3.6.54 \(\int \frac {1}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))} \, dx\) [554]

Optimal. Leaf size=164 \[ -\frac {(c-5 d) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} (c-d)^2 f}-\frac {2 d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{a^{3/2} (c-d)^2 \sqrt {c+d} f}-\frac {\cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.30, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2845, 3064, 2728, 212, 2852, 214} \begin {gather*} -\frac {2 d^{3/2} \tanh ^{-1}\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)^2 \sqrt {c+d}}-\frac {(c-5 d) \tanh ^{-1}\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)^2}-\frac {\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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*} \int \frac {1}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))} \, dx &=-\frac {\cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2}}-\frac {\int \frac {-\frac {1}{2} a (c-4 d)-\frac {1}{2} a d \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx}{2 a^2 (c-d)}\\ &=-\frac {\cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2}}+\frac {(c-5 d) \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx}{4 a (c-d)^2}+\frac {d^2 \int \frac {\sqrt {a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{a^2 (c-d)^2}\\ &=-\frac {\cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2}}-\frac {(c-5 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)^2 f}-\frac {\left (2 d^2\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)^2 f}\\ &=-\frac {(c-5 d) \tanh ^{-1}\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)^2 f}-\frac {2 d^{3/2} \tanh ^{-1}\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)^2 \sqrt {c+d} f}-\frac {\cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains complex when optimal does not.
time = 1.29, size = 385, normalized size = 2.35 \begin {gather*} \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (2 (c-d) \sin \left (\frac {1}{2} (e+f x)\right )-(c-d) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+(1+i) (-1)^{3/4} (c-5 d) \tanh ^{-1}\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-\frac {d^{3/2} \left (e+f x-2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )+2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right ) \left (\sqrt {c+d}+\sqrt {d} \cos \left (\frac {1}{2} (e+f x)\right )-\sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right )\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}{\sqrt {c+d}}+\frac {d^{3/2} \left (e+f x-2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )+2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right ) \left (\sqrt {c+d}-\sqrt {d} \cos \left (\frac {1}{2} (e+f x)\right )+\sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right )\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}{\sqrt {c+d}}\right )}{2 (c-d)^2 f (a (1+\sin (e+f x)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(2*(c - d)*Sin[(e + f*x)/2] - (c - d)*(Cos[(e + f*x)/2] + Sin[(e + f*x)
/2]) + (1 + I)*(-1)^(3/4)*(c - 5*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 - (d^(3/2)*(e + f*x - 2*Log[Sec[(e + f*x)/4]^2] + 2*Log[Sec[(e + f*x)/4]^2*(Sqrt[c + d]
+ Sqrt[d]*Cos[(e + f*x)/2] - Sqrt[d]*Sin[(e + f*x)/2])])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2)/Sqrt[c + d]
+ (d^(3/2)*(e + f*x - 2*Log[Sec[(e + f*x)/4]^2] + 2*Log[Sec[(e + f*x)/4]^2*(Sqrt[c + d] - Sqrt[d]*Cos[(e + f*x
)/2] + Sqrt[d]*Sin[(e + f*x)/2])])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2)/Sqrt[c + d]))/(2*(c - d)^2*f*(a*(1
+ Sin[e + f*x]))^(3/2))

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(337\) vs. \(2(135)=270\).
time = 4.43, size = 338, normalized size = 2.06

method	result	size
default	\(-\frac {\left (\sin \left (f x +e \right ) \left (8 d^{2} \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +d^{2} a}}\right ) a^{\frac {3}{2}}+\sqrt {a \left (c +d \right ) d}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a c -5 \sqrt {a \left (c +d \right ) d}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a d \right )+8 d^{2} \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {a c d +d^{2} a}}\right ) a^{\frac {3}{2}}+\sqrt {a \left (c +d \right ) d}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a c -5 \sqrt {a \left (c +d \right ) d}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a d +2 \sqrt {a \left (c +d \right ) d}\, \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a}\, c -2 \sqrt {a \left (c +d \right ) d}\, \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a}\, d \right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}}{4 a^{\frac {5}{2}} \sqrt {a \left (c +d \right ) d}\, \left (c -d \right )^{2} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(338\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/a^(5/2)*(sin(f*x+e)*(8*d^2*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(3/2)+(a*(c+d)*d)^(1/2
)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a*c-5*(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(a-a
*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a*d)+8*d^2*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(3/2)+(
a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a*c-5*(a*(c+d)*d)^(1/2)*2^(1/2)*a
rctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a*d+2*(a*(c+d)*d)^(1/2)*(a-a*sin(f*x+e))^(1/2)*a^(1/2)*c-2*
(a*(c+d)*d)^(1/2)*(a-a*sin(f*x+e))^(1/2)*a^(1/2)*d)*(-a*(sin(f*x+e)-1))^(1/2)/(a*(c+d)*d)^(1/2)/(c-d)^2/cos(f*
x+e)/(a+a*sin(f*x+e))^(1/2)/f

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 536 vs. \(2 (141) = 282\).
time = 0.55, size = 1371, normalized size = 8.36 \begin {gather*} \left [-\frac {\sqrt {2} {\left ({\left (c - 5 \, d\right )} \cos \left (f x + e\right )^{2} - {\left (c - 5 \, d\right )} \cos \left (f x + e\right ) - {\left ({\left (c - 5 \, d\right )} \cos \left (f x + e\right ) + 2 \, c - 10 \, d\right )} \sin \left (f x + e\right ) - 2 \, c + 10 \, d\right )} \sqrt {a} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\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 ) - 4 \, {\left (a d \cos \left (f x + e\right )^{2} - a d \cos \left (f x + e\right ) - 2 \, a d - {\left (a d \cos \left (f x + e\right ) + 2 \, a d\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{a c + a d}} \log \left (\frac {d^{2} \cos \left (f x + e\right )^{3} - {\left (6 \, c d + 7 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - c^{2} - 2 \, c d - d^{2} - 4 \, {\left ({\left (c d + d^{2}\right )} \cos \left (f x + e\right )^{2} - c^{2} - 4 \, c d - 3 \, d^{2} - {\left (c^{2} + 3 \, c d + 2 \, d^{2}\right )} \cos \left (f x + e\right ) + {\left (c^{2} + 4 \, c d + 3 \, d^{2} + {\left (c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {\frac {d}{a c + a d}} - {\left (c^{2} + 8 \, c d + 9 \, d^{2}\right )} \cos \left (f x + e\right ) + {\left (d^{2} \cos \left (f x + e\right )^{2} - c^{2} - 2 \, c d - d^{2} + 2 \, {\left (3 \, c d + 4 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{d^{2} \cos \left (f x + e\right )^{3} + {\left (2 \, c d + d^{2}\right )} \cos \left (f x + e\right )^{2} - c^{2} - 2 \, c d - d^{2} - {\left (c^{2} + d^{2}\right )} \cos \left (f x + e\right ) + {\left (d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \cos \left (f x + e\right ) - c^{2} - 2 \, c d - d^{2}\right )} \sin \left (f x + e\right )}\right ) - 4 \, {\left ({\left (c - d\right )} \cos \left (f x + e\right ) - {\left (c - d\right )} \sin \left (f x + e\right ) + c - d\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{8 \, {\left ({\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} f \cos \left (f x + e\right ) - 2 \, {\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} f - {\left ({\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} f \cos \left (f x + e\right ) + 2 \, {\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} f\right )} \sin \left (f x + e\right )\right )}}, -\frac {\sqrt {2} {\left ({\left (c - 5 \, d\right )} \cos \left (f x + e\right )^{2} - {\left (c - 5 \, d\right )} \cos \left (f x + e\right ) - {\left ({\left (c - 5 \, d\right )} \cos \left (f x + e\right ) + 2 \, c - 10 \, d\right )} \sin \left (f x + e\right ) - 2 \, c + 10 \, d\right )} \sqrt {a} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\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 ) + 8 \, {\left (a d \cos \left (f x + e\right )^{2} - a d \cos \left (f x + e\right ) - 2 \, a d - {\left (a d \cos \left (f x + e\right ) + 2 \, a d\right )} \sin \left (f x + e\right )\right )} \sqrt {-\frac {d}{a c + a d}} \arctan \left (\frac {\sqrt {a \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) - c - 2 \, d\right )} \sqrt {-\frac {d}{a c + a d}}}{2 \, d \cos \left (f x + e\right )}\right ) - 4 \, {\left ({\left (c - d\right )} \cos \left (f x + e\right ) - {\left (c - d\right )} \sin \left (f x + e\right ) + c - d\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{8 \, {\left ({\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} f \cos \left (f x + e\right ) - 2 \, {\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} f - {\left ({\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} f \cos \left (f x + e\right ) + 2 \, {\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} f\right )} \sin \left (f x + e\right )\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(sqrt(2)*((c - 5*d)*cos(f*x + e)^2 - (c - 5*d)*cos(f*x + e) - ((c - 5*d)*cos(f*x + e) + 2*c - 10*d)*sin(
f*x + e) - 2*c + 10*d)*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*(a*d*cos(f*x + e)^2 - a*d*cos(f*x + e) - 2*a*d - (a*d*cos(
f*x + e) + 2*a*d)*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 - d)*cos(f*x + e)
- (c - d)*sin(f*x + e) + c - d)*sqrt(a*sin(f*x + e) + a))/((a^2*c^2 - 2*a^2*c*d + a^2*d^2)*f*cos(f*x + e)^2 -
(a^2*c^2 - 2*a^2*c*d + a^2*d^2)*f*cos(f*x + e) - 2*(a^2*c^2 - 2*a^2*c*d + a^2*d^2)*f - ((a^2*c^2 - 2*a^2*c*d +
a^2*d^2)*f*cos(f*x + e) + 2*(a^2*c^2 - 2*a^2*c*d + a^2*d^2)*f)*sin(f*x + e)), -1/8*(sqrt(2)*((c - 5*d)*cos(f*
x + e)^2 - (c - 5*d)*cos(f*x + e) - ((c - 5*d)*cos(f*x + e) + 2*c - 10*d)*sin(f*x + e) - 2*c + 10*d)*sqrt(a)*l
og(-(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)) + 8*(a*d*cos(f*x + e)^2 - a*d*cos(f*x + e) - 2*a*d - (a*d*cos(f*x + e) + 2*a*d)*sin(f*x + e))*
sqrt(-d/(a*c + a*d))*arctan(1/2*sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) - c - 2*d)*sqrt(-d/(a*c + a*d))/(d*co
s(f*x + e))) - 4*((c - d)*cos(f*x + e) - (c - d)*sin(f*x + e) + c - d)*sqrt(a*sin(f*x + e) + a))/((a^2*c^2 - 2
*a^2*c*d + a^2*d^2)*f*cos(f*x + e)^2 - (a^2*c^2 - 2*a^2*c*d + a^2*d^2)*f*cos(f*x + e) - 2*(a^2*c^2 - 2*a^2*c*d
+ a^2*d^2)*f - ((a^2*c^2 - 2*a^2*c*d + a^2*d^2)*f*cos(f*x + e) + 2*(a^2*c^2 - 2*a^2*c*d + a^2*d^2)*f)*sin(f*x
+ e))]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \left (c + d \sin {\left (e + f x \right )}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 419 vs. \(2 (141) = 282\).
time = 0.56, size = 419, normalized size = 2.55 \begin {gather*} -\frac {\frac {8 \, \sqrt {a} d^{2} \arctan \left (\frac {\sqrt {2} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c d - d^{2}}}\right )}{{\left (a^{2} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 2 \, a^{2} c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + a^{2} d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {-c d - d^{2}}} - \frac {{\left (\sqrt {a} c - 5 \, \sqrt {a} d\right )} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{\sqrt {2} a^{2} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 2 \, \sqrt {2} a^{2} c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + \sqrt {2} a^{2} d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {{\left (\sqrt {a} c - 5 \, \sqrt {a} d\right )} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{\sqrt {2} a^{2} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 2 \, \sqrt {2} a^{2} c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + \sqrt {2} a^{2} d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {2 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (\sqrt {2} a^{2} c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - \sqrt {2} a^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} {\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}}}{4 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(8*sqrt(a)*d^2*arctan(sqrt(2)*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)/sqrt(-c*d - d^2))/((a^2*c^2*sgn(cos(-1/4*p
i + 1/2*f*x + 1/2*e)) - 2*a^2*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + a^2*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/
2*e)))*sqrt(-c*d - d^2)) - (sqrt(a)*c - 5*sqrt(a)*d)*log(sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(sqrt(2)*a^2*c^2*
sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 2*sqrt(2)*a^2*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + sqrt(2)*a^2*d^2*
sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) + (sqrt(a)*c - 5*sqrt(a)*d)*log(-sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(sqr
t(2)*a^2*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 2*sqrt(2)*a^2*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + sqr
t(2)*a^2*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) + 2*sqrt(a)*sin(-1/4*pi + 1/2*f*x + 1/2*e)/((sqrt(2)*a^2*c*s
gn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - sqrt(2)*a^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*(sin(-1/4*pi + 1/2*f*x
+ 1/2*e)^2 - 1)))/f

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________