\(\int \frac {1}{(3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2} \, dx\) [555]

Optimal result

Integrand size = 27, antiderivative size = 234 \[ \int \frac {1}{(3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2} \, dx=-\frac {(c-9 d) \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)}}\right )}{6 \sqrt {6} (c-d)^3 f}-\frac {d^{3/2} (5 c+3 d) \text {arctanh}\left (\frac {\sqrt {3} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {3+3 \sin (e+f x)}}\right )}{3 \sqrt {3} (c-d)^3 (c+d)^{3/2} f}-\frac {\cos (e+f x)}{2 (c-d) f (3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))}-\frac {d (c+3 d) \cos (e+f x)}{6 (c-d)^2 (c+d) f \sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))} \]

[Out]

-d^(3/2)*(5*c+3*d)*arctanh(cos(f*x+e)*a^(1/2)*d^(1/2)/(c+d)^(1/2)/(a+a*sin(f*x+e))^(1/2))/a^(3/2)/(c-d)^3/(c+d
)^(3/2)/f-1/2*cos(f*x+e)/(c-d)/f/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))-1/4*(c-9*d)*arctanh(1/2*cos(f*x+e)*a^
(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))/a^(3/2)/(c-d)^3/f*2^(1/2)-1/2*d*(c+3*d)*cos(f*x+e)/a/(c-d)^2/(c+d)/f/(c+
d*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2845, 3063, 3064, 2728, 212, 2852, 214} \[ \int \frac {1}{(3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2} \, dx=-\frac {d^{3/2} (5 c+3 d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{a^{3/2} f (c-d)^3 (c+d)^{3/2}}-\frac {(c-9 d) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{2 \sqrt {2} a^{3/2} f (c-d)^3}-\frac {d (c+3 d) \cos (e+f x)}{2 a f (c-d)^2 (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}-\frac {\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))} \]

[In]

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

[Out]

-1/2*((c - 9*d)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[2]*a^(3/2)*(c - d)^3
*f) - (d^(3/2)*(5*c + 3*d)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/(a^
(3/2)*(c - d)^3*(c + d)^(3/2)*f) - Cos[e + f*x]/(2*(c - d)*f*(a + a*Sin[e + f*x])^(3/2)*(c + d*Sin[e + f*x]))
- (d*(c + 3*d)*Cos[e + f*x])/(2*a*(c - d)^2*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*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 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 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 2845

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

Rule 2852

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

Rule 3063

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

Rule 3064

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[(
B*c - A*d)/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), 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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))}-\frac {\int \frac {-\frac {1}{2} a (c-6 d)-\frac {3}{2} a d \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2} \, dx}{2 a^2 (c-d)} \\ & = -\frac {\cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))}-\frac {d (c+3 d) \cos (e+f x)}{2 a (c-d)^2 (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))}+\frac {\int \frac {\frac {1}{2} a^2 \left (c^2-7 c d-6 d^2\right )+\frac {1}{2} a^2 d (c+3 d) \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx}{2 a^3 (c-d)^2 (c+d)} \\ & = -\frac {\cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))}-\frac {d (c+3 d) \cos (e+f x)}{2 a (c-d)^2 (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))}+\frac {(c-9 d) \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx}{4 a (c-d)^3}+\frac {\left (d^2 (5 c+3 d)\right ) \int \frac {\sqrt {a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{2 a^2 (c-d)^3 (c+d)} \\ & = -\frac {\cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))}-\frac {d (c+3 d) \cos (e+f x)}{2 a (c-d)^2 (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))}-\frac {(c-9 d) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{2 a (c-d)^3 f}-\frac {\left (d^2 (5 c+3 d)\right ) \text {Subst}\left (\int \frac {1}{a c+a d-d x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{a (c-d)^3 (c+d) f} \\ & = -\frac {(c-9 d) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} (c-d)^3 f}-\frac {d^{3/2} (5 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 )}{a^{3/2} (c-d)^3 (c+d)^{3/2} f}-\frac {\cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))}-\frac {d (c+3 d) \cos (e+f x)}{2 a (c-d)^2 (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 5.74 (sec) , antiderivative size = 856, normalized size of antiderivative = 3.66 \[ \int \frac {1}{(3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\frac {4 \sin \left (\frac {1}{2} (e+f x)\right )}{(c-d)^2}-\frac {2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}{(c-d)^2}+\frac {(2+2 i) (-1)^{3/4} (c-9 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 ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}{(c-d)^3}+\frac {d^{3/2} (5 c+3 d) \left (e+f x-2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )+\text {RootSum}\left [c+4 d \text {$\#$1}+2 c \text {$\#$1}^2-4 d \text {$\#$1}^3+c \text {$\#$1}^4\&,\frac {-d \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )+\sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )-c \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}+2 \sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}+3 d \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2-\sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2-c \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^3}{-d-c \text {$\#$1}+3 d \text {$\#$1}^2-c \text {$\#$1}^3}\&\right ]\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}{(-c+d)^3 (c+d)^{3/2}}+\frac {d^{3/2} (5 c+3 d) \left (e+f x-2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )+\text {RootSum}\left [c+4 d \text {$\#$1}+2 c \text {$\#$1}^2-4 d \text {$\#$1}^3+c \text {$\#$1}^4\&,\frac {-d \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )-\sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )-c \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}-2 \sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}+3 d \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2+\sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2-c \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^3}{-d-c \text {$\#$1}+3 d \text {$\#$1}^2-c \text {$\#$1}^3}\&\right ]\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}{(c-d)^3 (c+d)^{3/2}}-\frac {4 d^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}{(c-d)^2 (c+d) (c+d \sin (e+f x))}\right )}{12 \sqrt {3} f (1+\sin (e+f x))^{3/2}} \]

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*((4*Sin[(e + f*x)/2])/(c - d)^2 - (2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/
2]))/(c - d)^2 + ((2 + 2*I)*(-1)^(3/4)*(c - 9*d)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])]*(Cos[
(e + f*x)/2] + Sin[(e + f*x)/2])^2)/(c - d)^3 + (d^(3/2)*(5*c + 3*d)*(e + f*x - 2*Log[Sec[(e + f*x)/4]^2] + Ro
otSum[c + 4*d*#1 + 2*c*#1^2 - 4*d*#1^3 + c*#1^4 & , (-(d*Log[-#1 + Tan[(e + f*x)/4]]) + Sqrt[d]*Sqrt[c + d]*Lo
g[-#1 + Tan[(e + f*x)/4]] - c*Log[-#1 + Tan[(e + f*x)/4]]*#1 + 2*Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4
]]*#1 + 3*d*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 - Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 - c*Log[-#
1 + Tan[(e + f*x)/4]]*#1^3)/(-d - c*#1 + 3*d*#1^2 - c*#1^3) & ])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2)/((-c
+ d)^3*(c + d)^(3/2)) + (d^(3/2)*(5*c + 3*d)*(e + f*x - 2*Log[Sec[(e + f*x)/4]^2] + RootSum[c + 4*d*#1 + 2*c*
#1^2 - 4*d*#1^3 + c*#1^4 & , (-(d*Log[-#1 + Tan[(e + f*x)/4]]) - Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4
]] - c*Log[-#1 + Tan[(e + f*x)/4]]*#1 - 2*Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 + 3*d*Log[-#1 + T
an[(e + f*x)/4]]*#1^2 + Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 - c*Log[-#1 + Tan[(e + f*x)/4]]*#
1^3)/(-d - c*#1 + 3*d*#1^2 - c*#1^3) & ])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2)/((c - d)^3*(c + d)^(3/2)) -
(4*d^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2)/((c - d)^2*(c + d)*(c +
d*Sin[e + f*x]))))/(12*Sqrt[3]*f*(1 + Sin[e + f*x])^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(977\) vs. \(2(210)=420\).

Time = 1.64 (sec) , antiderivative size = 978, normalized size of antiderivative = 4.18

method	result	size
default	\(\text {Expression too large to display}\)	\(978\)

[In]

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

[Out]

-1/4/a^(5/2)*(2^(1/2)*arctanh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^(1/2)/a^(1/2))*(a*(c+d)*d)^(1/2)*sin(f*x+e)^2*a*
c^2*d-8*2^(1/2)*arctanh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^(1/2)/a^(1/2))*(a*(c+d)*d)^(1/2)*sin(f*x+e)^2*a*c*d^2-
9*2^(1/2)*arctanh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^(1/2)/a^(1/2))*(a*(c+d)*d)^(1/2)*sin(f*x+e)^2*a*d^3+20*a^(3/
2)*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*sin(f*x+e)^2*c*d^3+12*a^(3/2)*arctanh((-a*(sin(f*x+e
)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*sin(f*x+e)^2*d^4+2^(1/2)*arctanh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^(1/2)/a^(1/2
))*(a*(c+d)*d)^(1/2)*sin(f*x+e)*a*c^3-7*2^(1/2)*arctanh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^(1/2)/a^(1/2))*(a*(c+d
)*d)^(1/2)*sin(f*x+e)*a*c^2*d-17*2^(1/2)*arctanh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^(1/2)/a^(1/2))*(a*(c+d)*d)^(1
/2)*sin(f*x+e)*a*c*d^2-9*2^(1/2)*arctanh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^(1/2)/a^(1/2))*(a*(c+d)*d)^(1/2)*sin(
f*x+e)*a*d^3+20*a^(3/2)*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*sin(f*x+e)*c^2*d^2+32*a^(3/2)*a
rctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*sin(f*x+e)*c*d^3+12*a^(3/2)*arctanh((-a*(sin(f*x+e)-1))^
(1/2)*d/(a*(c+d)*d)^(1/2))*sin(f*x+e)*d^4+2*(-a*(sin(f*x+e)-1))^(1/2)*a^(1/2)*(a*(c+d)*d)^(1/2)*sin(f*x+e)*c^2
*d+4*(-a*(sin(f*x+e)-1))^(1/2)*a^(1/2)*(a*(c+d)*d)^(1/2)*sin(f*x+e)*c*d^2-6*(-a*(sin(f*x+e)-1))^(1/2)*a^(1/2)*
(a*(c+d)*d)^(1/2)*sin(f*x+e)*d^3+2^(1/2)*arctanh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^(1/2)/a^(1/2))*(a*(c+d)*d)^(1
/2)*a*c^3-8*2^(1/2)*arctanh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^(1/2)/a^(1/2))*(a*(c+d)*d)^(1/2)*a*c^2*d-9*2^(1/2)
*arctanh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^(1/2)/a^(1/2))*(a*(c+d)*d)^(1/2)*a*c*d^2+20*a^(3/2)*arctanh((-a*(sin(
f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*c^2*d^2+12*a^(3/2)*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2)
)*c*d^3+2*(-a*(sin(f*x+e)-1))^(1/2)*a^(1/2)*(a*(c+d)*d)^(1/2)*c^3+2*(-a*(sin(f*x+e)-1))^(1/2)*a^(1/2)*(a*(c+d)
*d)^(1/2)*c*d^2-4*(-a*(sin(f*x+e)-1))^(1/2)*a^(1/2)*(a*(c+d)*d)^(1/2)*d^3)*(-a*(sin(f*x+e)-1))^(1/2)/(a*(c+d)*
d)^(1/2)/(c+d*sin(f*x+e))/(c+d)/(c-d)^3/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. 1115 vs. \(2 (210) = 420\).

Time = 0.77 (sec) , antiderivative size = 2515, normalized size of antiderivative = 10.75 \[ \int \frac {1}{(3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[1/8*(sqrt(2)*((c^2*d - 8*c*d^2 - 9*d^3)*cos(f*x + e)^3 - 2*c^3 + 14*c^2*d + 34*c*d^2 + 18*d^3 + (c^3 - 6*c^2*
d - 25*c*d^2 - 18*d^3)*cos(f*x + e)^2 - (c^3 - 7*c^2*d - 17*c*d^2 - 9*d^3)*cos(f*x + e) - (2*c^3 - 14*c^2*d -
34*c*d^2 - 18*d^3 - (c^2*d - 8*c*d^2 - 9*d^3)*cos(f*x + e)^2 + (c^3 - 7*c^2*d - 17*c*d^2 - 9*d^3)*cos(f*x + e)
)*sin(f*x + e))*sqrt(a)*log(-(a*cos(f*x + e)^2 - 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sqrt(a)*(cos(f*x + e) - si
n(f*x + e) + 1) + 3*a*cos(f*x + e) - (a*cos(f*x + e) - 2*a)*sin(f*x + e) + 2*a)/(cos(f*x + e)^2 - (cos(f*x + e
) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) + 2*(10*a*c^2*d + 16*a*c*d^2 + 6*a*d^3 - (5*a*c*d^2 + 3*a*d^3)*cos(f*
x + e)^3 - (5*a*c^2*d + 13*a*c*d^2 + 6*a*d^3)*cos(f*x + e)^2 + (5*a*c^2*d + 8*a*c*d^2 + 3*a*d^3)*cos(f*x + e)
+ (10*a*c^2*d + 16*a*c*d^2 + 6*a*d^3 - (5*a*c*d^2 + 3*a*d^3)*cos(f*x + e)^2 + (5*a*c^2*d + 8*a*c*d^2 + 3*a*d^3
)*cos(f*x + e))*sin(f*x + e))*sqrt(d/(a*c + a*d))*log((d^2*cos(f*x + e)^3 - (6*c*d + 7*d^2)*cos(f*x + e)^2 - c
^2 - 2*c*d - d^2 + 4*((c*d + d^2)*cos(f*x + e)^2 - c^2 - 4*c*d - 3*d^2 - (c^2 + 3*c*d + 2*d^2)*cos(f*x + e) +
(c^2 + 4*c*d + 3*d^2 + (c*d + d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d/(a*c + a*d)) -
(c^2 + 8*c*d + 9*d^2)*cos(f*x + e) + (d^2*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 + 2*(3*c*d + 4*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)*sin(f*x + e))) + 4*(c^3 - c^2*d - c*d^2 + d
^3 + (c^2*d + 2*c*d^2 - 3*d^3)*cos(f*x + e)^2 + (c^3 + c*d^2 - 2*d^3)*cos(f*x + e) - (c^3 - c^2*d - c*d^2 + d^
3 - (c^2*d + 2*c*d^2 - 3*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((a^2*c^4*d - 2*a^2*c^3*d^
2 + 2*a^2*c*d^4 - a^2*d^5)*f*cos(f*x + e)^3 + (a^2*c^5 - 4*a^2*c^3*d^2 + 2*a^2*c^2*d^3 + 3*a^2*c*d^4 - 2*a^2*d
^5)*f*cos(f*x + e)^2 - (a^2*c^5 - a^2*c^4*d - 2*a^2*c^3*d^2 + 2*a^2*c^2*d^3 + a^2*c*d^4 - a^2*d^5)*f*cos(f*x +
e) - 2*(a^2*c^5 - a^2*c^4*d - 2*a^2*c^3*d^2 + 2*a^2*c^2*d^3 + a^2*c*d^4 - a^2*d^5)*f + ((a^2*c^4*d - 2*a^2*c^
3*d^2 + 2*a^2*c*d^4 - a^2*d^5)*f*cos(f*x + e)^2 - (a^2*c^5 - a^2*c^4*d - 2*a^2*c^3*d^2 + 2*a^2*c^2*d^3 + a^2*c
*d^4 - a^2*d^5)*f*cos(f*x + e) - 2*(a^2*c^5 - a^2*c^4*d - 2*a^2*c^3*d^2 + 2*a^2*c^2*d^3 + a^2*c*d^4 - a^2*d^5)
*f)*sin(f*x + e)), 1/8*(sqrt(2)*((c^2*d - 8*c*d^2 - 9*d^3)*cos(f*x + e)^3 - 2*c^3 + 14*c^2*d + 34*c*d^2 + 18*d
^3 + (c^3 - 6*c^2*d - 25*c*d^2 - 18*d^3)*cos(f*x + e)^2 - (c^3 - 7*c^2*d - 17*c*d^2 - 9*d^3)*cos(f*x + e) - (2
*c^3 - 14*c^2*d - 34*c*d^2 - 18*d^3 - (c^2*d - 8*c*d^2 - 9*d^3)*cos(f*x + e)^2 + (c^3 - 7*c^2*d - 17*c*d^2 - 9
*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a)*log(-(a*cos(f*x + e)^2 - 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sqrt(a)*
(cos(f*x + e) - sin(f*x + e) + 1) + 3*a*cos(f*x + e) - (a*cos(f*x + e) - 2*a)*sin(f*x + e) + 2*a)/(cos(f*x + e
)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) + 4*(10*a*c^2*d + 16*a*c*d^2 + 6*a*d^3 - (5*a*c*d^2
+ 3*a*d^3)*cos(f*x + e)^3 - (5*a*c^2*d + 13*a*c*d^2 + 6*a*d^3)*cos(f*x + e)^2 + (5*a*c^2*d + 8*a*c*d^2 + 3*a*
d^3)*cos(f*x + e) + (10*a*c^2*d + 16*a*c*d^2 + 6*a*d^3 - (5*a*c*d^2 + 3*a*d^3)*cos(f*x + e)^2 + (5*a*c^2*d + 8
*a*c*d^2 + 3*a*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(-d/(a*c + a*d))*arctan(1/2*sqrt(a*sin(f*x + e) + a)*(d*si
n(f*x + e) - c - 2*d)*sqrt(-d/(a*c + a*d))/(d*cos(f*x + e))) + 4*(c^3 - c^2*d - c*d^2 + d^3 + (c^2*d + 2*c*d^2
- 3*d^3)*cos(f*x + e)^2 + (c^3 + c*d^2 - 2*d^3)*cos(f*x + e) - (c^3 - c^2*d - c*d^2 + d^3 - (c^2*d + 2*c*d^2
- 3*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((a^2*c^4*d - 2*a^2*c^3*d^2 + 2*a^2*c*d^4 - a^2
*d^5)*f*cos(f*x + e)^3 + (a^2*c^5 - 4*a^2*c^3*d^2 + 2*a^2*c^2*d^3 + 3*a^2*c*d^4 - 2*a^2*d^5)*f*cos(f*x + e)^2
- (a^2*c^5 - a^2*c^4*d - 2*a^2*c^3*d^2 + 2*a^2*c^2*d^3 + a^2*c*d^4 - a^2*d^5)*f*cos(f*x + e) - 2*(a^2*c^5 - a^
2*c^4*d - 2*a^2*c^3*d^2 + 2*a^2*c^2*d^3 + a^2*c*d^4 - a^2*d^5)*f + ((a^2*c^4*d - 2*a^2*c^3*d^2 + 2*a^2*c*d^4 -
a^2*d^5)*f*cos(f*x + e)^2 - (a^2*c^5 - a^2*c^4*d - 2*a^2*c^3*d^2 + 2*a^2*c^2*d^3 + a^2*c*d^4 - a^2*d^5)*f*cos
(f*x + e) - 2*(a^2*c^5 - a^2*c^4*d - 2*a^2*c^3*d^2 + 2*a^2*c^2*d^3 + a^2*c*d^4 - a^2*d^5)*f)*sin(f*x + e))]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(3+3 \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{%%{[%%%{%%{[-384,0]:[1,0,-2]%%},[0]%%%},0]:[1,0,%%%{-1,[
1]%%%}]%%},

Mupad [F(-1)]

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

[In]

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

[Out]

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