\(\int \frac {(c+d \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx\) [597]

Optimal result

Integrand size = 29, antiderivative size = 188 \[ \int \frac {(c+d \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx=-\frac {(3 c-d) \sqrt {d} \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {a} f}-\frac {\sqrt {2} (c-d)^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {a} f}-\frac {d \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a+a \sin (e+f x)}} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 15.61 (sec) , antiderivative size = 1639, normalized size of antiderivative = 8.72 \[ \int \frac {(c+d \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-(d*Cos[(e + f*x)/2]) + d*Sin[(e + 
 f*x)/2])*Sqrt[c + d*Sin[e + f*x]])/(f*Sqrt[a*(1 + Sin[e + f*x])]) + ((Sqr 
t[2]*(c - d)^(3/2)*Log[1 + Tan[(e + f*x)/2]] - Sqrt[2]*(c - d)^(3/2)*Log[c 
 - d + 2*Sqrt[c - d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x] 
] + (-c + d)*Tan[(e + f*x)/2]] + (I/2)*Sqrt[d]*(-3*c + d)*(Log[((2*I)*(I*c 
 + d + (1 + I)*Sqrt[2]*Sqrt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Si 
n[e + f*x]] + (c + I*d)*Tan[(e + f*x)/2]))/(d^(3/2)*(-3*c + d)*(I + Tan[(e 
 + f*x)/2]))] - Log[(-2*(c + I*d + (1 + I)*Sqrt[2]*Sqrt[d]*Sqrt[(1 + Cos[e 
 + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (I*c + d)*Tan[(e + f*x)/2]))/(d^ 
(3/2)*(-3*c + d)*(-I + Tan[(e + f*x)/2]))]))*(Cos[(e + f*x)/2] + Sin[(e + 
f*x)/2])*(c^2/((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f* 
x]]) - (c*d)/(2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f 
*x]]) + d^2/(2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f* 
x]]) + (3*c*d*Sin[e + f*x])/(2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[ 
c + d*Sin[e + f*x]]) - (d^2*Sin[e + f*x])/(2*(Cos[(e + f*x)/2] + Sin[(e + 
f*x)/2])*Sqrt[c + d*Sin[e + f*x]])))/(f*Sqrt[a*(1 + Sin[e + f*x])]*(((c - 
d)^(3/2)*Sec[(e + f*x)/2]^2)/(Sqrt[2]*(1 + Tan[(e + f*x)/2])) - (Sqrt[2]*( 
c - d)^(3/2)*(((-c + d)*Sec[(e + f*x)/2]^2)/2 + (Sqrt[c - d]*d*Cos[e + f*x 
]*Sqrt[(1 + Cos[e + f*x])^(-1)])/Sqrt[c + d*Sin[e + f*x]] + Sqrt[c - d]*(( 
1 + Cos[e + f*x])^(-1))^(3/2)*Sin[e + f*x]*Sqrt[c + d*Sin[e + f*x]]))/(...
 

Rubi [A] (verified)

Time = 1.02 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 3257, 25, 3042, 3461, 3042, 3254, 218, 3261, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3257

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3461

\(\displaystyle \frac {2 a (c-d)^2 \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx+d (3 c-d) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx}{2 a}-\frac {d \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a (c-d)^2 \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx+d (3 c-d) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx}{2 a}-\frac {d \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3254

\(\displaystyle \frac {2 a (c-d)^2 \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx-\frac {2 a d (3 c-d) \int \frac {1}{\frac {a^2 d \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}+a}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f}}{2 a}-\frac {d \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 3261

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

((-2*Sqrt[a]*(3*c - d)*Sqrt[d]*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt 
[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/f - (2*Sqrt[2]*Sqrt[a]*(c 
 - d)^(3/2)*ArcTanh[(Sqrt[a]*Sqrt[c - d]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a 
*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/f)/(2*a) - (d*Cos[e + f*x]*Sqrt 
[c + d*Sin[e + f*x]])/(f*Sqrt[a + a*Sin[e + f*x]])
 

Defintions of rubi rules used

rule 25

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

rule 218

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

rule 221

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

rule 3042

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

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 3257

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_. 
) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*d*Cos[e + f*x]*((c + d*Sin[e + f*x]) 
^(n - 1)/(f*(2*n - 1)*Sqrt[a + b*Sin[e + f*x]])), x] - Simp[1/(b*(2*n - 1)) 
   Int[((c + d*Sin[e + f*x])^(n - 2)/Sqrt[a + b*Sin[e + f*x]])*Simp[a*c*d - 
 b*(2*d^2*(n - 1) + c^2*(2*n - 1)) + d*(a*d - b*c*(4*n - 3))*Sin[e + f*x], 
x], 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] && GtQ[n, 1] && IntegerQ[2*n]
 

rule 3261

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e 
_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f)   Subst[Int[1/(2*b^2 - (a*c 
 - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S 
in[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 3461

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + 
(f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Sim 
p[(A*b - a*B)/b   Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]) 
, x], x] + Simp[B/b   Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]] 
, x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[ 
a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

Maple [F(-1)]

Timed out.

hanged

Input:

int((c+d*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2),x)
 

Output:

int((c+d*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 428 vs. \(2 (157) = 314\).

Time = 0.48 (sec) , antiderivative size = 2621, normalized size of antiderivative = 13.94 \[ \int \frac {(c+d \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx=\text {Too large to display} \] Input:

integrate((c+d*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="fric 
as")
 

Output:

[-1/8*(4*sqrt(2)*(a*c - a*d + (a*c - a*d)*cos(f*x + e) + (a*c - a*d)*sin(f 
*x + e))*sqrt((c - d)/a)*log(-(2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sqrt(d*s 
in(f*x + e) + c)*sqrt((c - d)/a)*(cos(f*x + e) - sin(f*x + e) + 1) + (c - 
3*d)*cos(f*x + e)^2 + (3*c - d)*cos(f*x + e) - ((c - 3*d)*cos(f*x + e) - 2 
*c - 2*d)*sin(f*x + e) + 2*c + 2*d)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*s 
in(f*x + e) - cos(f*x + e) - 2)) + (3*a*c - a*d + (3*a*c - a*d)*cos(f*x + 
e) + (3*a*c - a*d)*sin(f*x + e))*sqrt(-d/a)*log((128*d^4*cos(f*x + e)^5 + 
128*(2*c*d^3 - d^4)*cos(f*x + e)^4 + c^4 + 4*c^3*d + 6*c^2*d^2 + 4*c*d^3 + 
 d^4 - 32*(5*c^2*d^2 - 14*c*d^3 + 13*d^4)*cos(f*x + e)^3 - 32*(c^3*d - 2*c 
^2*d^2 + 9*c*d^3 - 4*d^4)*cos(f*x + e)^2 + 8*(16*d^3*cos(f*x + e)^4 + 24*( 
c*d^2 - d^3)*cos(f*x + e)^3 - c^3 + 17*c^2*d - 59*c*d^2 + 51*d^3 - 2*(5*c^ 
2*d - 26*c*d^2 + 33*d^3)*cos(f*x + e)^2 - (c^3 - 7*c^2*d + 31*c*d^2 - 25*d 
^3)*cos(f*x + e) + (16*d^3*cos(f*x + e)^3 + c^3 - 17*c^2*d + 59*c*d^2 - 51 
*d^3 - 8*(3*c*d^2 - 5*d^3)*cos(f*x + e)^2 - 2*(5*c^2*d - 14*c*d^2 + 13*d^3 
)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) 
 + c)*sqrt(-d/a) + (c^4 - 28*c^3*d + 230*c^2*d^2 - 476*c*d^3 + 289*d^4)*co 
s(f*x + e) + (128*d^4*cos(f*x + e)^4 + c^4 + 4*c^3*d + 6*c^2*d^2 + 4*c*d^3 
 + d^4 - 256*(c*d^3 - d^4)*cos(f*x + e)^3 - 32*(5*c^2*d^2 - 6*c*d^3 + 5*d^ 
4)*cos(f*x + e)^2 + 32*(c^3*d - 7*c^2*d^2 + 15*c*d^3 - 9*d^4)*cos(f*x + e) 
)*sin(f*x + e))/(cos(f*x + e) + sin(f*x + e) + 1)) + 8*(d*cos(f*x + e) ...
 

Sympy [F]

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

integrate((c+d*sin(f*x+e))**(3/2)/(a+a*sin(f*x+e))**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate((c+d*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="maxi 
ma")
 

Output:

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

Giac [F(-1)]

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

integrate((c+d*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="giac 
")
 

Output:

Timed out
 

Mupad [F(-1)]

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

int((c + d*sin(e + f*x))^(3/2)/(a + a*sin(e + f*x))^(1/2),x)
 

Output:

int((c + d*sin(e + f*x))^(3/2)/(a + a*sin(e + f*x))^(1/2), x)
 

Reduce [F]

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

int((c+d*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2),x)
 

Output:

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