\(\int \frac {(3+3 \sin (e+f x))^2}{\sqrt {c-c \sin (e+f x)}} \, dx\) [302]

Optimal result

Integrand size = 28, antiderivative size = 104 \[ \int \frac {(3+3 \sin (e+f x))^2}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {36 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f}-\frac {6 c \cos ^3(e+f x)}{f (c-c \sin (e+f x))^{3/2}}-\frac {36 \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}} \]

[Out]

-2/3*a^2*c*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(3/2)+4*a^2*arctanh(1/2*cos(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))
^(1/2))*2^(1/2)/f/c^(1/2)-4*a^2*cos(f*x+e)/f/(c-c*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.11, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2815, 2758, 2728, 212} \[ \int \frac {(3+3 \sin (e+f x))^2}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {4 \sqrt {2} a^2 \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac {4 a^2 \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

(4*Sqrt[2]*a^2*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(Sqrt[c]*f) - (2*a^2*c*Cos[
e + f*x]^3)/(3*f*(c - c*Sin[e + f*x])^(3/2)) - (4*a^2*Cos[e + f*x])/(f*Sqrt[c - c*Sin[e + f*x]])

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2728

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

Rule 2758

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[g*(g*C
os[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + p))), x] + Dist[g^2*((p - 1)/(a*(m + p))), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p,
0] && IntegersQ[2*m, 2*p]

Rule 2815

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

Rubi steps \begin{align*} \text {integral}& = \left (a^2 c^2\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx \\ & = -\frac {2 a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}+\left (2 a^2 c\right ) \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx \\ & = -\frac {2 a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac {4 a^2 \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}}+\left (4 a^2\right ) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx \\ & = -\frac {2 a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac {4 a^2 \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}}-\frac {\left (8 a^2\right ) \text {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{f} \\ & = \frac {4 \sqrt {2} a^2 \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac {4 a^2 \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.98 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.20 \[ \int \frac {(3+3 \sin (e+f x))^2}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left ((24+24 i) \sqrt [4]{-1} \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right )+15 \cos \left (\frac {1}{2} (e+f x)\right )-\cos \left (\frac {3}{2} (e+f x)\right )+15 \sin \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {3}{2} (e+f x)\right )\right )}{f \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

(-3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*((24 + 24*I)*(-1)^(1/4)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e +
f*x)/4])] + 15*Cos[(e + f*x)/2] - Cos[(3*(e + f*x))/2] + 15*Sin[(e + f*x)/2] + Sin[(3*(e + f*x))/2]))/(f*Sqrt[
c - c*Sin[e + f*x]])

Maple [A] (verified)

Time = 2.32 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.08

method	result	size
default	\(-\frac {2 \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, a^{2} \left (6 c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )-\left (c \left (\sin \left (f x +e \right )+1\right )\right )^{\frac {3}{2}}-6 \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, c \right )}{3 c^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(112\)
parts	\(-\frac {a^{2} \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{\sqrt {c}\, \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {a^{2} \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \left (3 c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )-2 \left (c \left (\sin \left (f x +e \right )+1\right )\right )^{\frac {3}{2}}\right )}{3 c^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {2 a^{2} \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \left (-\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )+2 \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\right )}{c \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(270\)

[In]

int((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/3*(sin(f*x+e)-1)*(c*(sin(f*x+e)+1))^(1/2)*a^2*(6*c^(3/2)*2^(1/2)*arctanh(1/2*(c*(sin(f*x+e)+1))^(1/2)*2^(1/
2)/c^(1/2))-(c*(sin(f*x+e)+1))^(3/2)-6*(c*(sin(f*x+e)+1))^(1/2)*c)/c^2/cos(f*x+e)/(c-c*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. 238 vs. \(2 (100) = 200\).

Time = 0.28 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.29 \[ \int \frac {(3+3 \sin (e+f x))^2}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {2 \, {\left (\frac {3 \, \sqrt {2} {\left (a^{2} c \cos \left (f x + e\right ) - a^{2} c \sin \left (f x + e\right ) + a^{2} c\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) + \frac {2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt {c}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt {c}} + {\left (a^{2} \cos \left (f x + e\right )^{2} - 7 \, a^{2} \cos \left (f x + e\right ) - 8 \, a^{2} - {\left (a^{2} \cos \left (f x + e\right ) + 8 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}\right )}}{3 \, {\left (c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f\right )}} \]

[In]

integrate((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

2/3*(3*sqrt(2)*(a^2*c*cos(f*x + e) - a^2*c*sin(f*x + e) + a^2*c)*log(-(cos(f*x + e)^2 + (cos(f*x + e) - 2)*sin
(f*x + e) + 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*(cos(f*x + e) + sin(f*x + e) + 1)/sqrt(c) + 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(c) + (a^2*cos(f*x + e)^2 - 7*a^2
*cos(f*x + e) - 8*a^2 - (a^2*cos(f*x + e) + 8*a^2)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c*f*cos(f*x + e)
- c*f*sin(f*x + e) + c*f)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 210 vs. \(2 (100) = 200\).

Time = 0.34 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.02 \[ \int \frac {(3+3 \sin (e+f x))^2}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {2 \, {\left (\frac {3 \, \sqrt {2} a^{2} \log \left (-\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1}\right )}{\sqrt {c} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {8 \, \sqrt {2} {\left (2 \, a^{2} \sqrt {c} - \frac {3 \, a^{2} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {3 \, a^{2} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}\right )}}{c {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - 1\right )}^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )}}{3 \, f} \]

[In]

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

[Out]

2/3*(3*sqrt(2)*a^2*log(-(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1))/(sqrt(c)*sg
n(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - 8*sqrt(2)*(2*a^2*sqrt(c) - 3*a^2*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e)
- 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 3*a^2*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2/(cos(-1/4*pi
+ 1/2*f*x + 1/2*e) + 1)^2)/(c*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - 1)^
3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))))/f

Mupad [F(-1)]

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

[In]

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

[Out]

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