3.3.92 \(\int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{3/2}} \, dx\) [292]

3.3.92.1 Optimal result
3.3.92.2 Mathematica [C] (verified)
3.3.92.3 Rubi [A] (verified)
3.3.92.4 Maple [A] (verified)
3.3.92.5 Fricas [C] (verification not implemented)
3.3.92.6 Sympy [F(-1)]
3.3.92.7 Maxima [F]
3.3.92.8 Giac [F(-1)]
3.3.92.9 Mupad [F(-1)]

3.3.92.1 Optimal result

Integrand size = 27, antiderivative size = 239 \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{3/2}} \, dx=\frac {5 a^3 (e \cos (c+d x))^{3/2}}{d e^3 \sqrt {a+a \sin (c+d x)}}-\frac {5 a^2 \text {arcsinh}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d e^{3/2} (1+\cos (c+d x)+\sin (c+d x))}-\frac {5 a^2 \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d e^{3/2} (1+\cos (c+d x)+\sin (c+d x))}+\frac {4 a (a+a \sin (c+d x))^{3/2}}{d e \sqrt {e \cos (c+d x)}} \]

output
4*a*(a+a*sin(d*x+c))^(3/2)/d/e/(e*cos(d*x+c))^(1/2)+5*a^3*(e*cos(d*x+c))^( 
3/2)/d/e^3/(a+a*sin(d*x+c))^(1/2)-5*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^(3/2)/(1+cos(d*x+c)+si 
n(d*x+c))-5*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^(3/2)/(1+cos(d* 
x+c)+sin(d*x+c))
 
3.3.92.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.10 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.31 \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{3/2}} \, dx=\frac {8 \sqrt [4]{2} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},-\frac {1}{4},\frac {3}{4},\frac {1}{2} (1-\sin (c+d x))\right ) (a (1+\sin (c+d x)))^{5/2}}{d e \sqrt {e \cos (c+d x)} (1+\sin (c+d x))^{9/4}} \]

input
Integrate[(a + a*Sin[c + d*x])^(5/2)/(e*Cos[c + d*x])^(3/2),x]
 
output
(8*2^(1/4)*Hypergeometric2F1[-5/4, -1/4, 3/4, (1 - Sin[c + d*x])/2]*(a*(1 
+ Sin[c + d*x]))^(5/2))/(d*e*Sqrt[e*Cos[c + d*x]]*(1 + Sin[c + d*x])^(9/4) 
)
 
3.3.92.3 Rubi [A] (verified)

Time = 0.97 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.06, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.481, Rules used = {3042, 3155, 3042, 3157, 3042, 3163, 3042, 25, 3254, 216, 3312, 63, 222}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3155

\(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {5 a^2 \int \sqrt {e \cos (c+d x)} \sqrt {\sin (c+d x) a+a}dx}{e^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {5 a^2 \int \sqrt {e \cos (c+d x)} \sqrt {\sin (c+d x) a+a}dx}{e^2}\)

\(\Big \downarrow \) 3157

\(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {5 a^2 \left (\frac {1}{2} a \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {\sin (c+d x) a+a}}dx-\frac {a (e \cos (c+d x))^{3/2}}{d e \sqrt {a \sin (c+d x)+a}}\right )}{e^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {5 a^2 \left (\frac {1}{2} a \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {\sin (c+d x) a+a}}dx-\frac {a (e \cos (c+d x))^{3/2}}{d e \sqrt {a \sin (c+d x)+a}}\right )}{e^2}\)

\(\Big \downarrow \) 3163

\(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {5 a^2 \left (\frac {1}{2} a \left (\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\sqrt {\cos (c+d x)+1}}{\sqrt {e \cos (c+d x)}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}-\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {\cos (c+d x)+1}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}\right )-\frac {a (e \cos (c+d x))^{3/2}}{d e \sqrt {a \sin (c+d x)+a}}\right )}{e^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {5 a^2 \left (\frac {1}{2} a \left (\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}-\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int -\frac {\cos \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}\right )-\frac {a (e \cos (c+d x))^{3/2}}{d e \sqrt {a \sin (c+d x)+a}}\right )}{e^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {5 a^2 \left (\frac {1}{2} a \left (\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}+\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\sqrt {e \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )} \sqrt {\sin \left (\frac {1}{2} (2 c+\pi )+d x\right )+1}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}\right )-\frac {a (e \cos (c+d x))^{3/2}}{d e \sqrt {a \sin (c+d x)+a}}\right )}{e^2}\)

\(\Big \downarrow \) 3254

\(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {5 a^2 \left (\frac {1}{2} a \left (\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\sqrt {e \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )} \sqrt {\sin \left (\frac {1}{2} (2 c+\pi )+d x\right )+1}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}-\frac {2 e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x)}{\cos (c+d x)+1}+1}d\left (-\frac {\sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {\cos (c+d x)+1}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}\right )-\frac {a (e \cos (c+d x))^{3/2}}{d e \sqrt {a \sin (c+d x)+a}}\right )}{e^2}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {5 a^2 \left (\frac {1}{2} a \left (\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\sqrt {e \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )} \sqrt {\sin \left (\frac {1}{2} (2 c+\pi )+d x\right )+1}}dx}{a \sin (c+d x)+a \cos (c+d x)+a}+\frac {2 \sqrt {e} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}\right )-\frac {a (e \cos (c+d x))^{3/2}}{d e \sqrt {a \sin (c+d x)+a}}\right )}{e^2}\)

\(\Big \downarrow \) 3312

\(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {5 a^2 \left (\frac {1}{2} a \left (\frac {e \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {1}{\sqrt {e \cos (c+d x)} \sqrt {\cos (c+d x)+1}}d\cos (c+d x)}{d (a \sin (c+d x)+a \cos (c+d x)+a)}+\frac {2 \sqrt {e} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}\right )-\frac {a (e \cos (c+d x))^{3/2}}{d e \sqrt {a \sin (c+d x)+a}}\right )}{e^2}\)

\(\Big \downarrow \) 63

\(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {5 a^2 \left (\frac {1}{2} a \left (\frac {2 \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \int \frac {1}{\sqrt {\cos (c+d x)+1}}d\sqrt {e \cos (c+d x)}}{d (a \sin (c+d x)+a \cos (c+d x)+a)}+\frac {2 \sqrt {e} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}\right )-\frac {a (e \cos (c+d x))^{3/2}}{d e \sqrt {a \sin (c+d x)+a}}\right )}{e^2}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {4 a (a \sin (c+d x)+a)^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {5 a^2 \left (\frac {1}{2} a \left (\frac {2 \sqrt {e} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \text {arcsinh}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}+\frac {2 \sqrt {e} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d (a \sin (c+d x)+a \cos (c+d x)+a)}\right )-\frac {a (e \cos (c+d x))^{3/2}}{d e \sqrt {a \sin (c+d x)+a}}\right )}{e^2}\)

input
Int[(a + a*Sin[c + d*x])^(5/2)/(e*Cos[c + d*x])^(3/2),x]
 
output
(4*a*(a + a*Sin[c + d*x])^(3/2))/(d*e*Sqrt[e*Cos[c + d*x]]) - (5*a^2*(-((a 
*(e*Cos[c + d*x])^(3/2))/(d*e*Sqrt[a + a*Sin[c + d*x]])) + (a*((2*Sqrt[e]* 
ArcSinh[Sqrt[e*Cos[c + d*x]]/Sqrt[e]]*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a*Si 
n[c + d*x]])/(d*(a + a*Cos[c + d*x] + a*Sin[c + d*x])) + (2*Sqrt[e]*ArcTan 
[(Sqrt[e]*Sin[c + d*x])/(Sqrt[e*Cos[c + d*x]]*Sqrt[1 + Cos[c + d*x]])]*Sqr 
t[1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(d*(a + a*Cos[c + d*x] + a*S 
in[c + d*x]))))/2))/e^2
 

3.3.92.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 63
Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2/b   S 
ubst[Int[1/Sqrt[c + d*(x^2/b)], x], x, Sqrt[b*x]], x] /; FreeQ[{b, c, d}, x 
] && GtQ[c, 0]
 

rule 216
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 222
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 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3155
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] + Simp[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] && Int 
egersQ[2*m, 2*p]
 

rule 3157
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] + Simp[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] && Integers 
Q[2*m, 2*p]
 

rule 3163
Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.) 
*(x_)]], x_Symbol] :> Simp[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] - Simp[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*C 
os[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 3254
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]], x_Symbol] :> Simp[-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 3312
Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*(( 
c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b*f)   Su 
bst[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]
 
3.3.92.4 Maple [A] (verified)

Time = 6.72 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.28

method result size
default \(\frac {\left (5 \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )-5 \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+\cos ^{2}\left (d x +c \right )-\cos \left (d x +c \right ) \sin \left (d x +c \right )+5 \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-5 \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+9 \cos \left (d x +c \right )+8 \sin \left (d x +c \right )+8\right ) \sqrt {a \left (1+\sin \left (d x +c \right )\right )}\, a^{2}}{d \left (1+\cos \left (d x +c \right )+\sin \left (d x +c \right )\right ) e \sqrt {e \cos \left (d x +c \right )}}\) \(307\)

input
int((a+a*sin(d*x+c))^(5/2)/(e*cos(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 
output
1/d*(5*arctan((-cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*(-cos(d*x+c)/(1+cos(d*x+ 
c)))^(1/2)*cos(d*x+c)-5*arctanh(sin(d*x+c)/(1+cos(d*x+c))/(-cos(d*x+c)/(1+ 
cos(d*x+c)))^(1/2))*(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+cos(d*x+ 
c)^2-cos(d*x+c)*sin(d*x+c)+5*(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan((-c 
os(d*x+c)/(1+cos(d*x+c)))^(1/2))-5*(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arct 
anh(sin(d*x+c)/(1+cos(d*x+c))/(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+9*cos(d* 
x+c)+8*sin(d*x+c)+8)*(a*(1+sin(d*x+c)))^(1/2)*a^2/(1+cos(d*x+c)+sin(d*x+c) 
)/e/(e*cos(d*x+c))^(1/2)
 
3.3.92.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.46 (sec) , antiderivative size = 1093, normalized size of antiderivative = 4.57 \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

input
integrate((a+a*sin(d*x+c))^(5/2)/(e*cos(d*x+c))^(3/2),x, algorithm="fricas 
")
 
output
-1/4*(5*d*e^2*(-a^10/(d^4*e^6))^(1/4)*cos(d*x + c)*log(125/2*(2*(a^7*sin(d 
*x + c) + (a^2*d^2*e^3*cos(d*x + c) + a^2*d^2*e^3)*sqrt(-a^10/(d^4*e^6)))* 
sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a) + (2*d^3*e^5*cos(d*x + c)^2 
+ d^3*e^5*cos(d*x + c) - d^3*e^5*sin(d*x + c) - d^3*e^5)*(-a^10/(d^4*e^6)) 
^(3/4) + (a^5*d*e^2*cos(d*x + c) + a^5*d*e^2 + (2*a^5*d*e^2*cos(d*x + c) + 
 a^5*d*e^2)*sin(d*x + c))*(-a^10/(d^4*e^6))^(1/4))/(cos(d*x + c) + sin(d*x 
 + c) + 1)) - 5*d*e^2*(-a^10/(d^4*e^6))^(1/4)*cos(d*x + c)*log(125/2*(2*(a 
^7*sin(d*x + c) + (a^2*d^2*e^3*cos(d*x + c) + a^2*d^2*e^3)*sqrt(-a^10/(d^4 
*e^6)))*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a) - (2*d^3*e^5*cos(d*x 
 + c)^2 + d^3*e^5*cos(d*x + c) - d^3*e^5*sin(d*x + c) - d^3*e^5)*(-a^10/(d 
^4*e^6))^(3/4) - (a^5*d*e^2*cos(d*x + c) + a^5*d*e^2 + (2*a^5*d*e^2*cos(d* 
x + c) + a^5*d*e^2)*sin(d*x + c))*(-a^10/(d^4*e^6))^(1/4))/(cos(d*x + c) + 
 sin(d*x + c) + 1)) + 5*I*d*e^2*(-a^10/(d^4*e^6))^(1/4)*cos(d*x + c)*log(1 
25/2*(2*(a^7*sin(d*x + c) - (a^2*d^2*e^3*cos(d*x + c) + a^2*d^2*e^3)*sqrt( 
-a^10/(d^4*e^6)))*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a) - (2*I*d^3 
*e^5*cos(d*x + c)^2 + I*d^3*e^5*cos(d*x + c) - I*d^3*e^5*sin(d*x + c) - I* 
d^3*e^5)*(-a^10/(d^4*e^6))^(3/4) - (-I*a^5*d*e^2*cos(d*x + c) - I*a^5*d*e^ 
2 + (-2*I*a^5*d*e^2*cos(d*x + c) - I*a^5*d*e^2)*sin(d*x + c))*(-a^10/(d^4* 
e^6))^(1/4))/(cos(d*x + c) + sin(d*x + c) + 1)) - 5*I*d*e^2*(-a^10/(d^4*e^ 
6))^(1/4)*cos(d*x + c)*log(125/2*(2*(a^7*sin(d*x + c) - (a^2*d^2*e^3*co...
 
3.3.92.6 Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]

input
integrate((a+a*sin(d*x+c))**(5/2)/(e*cos(d*x+c))**(3/2),x)
 
output
Timed out
 
3.3.92.7 Maxima [F]

\[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]

input
integrate((a+a*sin(d*x+c))^(5/2)/(e*cos(d*x+c))^(3/2),x, algorithm="maxima 
")
 
output
integrate((a*sin(d*x + c) + a)^(5/2)/(e*cos(d*x + c))^(3/2), x)
 
3.3.92.8 Giac [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]

input
integrate((a+a*sin(d*x+c))^(5/2)/(e*cos(d*x+c))^(3/2),x, algorithm="giac")
 
output
Timed out
 
3.3.92.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{3/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

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