\(\int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{\sqrt {a+a \sin (e+f x)}} \, dx\) [309]

Optimal result

Integrand size = 35, antiderivative size = 130 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{\sqrt {a+a \sin (e+f x)}} \, dx=-\frac {\sqrt {2} (A-B) (c-d) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} f}-\frac {2 (3 B c+3 A d-2 B d) \cos (e+f x)}{3 f \sqrt {a+a \sin (e+f x)}}-\frac {2 B d \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{3 a f} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.94 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.04 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{\sqrt {a+a \sin (e+f x)}} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left ((-6-6 i) (-1)^{3/4} (A-B) (c-d) \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right )+2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (3 B c+3 A d-B d+B d \sin (e+f x))\right )}{3 f \sqrt {a (1+\sin (e+f x))}} \] Input:

Integrate[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x]))/Sqrt[a + a*Sin[e + f 
*x]],x]
 

Output:

-1/3*((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*((-6 - 6*I)*(-1)^(3/4)*(A - B) 
*(c - d)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])] + 2*(Cos[ 
(e + f*x)/2] - Sin[(e + f*x)/2])*(3*B*c + 3*A*d - B*d + B*d*Sin[e + f*x])) 
)/(f*Sqrt[a*(1 + Sin[e + f*x])])
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3447, 3042, 3502, 27, 3042, 3230, 3042, 3128, 219}

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.

\(\displaystyle \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{\sqrt {a \sin (e+f x)+a}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{\sqrt {a \sin (e+f x)+a}}dx\)

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3230

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

\(\displaystyle \frac {-\frac {6 a (A-B) (c-d) \int \frac {1}{2 a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f}-\frac {2 a (3 A d+3 B c-2 B d) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}}{3 a}-\frac {2 B d \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 a f}\)

\(\Big \downarrow \) 219

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

Input:

Int[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x]))/Sqrt[a + a*Sin[e + f*x]],x 
]
 

Output:

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

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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 3128

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d 
Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
 

rule 3230

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

rule 3447

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

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(231\) vs. \(2(113)=226\).

Time = 0.73 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.78

method	result	size
default	\(-\frac {\left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (3 A \sqrt {2}\, a^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) c -3 A \sqrt {2}\, a^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) d -3 B \sqrt {2}\, a^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) c +3 B \sqrt {2}\, a^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) d -2 B \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} d +6 A a d \sqrt {a -a \sin \left (f x +e \right )}+6 B a c \sqrt {a -a \sin \left (f x +e \right )}\right )}{3 a^{2} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(232\)
parts	\(-\frac {A c \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{\sqrt {a}\, \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {\left (A d +B c \right ) \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (\sqrt {a}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-2 \sqrt {a -a \sin \left (f x +e \right )}\right )}{a \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}-\frac {B d \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (3 a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-2 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}}\right )}{3 a^{2} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(275\)

Input:

int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2),x,method=_RET 
URNVERBOSE)
 

Output:

-1/3*(1+sin(f*x+e))*(-a*(sin(f*x+e)-1))^(1/2)*(3*A*2^(1/2)*a^(3/2)*arctanh 
(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*c-3*A*2^(1/2)*a^(3/2)*arctanh 
(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*d-3*B*2^(1/2)*a^(3/2)*arctanh 
(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*c+3*B*2^(1/2)*a^(3/2)*arctanh 
(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*d-2*B*(a-a*sin(f*x+e))^(3/2)* 
d+6*A*a*d*(a-a*sin(f*x+e))^(1/2)+6*B*a*c*(a-a*sin(f*x+e))^(1/2))/a^2/cos(f 
*x+e)/(a+a*sin(f*x+e))^(1/2)/f
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 303 vs. \(2 (113) = 226\).

Time = 0.10 (sec) , antiderivative size = 303, normalized size of antiderivative = 2.33 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{\sqrt {a+a \sin (e+f x)}} \, dx=\frac {\frac {3 \, \sqrt {2} {\left ({\left (A - B\right )} a c - {\left (A - B\right )} a d + {\left ({\left (A - B\right )} a c - {\left (A - B\right )} a d\right )} \cos \left (f x + e\right ) + {\left ({\left (A - B\right )} a c - {\left (A - B\right )} a d\right )} \sin \left (f x + e\right )\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) - \frac {2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt {a}} - 4 \, {\left (B d \cos \left (f x + e\right )^{2} + 3 \, B c + {\left (3 \, A - 2 \, B\right )} d + {\left (3 \, B c + {\left (3 \, A - B\right )} d\right )} \cos \left (f x + e\right ) + {\left (B d \cos \left (f x + e\right ) - 3 \, B c - {\left (3 \, A - 2 \, B\right )} d\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{6 \, {\left (a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f\right )}} \] Input:

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2),x, algo 
rithm="fricas")
 

Output:

1/6*(3*sqrt(2)*((A - B)*a*c - (A - B)*a*d + ((A - B)*a*c - (A - B)*a*d)*co 
s(f*x + e) + ((A - B)*a*c - (A - B)*a*d)*sin(f*x + e))*log(-(cos(f*x + e)^ 
2 - (cos(f*x + e) - 2)*sin(f*x + e) - 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*( 
cos(f*x + e) - sin(f*x + e) + 1)/sqrt(a) + 3*cos(f*x + e) + 2)/(cos(f*x + 
e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2))/sqrt(a) - 4*(B 
*d*cos(f*x + e)^2 + 3*B*c + (3*A - 2*B)*d + (3*B*c + (3*A - B)*d)*cos(f*x 
+ e) + (B*d*cos(f*x + e) - 3*B*c - (3*A - 2*B)*d)*sin(f*x + e))*sqrt(a*sin 
(f*x + e) + a))/(a*f*cos(f*x + e) + a*f*sin(f*x + e) + a*f)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2),x, algo 
rithm="maxima")
 

Output:

integrate((B*sin(f*x + e) + A)*(d*sin(f*x + e) + c)/sqrt(a*sin(f*x + e) + 
a), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{\sqrt {a+a \sin (e+f x)}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2),x, algo 
rithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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