\(\int \frac {(3+3 \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{5/2}} \, dx\) [584]

Optimal result

Integrand size = 29, antiderivative size = 173 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{5/2}} \, dx=-\frac {18 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt {d} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{d^{5/2} f}+\frac {6 (c-d) \cos (e+f x) \sqrt {3+3 \sin (e+f x)}}{d (c+d) f (c+d \sin (e+f x))^{3/2}}+\frac {18 (c-d) (3 c+7 d) \cos (e+f x)}{d^2 (c+d)^2 f \sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \]

[Out]

-2*a^(5/2)*arctan(cos(f*x+e)*a^(1/2)*d^(1/2)/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2))/d^(5/2)/f+2/3*a^2*
(c-d)*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/d/(c+d)/f/(c+d*sin(f*x+e))^(3/2)+2/3*a^3*(c-d)*(3*c+7*d)*cos(f*x+e)/d^
2/(c+d)^2/f/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2841, 3059, 2854, 211} \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{5/2}} \, dx=-\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{d^{5/2} f}+\frac {2 a^3 (c-d) (3 c+7 d) \cos (e+f x)}{3 d^2 f (c+d)^2 \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}+\frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}} \]

[In]

Int[(a + a*Sin[e + f*x])^(5/2)/(c + d*Sin[e + f*x])^(5/2),x]

[Out]

(-2*a^(5/2)*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/(d^(5/
2)*f) + (2*a^2*(c - d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(3*d*(c + d)*f*(c + d*Sin[e + f*x])^(3/2)) + (2*
a^3*(c - d)*(3*c + 7*d)*Cos[e + f*x])/(3*d^2*(c + d)^2*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2841

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-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] + Dist[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 2854

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

Rule 3059

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n +
1)*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{3 d (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac {(2 a) \int \frac {\sqrt {a+a \sin (e+f x)} \left (\frac {1}{2} a (c-7 d)-\frac {3}{2} a (c+d) \sin (e+f x)\right )}{(c+d \sin (e+f x))^{3/2}} \, dx}{3 d (c+d)} \\ & = \frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{3 d (c+d) f (c+d \sin (e+f x))^{3/2}}+\frac {2 a^3 (c-d) (3 c+7 d) \cos (e+f x)}{3 d^2 (c+d)^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}+\frac {a^2 \int \frac {\sqrt {a+a \sin (e+f x)}}{\sqrt {c+d \sin (e+f x)}} \, dx}{d^2} \\ & = \frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{3 d (c+d) f (c+d \sin (e+f x))^{3/2}}+\frac {2 a^3 (c-d) (3 c+7 d) \cos (e+f x)}{3 d^2 (c+d)^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}-\frac {\left (2 a^3\right ) \text {Subst}\left (\int \frac {1}{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 )}{d^2 f} \\ & = -\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{d^{5/2} f}+\frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{3 d (c+d) f (c+d \sin (e+f x))^{3/2}}+\frac {2 a^3 (c-d) (3 c+7 d) \cos (e+f x)}{3 d^2 (c+d)^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 7.16 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.53 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{5/2}} \, dx=\frac {9 \sqrt {3} (1+\sin (e+f x))^{5/2} \left (\frac {2 \arctan \left (\frac {\sqrt {2} \sqrt {d} \sin \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{\sqrt {c+d \sin (e+f x)}}\right )+\text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{\sqrt {c+d \sin (e+f x)}}\right )-\log \left (\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e-\pi +2 f x)\right )+\sqrt {c+d \sin (e+f x)}\right )}{d^{5/2}}+\frac {2 (c-d) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (3 c^2+8 c d+d^2+4 d (c+2 d) \sin (e+f x)\right )}{3 d^2 (c+d)^2 (c+d \sin (e+f x))^{3/2}}\right )}{f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \]

[In]

Integrate[(3 + 3*Sin[e + f*x])^(5/2)/(c + d*Sin[e + f*x])^(5/2),x]

[Out]

(9*Sqrt[3]*(1 + Sin[e + f*x])^(5/2)*((2*ArcTan[(Sqrt[2]*Sqrt[d]*Sin[(2*e - Pi + 2*f*x)/4])/Sqrt[c + d*Sin[e +
f*x]]] + ArcTanh[(Sqrt[2]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4])/Sqrt[c + d*Sin[e + f*x]]] - Log[Sqrt[2]*Sqrt[d]*C
os[(2*e - Pi + 2*f*x)/4] + Sqrt[c + d*Sin[e + f*x]]])/d^(5/2) + (2*(c - d)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2
])*(3*c^2 + 8*c*d + d^2 + 4*d*(c + 2*d)*Sin[e + f*x]))/(3*d^2*(c + d)^2*(c + d*Sin[e + f*x])^(3/2))))/(f*(Cos[
(e + f*x)/2] + Sin[(e + f*x)/2])^5)

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(13121\) vs. \(2(159)=318\).

Time = 1.29 (sec) , antiderivative size = 13122, normalized size of antiderivative = 75.85

\[\text {output too large to display}\]

[In]

int((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(5/2),x)

[Out]

result too large to display

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 919 vs. \(2 (159) = 318\).

Time = 0.67 (sec) , antiderivative size = 2297, normalized size of antiderivative = 13.28 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*(a^2*c^4 + 4*a^2*c^3*d + 6*a^2*c^2*d^2 + 4*a^2*c*d^3 + a^2*d^4 - (a^2*c^2*d^2 + 2*a^2*c*d^3 + a^2*d^
4)*cos(f*x + e)^3 - (2*a^2*c^3*d + 5*a^2*c^2*d^2 + 4*a^2*c*d^3 + a^2*d^4)*cos(f*x + e)^2 + (a^2*c^4 + 2*a^2*c^
3*d + 2*a^2*c^2*d^2 + 2*a^2*c*d^3 + a^2*d^4)*cos(f*x + e) + (a^2*c^4 + 4*a^2*c^3*d + 6*a^2*c^2*d^2 + 4*a^2*c*d
^3 + a^2*d^4 - (a^2*c^2*d^2 + 2*a^2*c*d^3 + a^2*d^4)*cos(f*x + e)^2 + 2*(a^2*c^3*d + 2*a^2*c^2*d^2 + a^2*c*d^3
)*cos(f*x + e))*sin(f*x + e))*sqrt(-a/d)*log((128*a*d^4*cos(f*x + e)^5 + a*c^4 + 4*a*c^3*d + 6*a*c^2*d^2 + 4*a
*c*d^3 + a*d^4 + 128*(2*a*c*d^3 - a*d^4)*cos(f*x + e)^4 - 32*(5*a*c^2*d^2 - 14*a*c*d^3 + 13*a*d^4)*cos(f*x + e
)^3 - 32*(a*c^3*d - 2*a*c^2*d^2 + 9*a*c*d^3 - 4*a*d^4)*cos(f*x + e)^2 - 8*(16*d^4*cos(f*x + e)^4 - c^3*d + 17*
c^2*d^2 - 59*c*d^3 + 51*d^4 + 24*(c*d^3 - d^4)*cos(f*x + e)^3 - 2*(5*c^2*d^2 - 26*c*d^3 + 33*d^4)*cos(f*x + e)
^2 - (c^3*d - 7*c^2*d^2 + 31*c*d^3 - 25*d^4)*cos(f*x + e) + (16*d^4*cos(f*x + e)^3 + c^3*d - 17*c^2*d^2 + 59*c
*d^3 - 51*d^4 - 8*(3*c*d^3 - 5*d^4)*cos(f*x + e)^2 - 2*(5*c^2*d^2 - 14*c*d^3 + 13*d^4)*cos(f*x + e))*sin(f*x +
e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)*sqrt(-a/d) + (a*c^4 - 28*a*c^3*d + 230*a*c^2*d^2 - 476*
a*c*d^3 + 289*a*d^4)*cos(f*x + e) + (128*a*d^4*cos(f*x + e)^4 + a*c^4 + 4*a*c^3*d + 6*a*c^2*d^2 + 4*a*c*d^3 +
a*d^4 - 256*(a*c*d^3 - a*d^4)*cos(f*x + e)^3 - 32*(5*a*c^2*d^2 - 6*a*c*d^3 + 5*a*d^4)*cos(f*x + e)^2 + 32*(a*c
^3*d - 7*a*c^2*d^2 + 15*a*c*d^3 - 9*a*d^4)*cos(f*x + e))*sin(f*x + e))/(cos(f*x + e) + sin(f*x + e) + 1)) + 8*
(3*a^2*c^3 + a^2*c^2*d - 11*a^2*c*d^2 + 7*a^2*d^3 + 4*(a^2*c^2*d + a^2*c*d^2 - 2*a^2*d^3)*cos(f*x + e)^2 + (3*
a^2*c^3 + 5*a^2*c^2*d - 7*a^2*c*d^2 - a^2*d^3)*cos(f*x + e) - (3*a^2*c^3 + a^2*c^2*d - 11*a^2*c*d^2 + 7*a^2*d^
3 - 4*(a^2*c^2*d + a^2*c*d^2 - 2*a^2*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x
+ e) + c))/((c^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x + e)^3 + (2*c^3*d^3 + 5*c^2*d^4 + 4*c*d^5 + d^6)*f*cos(f*x + e
)^2 - (c^4*d^2 + 2*c^3*d^3 + 2*c^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x + e) - (c^4*d^2 + 4*c^3*d^3 + 6*c^2*d^4 + 4*
c*d^5 + d^6)*f + ((c^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x + e)^2 - 2*(c^3*d^3 + 2*c^2*d^4 + c*d^5)*f*cos(f*x + e)
- (c^4*d^2 + 4*c^3*d^3 + 6*c^2*d^4 + 4*c*d^5 + d^6)*f)*sin(f*x + e)), -1/6*(3*(a^2*c^4 + 4*a^2*c^3*d + 6*a^2*c
^2*d^2 + 4*a^2*c*d^3 + a^2*d^4 - (a^2*c^2*d^2 + 2*a^2*c*d^3 + a^2*d^4)*cos(f*x + e)^3 - (2*a^2*c^3*d + 5*a^2*c
^2*d^2 + 4*a^2*c*d^3 + a^2*d^4)*cos(f*x + e)^2 + (a^2*c^4 + 2*a^2*c^3*d + 2*a^2*c^2*d^2 + 2*a^2*c*d^3 + a^2*d^
4)*cos(f*x + e) + (a^2*c^4 + 4*a^2*c^3*d + 6*a^2*c^2*d^2 + 4*a^2*c*d^3 + a^2*d^4 - (a^2*c^2*d^2 + 2*a^2*c*d^3
+ a^2*d^4)*cos(f*x + e)^2 + 2*(a^2*c^3*d + 2*a^2*c^2*d^2 + a^2*c*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a/d)*ar
ctan(1/4*(8*d^2*cos(f*x + e)^2 - c^2 + 6*c*d - 9*d^2 - 8*(c*d - d^2)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sq
rt(d*sin(f*x + e) + c)*sqrt(a/d)/(2*a*d^2*cos(f*x + e)^3 - (3*a*c*d - a*d^2)*cos(f*x + e)*sin(f*x + e) - (a*c^
2 - a*c*d + 2*a*d^2)*cos(f*x + e))) + 4*(3*a^2*c^3 + a^2*c^2*d - 11*a^2*c*d^2 + 7*a^2*d^3 + 4*(a^2*c^2*d + a^2
*c*d^2 - 2*a^2*d^3)*cos(f*x + e)^2 + (3*a^2*c^3 + 5*a^2*c^2*d - 7*a^2*c*d^2 - a^2*d^3)*cos(f*x + e) - (3*a^2*c
^3 + a^2*c^2*d - 11*a^2*c*d^2 + 7*a^2*d^3 - 4*(a^2*c^2*d + a^2*c*d^2 - 2*a^2*d^3)*cos(f*x + e))*sin(f*x + e))*
sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c))/((c^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x + e)^3 + (2*c^3*d^3 +
5*c^2*d^4 + 4*c*d^5 + d^6)*f*cos(f*x + e)^2 - (c^4*d^2 + 2*c^3*d^3 + 2*c^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x + e)
- (c^4*d^2 + 4*c^3*d^3 + 6*c^2*d^4 + 4*c*d^5 + d^6)*f + ((c^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x + e)^2 - 2*(c^3*
d^3 + 2*c^2*d^4 + c*d^5)*f*cos(f*x + e) - (c^4*d^2 + 4*c^3*d^3 + 6*c^2*d^4 + 4*c*d^5 + d^6)*f)*sin(f*x + e))]

Sympy [F(-1)]

Timed out. \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(f*x+e))**(5/2)/(c+d*sin(f*x+e))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(5/2)/(d*sin(f*x + e) + c)^(5/2), x)

Giac [F(-1)]

Timed out. \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{5/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

[In]

int((a + a*sin(e + f*x))^(5/2)/(c + d*sin(e + f*x))^(5/2),x)

[Out]

int((a + a*sin(e + f*x))^(5/2)/(c + d*sin(e + f*x))^(5/2), x)