\(\int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^2} \, dx\) [542]

Optimal result

Integrand size = 27, antiderivative size = 119 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^2} \, dx=-\frac {a^{3/2} (c+3 d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{d^{3/2} (c+d)^{3/2} f}+\frac {a^2 (c-d) \cos (e+f x)}{d (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \] Output:

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

Mathematica [C] (warning: unable to verify)

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

Time = 8.20 (sec) , antiderivative size = 759, normalized size of antiderivative = 6.38 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

((a*(1 + Sin[e + f*x]))^(3/2)*((c + 3*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)*L 
og[-#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 + 3*d)*R 
ootSum[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^2 + d^2)*(Cos[(e + f*x)/2] - Sin[(e + 
f*x)/2]))/(c + d*Sin[e + f*x])))/(4*d^(3/2)*(c + d)^2*f*(Cos[(e + f*x)/2] 
+ Sin[(e + f*x)/2])^3)
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 119, 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, 3241, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3241

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3252

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

-((a^(3/2)*(c + 3*d)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*S 
qrt[a + a*Sin[e + f*x]])])/(d^(3/2)*(c + d)^(3/2)*f)) + (a^2*(c - d)*Cos[e 
 + f*x])/(d*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*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 3241

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

Leaf count of result is larger than twice the leaf count of optimal. \(232\) vs. \(2(103)=206\).

Time = 0.64 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.96

method	result	size
default	\(\frac {a \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \left (-\sin \left (f x +e \right ) \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a d \left (c +3 d \right )-\operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a \,c^{2}-3 \,\operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, d}{\sqrt {a c d +a \,d^{2}}}\right ) a c d +\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {a \left (c +d \right ) d}\, c -\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {a \left (c +d \right ) d}\, d \right )}{d \left (c +d \right ) \left (c +d \sin \left (f x +e \right )\right ) \sqrt {a \left (c +d \right ) d}\, \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\)	\(233\)

Input:

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

Output:

a*(1+sin(f*x+e))*(-a*(-1+sin(f*x+e)))^(1/2)*(-sin(f*x+e)*arctanh((a-sin(f* 
x+e)*a)^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a*d*(c+3*d)-arctanh((a-sin(f*x+e)*a)^ 
(1/2)*d/(a*c*d+a*d^2)^(1/2))*a*c^2-3*arctanh((a-sin(f*x+e)*a)^(1/2)*d/(a*c 
*d+a*d^2)^(1/2))*a*c*d+(a-sin(f*x+e)*a)^(1/2)*(a*(c+d)*d)^(1/2)*c-(a-sin(f 
*x+e)*a)^(1/2)*(a*(c+d)*d)^(1/2)*d)/d/(c+d)/(c+d*sin(f*x+e))/(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. 327 vs. \(2 (103) = 206\).

Time = 0.18 (sec) , antiderivative size = 970, normalized size of antiderivative = 8.15 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/4*((a*c^2 + 4*a*c*d + 3*a*d^2 - (a*c*d + 3*a*d^2)*cos(f*x + e)^2 + (a* 
c^2 + 3*a*c*d)*cos(f*x + e) + (a*c^2 + 4*a*c*d + 3*a*d^2 + (a*c*d + 3*a*d^ 
2)*cos(f*x + e))*sin(f*x + e))*sqrt(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)*cos(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)*si 
n(f*x + e))) + 4*(a*c - a*d + (a*c - a*d)*cos(f*x + e) - (a*c - a*d)*sin(f 
*x + e))*sqrt(a*sin(f*x + e) + a))/((c*d^2 + d^3)*f*cos(f*x + e)^2 - (c^2* 
d + c*d^2)*f*cos(f*x + e) - (c^2*d + 2*c*d^2 + d^3)*f - ((c*d^2 + d^3)*f*c 
os(f*x + e) + (c^2*d + 2*c*d^2 + d^3)*f)*sin(f*x + e)), 1/2*((a*c^2 + 4*a* 
c*d + 3*a*d^2 - (a*c*d + 3*a*d^2)*cos(f*x + e)^2 + (a*c^2 + 3*a*c*d)*cos(f 
*x + e) + (a*c^2 + 4*a*c*d + 3*a*d^2 + (a*c*d + 3*a*d^2)*cos(f*x + e))*sin 
(f*x + e))*sqrt(-a/(c*d + d^2))*arctan(1/2*sqrt(a*sin(f*x + e) + a)*(d*sin 
(f*x + e) - c - 2*d)*sqrt(-a/(c*d + d^2))/(a*cos(f*x + e))) - 2*(a*c - a*d 
 + (a*c - a*d)*cos(f*x + e) - (a*c - a*d)*sin(f*x + e))*sqrt(a*sin(f*x ...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.65 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^2} \, dx=-\frac {\sqrt {2} \sqrt {a} {\left (\frac {\sqrt {2} {\left (a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \arctan \left (\frac {\sqrt {2} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c d - d^{2}}}\right )}{{\left (c d + d^{2}\right )} \sqrt {-c d - d^{2}}} - \frac {2 \, {\left (a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (2 \, d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c - d\right )} {\left (c d + d^{2}\right )}}\right )}}{2 \, f} \] Input:

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

Output:

-1/2*sqrt(2)*sqrt(a)*(sqrt(2)*(a*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 3 
*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))/((c*d + d^2)*sqrt(-c*d - d^2)) - 2*(a*c* 
sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e) - a*d*s 
gn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e))/((2*d*s 
in(-1/4*pi + 1/2*f*x + 1/2*e)^2 - c - d)*(c*d + d^2)))/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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