\(\int \frac {(a+a \sin (e+f x))^{3/2}}{\sqrt {c+d \sin (e+f x)}} \, dx\) [582]

Optimal result

Integrand size = 29, antiderivative size = 111 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{\sqrt {c+d \sin (e+f x)}} \, dx=\frac {a^{3/2} (c-3 d) \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{d^{3/2} f}-\frac {a^2 \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{d f \sqrt {a+a \sin (e+f x)}} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(301\) vs. \(2(111)=222\).

Time = 1.79 (sec) , antiderivative size = 301, normalized size of antiderivative = 2.71 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{\sqrt {c+d \sin (e+f x)}} \, dx=\frac {(a (1+\sin (e+f x)))^{3/2} \left (-2 (c-3 d) \arctan \left (\frac {\sqrt {2} \sqrt {d} \sin \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{\sqrt {c+d \sin (e+f x)}}\right )-(c-3 d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{\sqrt {c+d \sin (e+f x)}}\right )+c \log \left (\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right )+\sqrt {c+d \sin (e+f x)}\right )-3 d \log \left (\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right )+\sqrt {c+d \sin (e+f x)}\right )-2 \sqrt {d} \cos \left (\frac {1}{2} (e+f x)\right ) \sqrt {c+d \sin (e+f x)}+2 \sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right ) \sqrt {c+d \sin (e+f x)}\right )}{2 d^{3/2} f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \] Input:

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

Output:

((a*(1 + Sin[e + f*x]))^(3/2)*(-2*(c - 3*d)*ArcTan[(Sqrt[2]*Sqrt[d]*Sin[(2 
*e - Pi + 2*f*x)/4])/Sqrt[c + d*Sin[e + f*x]]] - (c - 3*d)*ArcTanh[(Sqrt[2 
]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4])/Sqrt[c + d*Sin[e + f*x]]] + c*Log[Sqr 
t[2]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4] + Sqrt[c + d*Sin[e + f*x]]] - 3*d*L 
og[Sqrt[2]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4] + Sqrt[c + d*Sin[e + f*x]]] - 
 2*Sqrt[d]*Cos[(e + f*x)/2]*Sqrt[c + d*Sin[e + f*x]] + 2*Sqrt[d]*Sin[(e + 
f*x)/2]*Sqrt[c + d*Sin[e + f*x]]))/(2*d^(3/2)*f*(Cos[(e + f*x)/2] + Sin[(e 
 + f*x)/2])^3)
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {3042, 3242, 27, 2011, 3042, 3254, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3242

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3254

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + b*x])
 

rule 3042

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

rule 3242

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x 
])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m 
+ n))   Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c* 
(m - 2) + b^2*d*(n + 1) + a^2*d*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 
 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c 
 - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] &&  !LtQ[ 
n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[ 
c, 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]
 

Maple [F(-1)]

Timed out.

hanged

Input:

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

Output:

int((a+a*sin(f*x+e))^(3/2)/(c+d*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. 265 vs. \(2 (95) = 190\).

Time = 0.39 (sec) , antiderivative size = 989, normalized size of antiderivative = 8.91 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{\sqrt {c+d \sin (e+f x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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