\(\int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{3/2}} \, dx\) [284]

Optimal result

Integrand size = 27, antiderivative size = 210 \[ \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{3/2}} \, dx=\frac {4 a \sqrt {a+a \sin (c+d x)}}{d e \sqrt {e \cos (c+d x)}}-\frac {2 a^2 \text {arcsinh}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d e^{3/2} (a+a \cos (c+d x)+a \sin (c+d x))}-\frac {2 a^2 \arctan \left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d e^{3/2} (a+a \cos (c+d x)+a \sin (c+d x))} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.36 \[ \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{3/2}} \, dx=\frac {4 \sqrt [4]{2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},-\frac {1}{4},\frac {3}{4},\frac {1}{2} (1-\sin (c+d x))\right ) (a (1+\sin (c+d x)))^{3/2}}{d e \sqrt {e \cos (c+d x)} (1+\sin (c+d x))^{5/4}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.01, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {3042, 3155, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3155

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3163

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3254

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 3312

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

\(\Big \downarrow \) 63

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

\(\Big \downarrow \) 222

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

Input:

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

Output:

(4*a*Sqrt[a + a*Sin[c + d*x]])/(d*e*Sqrt[e*Cos[c + d*x]]) - (a^2*((2*Sqrt[ 
e]*ArcSinh[Sqrt[e*Cos[c + d*x]]/Sqrt[e]]*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a 
*Sin[c + d*x]])/(d*(a + a*Cos[c + d*x] + a*Sin[c + d*x])) + (2*Sqrt[e]*Arc 
Tan[(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 + a*Cos[c + d*x] + 
a*Sin[c + d*x]))))/e^2
 

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 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]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 4.43 (sec) , antiderivative size = 857, normalized size of antiderivative = 4.08

method	result	size
default	\(\text {Expression too large to display}\)	\(857\)

Input:

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

Output:

1/d/(1+2^(1/2))/(2*2^(1/2)+3)^(1/2)/e^(7/2)*(((-4*cos(1/2*d*x+1/2*c)^2-8*c 
os(1/2*d*x+1/2*c)-4)*2^(1/2)-8*cos(1/2*d*x+1/2*c)^2-16*cos(1/2*d*x+1/2*c)- 
8)*((2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)+2*cos(1/2*d*x+1/2*c)-1)/(cos(1/2*d 
*x+1/2*c)+1))^(1/2)*e^(5/2)*(e*(2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2 
*c)+1)^2)^(1/2)*(-2*(2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)-2*cos(1/2*d*x+1/2* 
c)+1)/(cos(1/2*d*x+1/2*c)+1))^(1/2)*EllipticPi((2*2^(1/2)+3)^(1/2)*(csc(1/ 
2*d*x+1/2*c)-cot(1/2*d*x+1/2*c)),-1/(2*2^(1/2)+3),(-2*2^(1/2)+3)^(1/2)/(2* 
2^(1/2)+3)^(1/2))+((-6*cos(1/2*d*x+1/2*c)^2+3)*2^(1/2)-8*cos(1/2*d*x+1/2*c 
)^2+4)*e^3*arctanh(2^(1/2)*e^(1/2)*cos(1/2*d*x+1/2*c)/(cos(1/2*d*x+1/2*c)+ 
1)/(e*(2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2))+((2*cos( 
1/2*d*x+1/2*c)^2+4*cos(1/2*d*x+1/2*c)+2)*2^(1/2)+4*cos(1/2*d*x+1/2*c)^2+8* 
cos(1/2*d*x+1/2*c)+4)*((2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)+2*cos(1/2*d*x+1 
/2*c)-1)/(cos(1/2*d*x+1/2*c)+1))^(1/2)*e^(5/2)*EllipticF((1+2^(1/2))*(csc( 
1/2*d*x+1/2*c)-cot(1/2*d*x+1/2*c)),-2*2^(1/2)+3)*(e*(2*cos(1/2*d*x+1/2*c)^ 
2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)*(-2*(2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/ 
2)-2*cos(1/2*d*x+1/2*c)+1)/(cos(1/2*d*x+1/2*c)+1))^(1/2)+(((8*cos(1/2*d*x+ 
1/2*c)+8)*sin(1/2*d*x+1/2*c)+8*cos(1/2*d*x+1/2*c)*(cos(1/2*d*x+1/2*c)+1))* 
2^(1/2)+(12*cos(1/2*d*x+1/2*c)+12)*sin(1/2*d*x+1/2*c)+12*cos(1/2*d*x+1/2*c 
)*(cos(1/2*d*x+1/2*c)+1))*(e*(2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c 
)+1)^2)^(1/2)*e^(5/2))*((1+2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c))*a)^...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 380 vs. \(2 (184) = 368\).

Time = 0.16 (sec) , antiderivative size = 380, normalized size of antiderivative = 1.81 \[ \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{3/2}} \, dx=-\frac {4 \, \sqrt {2} a e \sqrt {\frac {a}{e}} \arctan \left (\frac {\sqrt {2} \sqrt {e \cos \left (d x + c\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {\frac {a}{e}} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right ) + \sqrt {2} a e \sqrt {\frac {a}{e}} \cos \left (d x + c\right ) \log \left (\frac {2 \, \sqrt {2} \sqrt {e \cos \left (d x + c\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {\frac {a}{e}} {\left (\cos \left (d x + c\right ) + 1\right )} + 2 \, a \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) + {\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + a}{\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1}\right ) - \sqrt {2} a e \sqrt {\frac {a}{e}} \cos \left (d x + c\right ) \log \left (-\frac {2 \, \sqrt {2} \sqrt {e \cos \left (d x + c\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {\frac {a}{e}} {\left (\cos \left (d x + c\right ) + 1\right )} - 2 \, a \cos \left (d x + c\right )^{2} - 3 \, a \cos \left (d x + c\right ) - {\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1}\right ) - 16 \, \sqrt {e \cos \left (d x + c\right )} \sqrt {a \sin \left (d x + c\right ) + a} a}{4 \, d e^{2} \cos \left (d x + c\right )} \] Input:

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

Output:

-1/4*(4*sqrt(2)*a*e*sqrt(a/e)*arctan(sqrt(2)*sqrt(e*cos(d*x + c))*sqrt(a*s 
in(d*x + c) + a)*sqrt(a/e)*sin(d*x + c)/(a*cos(d*x + c)^2 + a*cos(d*x + c) 
*sin(d*x + c) + a*cos(d*x + c)))*cos(d*x + c) + sqrt(2)*a*e*sqrt(a/e)*cos( 
d*x + c)*log((2*sqrt(2)*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a)*sqrt 
(a/e)*(cos(d*x + c) + 1) + 2*a*cos(d*x + c)^2 + 3*a*cos(d*x + c) + (2*a*co 
s(d*x + c) + a)*sin(d*x + c) + a)/(cos(d*x + c) + sin(d*x + c) + 1)) - sqr 
t(2)*a*e*sqrt(a/e)*cos(d*x + c)*log(-(2*sqrt(2)*sqrt(e*cos(d*x + c))*sqrt( 
a*sin(d*x + c) + a)*sqrt(a/e)*(cos(d*x + c) + 1) - 2*a*cos(d*x + c)^2 - 3* 
a*cos(d*x + c) - (2*a*cos(d*x + c) + a)*sin(d*x + c) - a)/(cos(d*x + c) + 
sin(d*x + c) + 1)) - 16*sqrt(e*cos(d*x + c))*sqrt(a*sin(d*x + c) + a)*a)/( 
d*e^2*cos(d*x + c))
 

Sympy [F]

\[ \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{3/2}} \, dx=\int \frac {\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}{\left (e \cos {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {e}\, \sqrt {a}\, a \left (\int \frac {\sqrt {\sin \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sin \left (d x +c \right )}{\cos \left (d x +c \right )^{2}}d x +\int \frac {\sqrt {\sin \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right )}{e^{2}} \] Input:

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

Output:

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