\(\int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx\) [592]

Optimal result

Integrand size = 29, antiderivative size = 125 \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx=-\frac {\sqrt {\frac {2}{3}} \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} \sqrt {c-d} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{(c-d)^{3/2} f}+\frac {2 d \cos (e+f x)}{\left (c^2-d^2\right ) f \sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2858, 12, 2861, 214} \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx=\frac {2 d \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {a} f (c-d)^{3/2}} \]

[In]

Int[1/(Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(3/2)),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 214

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 2858

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp
[(-d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] - Dist[
1/(2*b*(n + 1)*(c^2 - d^2)), Int[(c + d*Sin[e + f*x])^(n + 1)*(Simp[a*d - 2*b*c*(n + 1) + b*d*(2*n + 3)*Sin[e
+ f*x], x]/Sqrt[a + b*Sin[e + f*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] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2861

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[-2*(a/f), Subst[Int[1/(2*b^2 - (a*c - 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]

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(309\) vs. \(2(125)=250\).

Time = 3.26 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.47 \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx=\frac {\frac {2 d \cos (e+f x)}{c+d}+\frac {\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (c-d+2 \sqrt {c-d} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )\right )}{\frac {\sec ^2\left (\frac {1}{2} (e+f x)\right )}{2+2 \tan \left (\frac {1}{2} (e+f x)\right )}-\frac {-\frac {1}{2} (c-d) \sec ^2\left (\frac {1}{2} (e+f x)\right )+\frac {\sqrt {c-d} \left (\frac {1}{1+\cos (e+f x)}\right )^{3/2} (d+d \cos (e+f x)+c \sin (e+f x))}{\sqrt {c+d \sin (e+f x)}}}{c-d+2 \sqrt {c-d} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )}}}{\sqrt {3} (c-d) f \sqrt {1+\sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \]

[In]

Integrate[1/(Sqrt[3 + 3*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(3/2)),x]

[Out]

((2*d*Cos[e + f*x])/(c + d) + (Log[1 + Tan[(e + f*x)/2]] - Log[c - d + 2*Sqrt[c - d]*Sqrt[(1 + Cos[e + f*x])^(
-1)]*Sqrt[c + d*Sin[e + f*x]] + (-c + d)*Tan[(e + f*x)/2]])/(Sec[(e + f*x)/2]^2/(2 + 2*Tan[(e + f*x)/2]) - (-1
/2*((c - d)*Sec[(e + f*x)/2]^2) + (Sqrt[c - d]*((1 + Cos[e + f*x])^(-1))^(3/2)*(d + d*Cos[e + f*x] + c*Sin[e +
f*x]))/Sqrt[c + d*Sin[e + f*x]])/(c - d + 2*Sqrt[c - d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]
] + (-c + d)*Tan[(e + f*x)/2])))/(Sqrt[3]*(c - d)*f*Sqrt[1 + Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(443\) vs. \(2(112)=224\).

Time = 5.36 (sec) , antiderivative size = 444, normalized size of antiderivative = 3.55

method	result	size
default	\(-\frac {2 \left (-\cos \left (f x +e \right )+1+\sin \left (f x +e \right )\right ) \sqrt {c +d \sin \left (f x +e \right )}\, \left (\ln \left (\frac {2 \sqrt {2 c -2 d}\, \sqrt {2}\, \sqrt {\frac {c +d \sin \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right )+2 c \sin \left (f x +e \right )-2 d \sin \left (f x +e \right )+2 c \cos \left (f x +e \right )-2 d \cos \left (f x +e \right )-2 c +2 d}{-\cos \left (f x +e \right )+1+\sin \left (f x +e \right )}\right ) \sqrt {2}\, \sqrt {\frac {c +d \sin \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, c \sin \left (f x +e \right )+\ln \left (\frac {2 \sqrt {2 c -2 d}\, \sqrt {2}\, \sqrt {\frac {c +d \sin \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right )+2 c \sin \left (f x +e \right )-2 d \sin \left (f x +e \right )+2 c \cos \left (f x +e \right )-2 d \cos \left (f x +e \right )-2 c +2 d}{-\cos \left (f x +e \right )+1+\sin \left (f x +e \right )}\right ) \sqrt {2}\, \sqrt {\frac {c +d \sin \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, d \sin \left (f x +e \right )-d \sqrt {2 c -2 d}\, \sin \left (f x +e \right )-d \sqrt {2 c -2 d}\, \cos \left (f x +e \right )+d \sqrt {2 c -2 d}\right )}{f \sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, \left (c \left (\cos ^{2}\left (f x +e \right )\right )-2 d \sin \left (f x +e \right ) \cos \left (f x +e \right )+c \left (\sin ^{2}\left (f x +e \right )\right )-2 c \cos \left (f x +e \right )+2 d \sin \left (f x +e \right )+c \right ) \left (c +d \right ) \sqrt {2 c -2 d}\, \left (c -d \right )}\)	\(444\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 378 vs. \(2 (112) = 224\).

Time = 0.49 (sec) , antiderivative size = 1016, normalized size of antiderivative = 8.13 \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[-1/4*(8*(d*cos(f*x + e) - d*sin(f*x + e) + d)*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c) - sqrt(2)*(a*
c^2 + 2*a*c*d + a*d^2 - (a*c*d + a*d^2)*cos(f*x + e)^2 + (a*c^2 + a*c*d)*cos(f*x + e) + (a*c^2 + 2*a*c*d + a*d
^2 + (a*c*d + a*d^2)*cos(f*x + e))*sin(f*x + e))*log(((c^2 - 14*c*d + 17*d^2)*cos(f*x + e)^3 - (13*c^2 - 22*c*
d - 3*d^2)*cos(f*x + e)^2 + 4*sqrt(2)*((c^2 - 4*c*d + 3*d^2)*cos(f*x + e)^2 - 4*c^2 + 8*c*d - 4*d^2 - (3*c^2 -
4*c*d + d^2)*cos(f*x + e) + (4*c^2 - 8*c*d + 4*d^2 + (c^2 - 4*c*d + 3*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a
*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)/sqrt(a*c - a*d) - 4*c^2 - 8*c*d - 4*d^2 - 2*(9*c^2 - 14*c*d + 9*d^
2)*cos(f*x + e) + ((c^2 - 14*c*d + 17*d^2)*cos(f*x + e)^2 - 4*c^2 - 8*c*d - 4*d^2 + 2*(7*c^2 - 18*c*d + 7*d^2)
*cos(f*x + e))*sin(f*x + e))/(cos(f*x + e)^3 + 3*cos(f*x + e)^2 + (cos(f*x + e)^2 - 2*cos(f*x + e) - 4)*sin(f*
x + e) - 2*cos(f*x + e) - 4))/sqrt(a*c - a*d))/((a*c^2*d - a*d^3)*f*cos(f*x + e)^2 - (a*c^3 - a*c*d^2)*f*cos(f
*x + e) - (a*c^3 + a*c^2*d - a*c*d^2 - a*d^3)*f - ((a*c^2*d - a*d^3)*f*cos(f*x + e) + (a*c^3 + a*c^2*d - a*c*d
^2 - a*d^3)*f)*sin(f*x + e)), -1/2*(sqrt(2)*(a*c^2 + 2*a*c*d + a*d^2 - (a*c*d + a*d^2)*cos(f*x + e)^2 + (a*c^2
+ a*c*d)*cos(f*x + e) + (a*c^2 + 2*a*c*d + a*d^2 + (a*c*d + a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(-1/(a*c -
a*d))*arctan(-1/4*sqrt(2)*sqrt(a*sin(f*x + e) + a)*((c - 3*d)*sin(f*x + e) - 3*c + d)*sqrt(d*sin(f*x + e) + c
)*sqrt(-1/(a*c - a*d))/(d*cos(f*x + e)*sin(f*x + e) + c*cos(f*x + e))) + 4*(d*cos(f*x + e) - d*sin(f*x + e) +
d)*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c))/((a*c^2*d - a*d^3)*f*cos(f*x + e)^2 - (a*c^3 - a*c*d^2)*
f*cos(f*x + e) - (a*c^3 + a*c^2*d - a*c*d^2 - a*d^3)*f - ((a*c^2*d - a*d^3)*f*cos(f*x + e) + (a*c^3 + a*c^2*d
- a*c*d^2 - a*d^3)*f)*sin(f*x + e))]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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