3.3.93 \(\int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{5/2}} \, dx\) [293]

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

[Out]

4/3*a*(a+a*sin(d*x+c))^(3/2)/d/e/(e*cos(d*x+c))^(3/2)+2*a^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/e^(5/2)/(1+cos(d*x+c)+sin(d*x+c))-2*a^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/e^(5/2)/(1+cos(d*x+c)+sin(d*x+c
))

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Rubi [A]
time = 0.20, antiderivative size = 204, 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 = {2755, 2756, 2854, 209, 2912, 65, 221} \begin {gather*} -\frac {2 a^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 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}} \end {gather*}

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 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 2755

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 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 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 {(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 ) \text {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 ) \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 )}{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 ) \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.19, size = 77, normalized size = 0.38 \begin {gather*} \frac {4\ 2^{3/4} \, _2F_1\left (-\frac {3}{4},-\frac {3}{4};\frac {1}{4};\frac {1}{2} (1-\sin (c+d x))\right ) (a (1+\sin (c+d x)))^{5/2}}{3 d e (e \cos (c+d x))^{3/2} (1+\sin (c+d x))^{7/4}} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(544\) vs. \(2(176)=352\).
time = 0.17, size = 545, normalized size = 2.67

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

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,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
tanh(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)*a
rctan(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+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)+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)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*cos(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)))*(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 = 0.00, 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((a+a*sin(d*x+c))^(5/2)/(e*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 3497 vs. \(2 (160) = 320\).
time = 178.44, size = 3497, normalized size = 17.14 \begin {gather*} \text {Too large to display} \end {gather*}

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]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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