 4*sqrt(2)*a^3*d + (sqrt(2)*a^3*d*cos(d*x + c)^2 - 4*sqrt(2)*a^3*d)*
sin(d*x + c))*(1/(a^10*d^4))^(1/4)*arctan(-1/4*(sqrt(2)*((sqrt(2)*a^8*d^3*cos(d*x + c)^6 - 3*sqrt(2)*a^8*d^3*c
os(d*x + c)^5 - 8*sqrt(2)*a^8*d^3*cos(d*x + c)^4 + 4*sqrt(2)*a^8*d^3*cos(d*x + c)^3 + 8*sqrt(2)*a^8*d^3*cos(d*
x + c)^2 - (sqrt(2)*a^8*d^3*cos(d*x + c)^5 + 4*sqrt(2)*a^8*d^3*cos(d*x + c)^4 - 4*sqrt(2)*a^8*d^3*cos(d*x + c)
^3 - 8*sqrt(2)*a^8*d^3*cos(d*x + c)^2)*sin(d*x + c))*(1/(a^10*d^4))^(3/4)*e^(15/2) + (sqrt(2)*a^3*d*cos(d*x +
c)^6*e^5 + 5*sqrt(2)*a^3*d*cos(d*x + c)^5*e^5 - 8*sqrt(2)*a^3*d*cos(d*x + c)^4*e^5 - 20*sqrt(2)*a^3*d*cos(d*x
+ c)^3*e^5 + 8*sqrt(2)*a^3*d*cos(d*x + c)^2*e^5 + 16*sqrt(2)*a^3*d*cos(d*x + c)*e^5 + (sqrt(2)*a^3*d*cos(d*x +
c)^5*e^5 - 4*sqrt(2)*a^3*d*cos(d*x + c)^4*e^5 - 12*sqrt(2)*a^3*d*cos(d*x + c)^3*e^5 + 8*sqrt(2)*a^3*d*cos(d*x
+ c)^2*e^5 + 16*sqrt(2)*a^3*d*cos(d*x + c)*e^5)*sin(d*x + c))*(1/(a^10*d^4))^(1/4)*e^(5/2) - (cos(d*x + c)^4*
e^(15/2) - 3*cos(d*x + c)^3*e^(15/2) - 8*cos(d*x + c)^2*e^(15/2) + (2*a^5*d^2*cos(d*x + c)^5*e^(5/2) - 5*a^5*d
^2*cos(d*x + c)^4*e^(5/2) - 19*a^5*d^2*cos(d*x + c)^3*e^(5/2) + 20*a^5*d^2*cos(d*x + c)*e^(5/2) + 8*a^5*d^2*e^
(5/2) - (2*a^5*d^2*cos(d*x + c)^4*e^(5/2) + 9*a^5*d^2*cos(d*x + c)^3*e^(5/2) - 4*a^5*d^2*cos(d*x + c)^2*e^(5/2
) - 20*a^5*d^2*cos(d*x + c)*e^(5/2) - 8*a^5*d^2*e^(5/2))*sin(d*x + c))*sqrt(1/(a^10*d^4))*e^5 + 4*cos(d*x + c)
*e^(15/2) - (cos(d*x + c)^3*e^(15/2) + 4*cos(d*x + c)^2*e^(15/2) - 4*cos(d*x + c)*e^(15/2) - 8*e^(15/2))*sin(d
*x + c) + 8*e^(15/2))*sqrt(a*sin(d*x + c) + a)*sqrt(cos(d*x + c)))*sqrt((2*a*cos(d*x + c)*e^15*sin(d*x + c) +
2*a*cos(d*x + c)*e^15 + (a^6*d^2*e^10*sin(d*x + c) + a^6*d^2*e^10)*sqrt(1/(a^10*d^4))*e^5 + (sqrt(2)*a^3*d*(1/
(a^10*d^4))^(1/4)*cos(d*x + c)*e^15 + (sqrt(2)*a^8*d^3*e^(15/2)*sin(d*x + c) + sqrt(2)*a^8*d^3*e^(15/2))*(1/(a
^10*d^4))^(3/4)*e^(15/2))*sqrt(a*sin(d*x + c) + a)*sqrt(cos(d*x + c)))/(sin(d*x + c) + 1)) + ((2*sqrt(2)*a^8*d
^3*cos(d*x + c)^5*e^(15/2) + sqrt(2)*a^8*d^3*cos(d*x + c)^4*e^(15/2) - 13*sqrt(2)*a^8*d^3*cos(d*x + c)^3*e^(15
/2) - 8*sqrt(2)*a^8*d^3*cos(d*x + c)^2*e^(15/2) + 12*sqrt(2)*a^8*d^3*cos(d*x + c)*e^(15/2) + 8*sqrt(2)*a^8*d^3
*e^(15/2) - (7*sqrt(2)*a^8*d^3*cos(d*x + c)^3*e^(15/2) + 4*sqrt(2)*a^8*d^3*cos(d*x + c)^2*e^(15/2) - 12*sqrt(2
)*a^8*d^3*cos(d*x + c)*e^(15/2) - 8*sqrt(2)*a^8*d^3*e^(15/2))*sin(d*x + c))*(1/(a^10*d^4))^(3/4)*e^(15/2) + (7
*sqrt(2)*a^3*d*cos(d*x + c)^4*e^(25/2) + 3*sqrt(2)*a^3*d*cos(d*x + c)^3*e^(25/2) - 16*sqrt(2)*a^3*d*cos(d*x +
c)^2*e^(25/2) - 4*sqrt(2)*a^3*d*cos(d*x + c)*e^(25/2) + 8*sqrt(2)*a^3*d*e^(25/2) + (2*sqrt(2)*a^3*d*cos(d*x +
c)^4*e^(25/2) + sqrt(2)*a^3*d*cos(d*x + c)^3*e^(25/2) - 12*sqrt(2)*a^3*d*cos(d*x + c)^2*e^(25/2) - 4*sqrt(2)*a
^3*d*cos(d*x + c)*e^(25/2) + 8*sqrt(2)*a^3*d*e^(25/2))*sin(d*x + c))*(1/(a^10*d^4))^(1/4)*e^(5/2))*sqrt(a*sin(
d*x + c) + a)*sqrt(cos(d*x + c)))/(a*cos(d*x + c)^6*e^15 + a*cos(d*x + c)^5*e^15 - 8*a*cos(d*x + c)^4*e^15 - 8
*a*cos(d*x + c)^3*e^15 + 8*a*cos(d*x + c)^2*e^15 + 8*a*cos(d*x + c)*e^15 - 4*(a*cos(d*x + c)^4*e^15 + a*cos(d*
x + c)^3*e^15 - 2*a*cos(d*x + c)^2*e^15 - 2*a*cos(d*x + c)*e^15)*sin(d*x + c)))*e^(5/2) - 12*(3*sqrt(2)*a^3*d*
cos(d*x + c)^2 - 4*sqrt(2)*a^3*d + (sqrt(2)*a^3*d*cos(d*x + c)^2 - 4*sqrt(2)*a^3*d)*sin(d*x + c))*(1/(a^10*d^4
))^(1/4)*arctan(1/4*(sqrt(2)*((sqrt(2)*a^8*d^3*cos(d*x + c)^6 - 3*sqrt(2)*a^8*d^3*cos(d*x + c)^5 - 8*sqrt(2)*a
^8*d^3*cos(d*x + c)^4 + 4*sqrt(2)*a^8*d^3*cos(d*x + c)^3 + 8*sqrt(2)*a^8*d^3*cos(d*x + c)^2 - (sqrt(2)*a^8*d^3
*cos(d*x + c)^5 + 4*sqrt(2)*a^8*d^3*cos(d*x + c)^4 - 4*sqrt(2)*a^8*d^3*cos(d*x + c)^3 - 8*sqrt(2)*a^8*d^3*cos(
d*x + c)^2)*sin(d*x + c))*(1/(a^10*d^4))^(3/4)*e^(15/2) + (sqrt(2)*a^3*d*cos(d*x + c)^6*e^5 + 5*sqrt(2)*a^3*d*
cos(d*x + c)^5*e^5 - 8*sqrt(2)*a^3*d*cos(d*x + c)^4*e^5 - 20*sqrt(2)*a^3*d*cos(d*x + c)^3*e^5 + 8*sqrt(2)*a^3*
d*cos(d*x + c)^2*e^5 + 16*sqrt(2)*a^3*d*cos(d*x + c)*e^5 + (sqrt(2)*a^3*d*cos(d*x + c)^5*e^5 - 4*sqrt(2)*a^3*d
*cos(d*x + c)^4*e^5 - 12*sqrt(2)*a^3*d*cos(d*x + c)^3*e^5 + 8*sqrt(2)*a^3*d*cos(d*x + c)^2*e^5 + 16*sqrt(2)*a^
3*d*cos(d*x + c)*e^5)*sin(d*x + c))*(1/(a^10*d^4))^(1/4)*e^(5/2) + (cos(d*x + c)^4*e^(15/2) - 3*cos(d*x + c)^3
*e^(15/2) - 8*cos(d*x + c)^2*e^(15/2) + (2*a^5*d^2*cos(d*x + c)^5*e^(5/2) - 5*a^5*d^2*cos(d*x + c)^4*e^(5/2) -
19*a^5*d^2*cos(d*x + c)^3*e^(5/2) + 20*a^5*d^2*cos(d*x + c)*e^(5/2) + 8*a^5*d^2*e^(5/2) - (2*a^5*d^2*cos(d*x
+ c)^4*e^(5/2) + 9*a^5*d^2*cos(d*x + c)^3*e^(5/2) - 4*a^5*d^2*cos(d*x + c)^2*e^(5/2) - 20*a^5*d^2*cos(d*x + c)
*e^(5/2) - 8*a^5*d^2*e^(5/2))*sin(d*x + c))*sqrt(1/(a^10*d^4))*e^5 + 4*cos(d*x + c)*e^(15/2) - (cos(d*x + c)^3
*e^(15/2) + 4*cos(d*x + c)^2*e^(15/2) - 4*cos(d*x + c)*e^(15/2) - 8*e^(15/2))*sin(d*x + c) + 8*e^(15/2))*sqrt(
a*sin(d*x + c) + a)*sqrt(cos(d*x + c)))*sqrt((2*a*cos(d*x + c)*e^15*sin(d*x + c) + 2*a*cos(d*x + c)*e^15 + (a^
6*d^2*e^10*sin(d*x + c) + a^6*d^2*e^10)*sqrt(1/(a^10*d^4))*e^5 - (sqrt(2)*a^3*d*(1/(a^10*d^4))^(1/4)*cos(d*x +
c)*e^15 + (sqrt(2)*a^8*d^3*e^(15/2)*sin(d*x + ...

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4370 deep

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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))^(5/2)/(a+a*sin(d*x+c))^(5/2),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 )}^{5/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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