\(\int \frac {(a+a \sin (e+f x))^{3/2}}{c+d \sin (e+f x)} \, dx\) [541]

Optimal result

Integrand size = 27, antiderivative size = 98 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{c+d \sin (e+f x)} \, dx=\frac {2 a^{3/2} (c-d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{d^{3/2} \sqrt {c+d} f}-\frac {2 a^2 \cos (e+f x)}{d f \sqrt {a+a \sin (e+f x)}} \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 7.83 (sec) , antiderivative size = 741, normalized size of antiderivative = 7.56 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{c+d \sin (e+f x)} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/2*(((c - d)*RootSum[c + 4*d*#1 + 2*c*#1^2 - 4*d*#1^3 + c*#1^4 & , (-(c* 
Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]) - d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]] 
 - d*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]] - 2*c*Sqrt[d]*Log[-#1 + Tan[( 
e + f*x)/4]]*#1 - 2*d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]]*#1 - c*Sqrt[c + d] 
*Log[-#1 + Tan[(e + f*x)/4]]*#1 + c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 
^2 + d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 + 3*d*Sqrt[c + d]*Log[-#1 + 
Tan[(e + f*x)/4]]*#1^2 - c*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^3)/( 
-d - c*#1 + 3*d*#1^2 - c*#1^3) & ] + (c - d)*RootSum[c + 4*d*#1 + 2*c*#1^2 
 - 4*d*#1^3 + c*#1^4 & , (-(c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]) - d^(3/ 
2)*Log[-#1 + Tan[(e + f*x)/4]] + d*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]] 
 - 2*c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 - 2*d^(3/2)*Log[-#1 + Tan[(e 
 + f*x)/4]]*#1 + c*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 + c*Sqrt[d]* 
Log[-#1 + Tan[(e + f*x)/4]]*#1^2 + d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]]*#1^ 
2 - 3*d*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 + c*Sqrt[c + d]*Log[- 
#1 + Tan[(e + f*x)/4]]*#1^3)/(-d - c*#1 + 3*d*#1^2 - c*#1^3) & ] + 4*Sqrt[ 
d]*(c + d)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))*(a*(1 + Sin[e + f*x]))^( 
3/2))/(d^(3/2)*(c + d)*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3042, 3242, 27, 2011, 3042, 3252, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3242

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3252

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(2*a^(3/2)*(c - d)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqr 
t[a + a*Sin[e + f*x]])])/(d^(3/2)*Sqrt[c + d]*f) - (2*a^2*Cos[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 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 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 3252

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

Time = 0.62 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.40

method	result	size
default	\(-\frac {2 a \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \left (-\operatorname {arctanh}\left (\frac {\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) a c +a \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) d +\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \sqrt {a \left (c +d \right ) d}\right )}{d \sqrt {a \left (c +d \right ) d}\, \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\)	\(137\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 167 vs. \(2 (82) = 164\).

Time = 0.16 (sec) , antiderivative size = 651, normalized size of antiderivative = 6.64 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{c+d \sin (e+f x)} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/2*((a*c - a*d + (a*c - a*d)*cos(f*x + e) + (a*c - a*d)*sin(f*x + e))*s 
qrt(a/(c*d + d^2))*log((a*d^2*cos(f*x + e)^3 - a*c^2 - 2*a*c*d - a*d^2 - ( 
6*a*c*d + 7*a*d^2)*cos(f*x + e)^2 + 4*(c^2*d + 4*c*d^2 + 3*d^3 - (c*d^2 + 
d^3)*cos(f*x + e)^2 + (c^2*d + 3*c*d^2 + 2*d^3)*cos(f*x + e) - (c^2*d + 4* 
c*d^2 + 3*d^3 + (c*d^2 + d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + 
 e) + a)*sqrt(a/(c*d + d^2)) - (a*c^2 + 8*a*c*d + 9*a*d^2)*cos(f*x + e) + 
(a*d^2*cos(f*x + e)^2 - a*c^2 - 2*a*c*d - a*d^2 + 2*(3*a*c*d + 4*a*d^2)*co 
s(f*x + e))*sin(f*x + e))/(d^2*cos(f*x + e)^3 + (2*c*d + d^2)*cos(f*x + e) 
^2 - c^2 - 2*c*d - d^2 - (c^2 + d^2)*cos(f*x + e) + (d^2*cos(f*x + e)^2 - 
2*c*d*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))) + 4*(a*cos(f*x + e) 
 - a*sin(f*x + e) + a)*sqrt(a*sin(f*x + e) + a))/(d*f*cos(f*x + e) + d*f*s 
in(f*x + e) + d*f), ((a*c - a*d + (a*c - a*d)*cos(f*x + e) + (a*c - a*d)*s 
in(f*x + e))*sqrt(-a/(c*d + d^2))*arctan(1/2*sqrt(a*sin(f*x + e) + a)*(d*s 
in(f*x + e) - c - 2*d)*sqrt(-a/(c*d + d^2))/(a*cos(f*x + e))) - 2*(a*cos(f 
*x + e) - a*sin(f*x + e) + a)*sqrt(a*sin(f*x + e) + a))/(d*f*cos(f*x + e) 
+ d*f*sin(f*x + e) + d*f)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.28 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{c+d \sin (e+f x)} \, dx=\frac {\sqrt {2} {\left (\frac {2 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{d} + \frac {\sqrt {2} {\left (a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \arctan \left (\frac {\sqrt {2} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c d - d^{2}}}\right )}{\sqrt {-c d - d^{2}} d}\right )} \sqrt {a}}{f} \] Input:

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

Output:

sqrt(2)*(2*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1 
/2*e)/d + sqrt(2)*(a*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - a*d*sgn(cos(- 
1/4*pi + 1/2*f*x + 1/2*e)))*arctan(sqrt(2)*d*sin(-1/4*pi + 1/2*f*x + 1/2*e 
)/sqrt(-c*d - d^2))/(sqrt(-c*d - d^2)*d))*sqrt(a)/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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