3.4.18 \(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))} \, dx\) [318]

Optimal. Leaf size=187 \[ -\frac {(A (c-5 d)+B (3 c+d)) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} (c-d)^2 f}+\frac {2 \sqrt {d} (B c-A d) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{a^{3/2} (c-d)^2 \sqrt {c+d} f}-\frac {(A-B) \cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.40, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {3057, 3064, 2728, 212, 2852, 214} \begin {gather*} -\frac {(A (c-5 d)+B (3 c+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 {2 \sqrt {d} (B c-A d) \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 {(A-B) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((A*(c - 5*d) + B*(3*c + 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*Sqrt[d]*(B*c - A*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)^2*Sqrt[c + d]*f) - ((A - B)*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 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 3057

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*(A*b - a*B)*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] + Dist[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*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
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 {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))} \, dx &=-\frac {(A-B) \cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2}}-\frac {\int \frac {-\frac {1}{2} a (3 B c+A (c-4 d))-\frac {1}{2} a (A-B) 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 {(A-B) \cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2}}-\frac {(d (B c-A d)) \int \frac {\sqrt {a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{a^2 (c-d)^2}+\frac {(A (c-5 d)+B (3 c+d)) \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx}{4 a (c-d)^2}\\ &=-\frac {(A-B) \cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2}}+\frac {(2 d (B c-A d)) \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 {(A (c-5 d)+B (3 c+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 {(A (c-5 d)+B (3 c+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 \sqrt {d} (B c-A d) \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 {(A-B) \cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 2.22, size = 419, normalized size = 2.24 \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 (A-B) (c-d) \sin \left (\frac {1}{2} (e+f x)\right )+(-A+B) (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} (A (c-5 d)+B (3 c+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 {\sqrt {d} (B c-A d) \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 {\sqrt {d} (-B c+A d) \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[(A + B*Sin[e + f*x])/((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*(A - B)*(c - d)*Sin[(e + f*x)/2] + (-A + B)*(c - d)*(Cos[(e + f*x)/2
] + Sin[(e + f*x)/2]) + (1 + I)*(-1)^(3/4)*(A*(c - 5*d) + B*(3*c + d))*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Ta
n[(e + f*x)/4])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + (Sqrt[d]*(B*c - A*d)*(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] + (Sqrt[d]*(-(B*c) + A*d)*(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))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(623\) vs. \(2(158)=316\).
time = 9.64, size = 624, normalized size = 3.34

method	result	size
default	\(-\frac {\left (\sin \left (f x +e \right ) \left (8 A \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +a \,d^{2}}}\right ) a^{\frac {3}{2}} d^{2}-8 B \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +a \,d^{2}}}\right ) a^{\frac {3}{2}} c d +A \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 A \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 +3 B \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 +B \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 A \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +a \,d^{2}}}\right ) a^{\frac {3}{2}} d^{2}-8 B \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +a \,d^{2}}}\right ) a^{\frac {3}{2}} c d +A \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 A \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 +3 B \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 +B \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 A \sqrt {a \left (c +d \right ) d}\, \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a}\, c -2 A \sqrt {a \left (c +d \right ) d}\, \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a}\, d -2 B \sqrt {a \left (c +d \right ) d}\, \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a}\, c +2 B \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}\)	\(624\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))/(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*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(3/2)*d^2-8*B*arctanh((a
-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(3/2)*c*d+A*(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*(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+3*B*(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+B*(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*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d
+a*d^2)^(1/2))*a^(3/2)*d^2-8*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(3/2)*c*d+A*(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*(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+3*B*(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+B*(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+2*A
*(a*(c+d)*d)^(1/2)*(a-a*sin(f*x+e))^(1/2)*a^(1/2)*c-2*A*(a*(c+d)*d)^(1/2)*(a-a*sin(f*x+e))^(1/2)*a^(1/2)*d-2*B
*(a*(c+d)*d)^(1/2)*(a-a*sin(f*x+e))^(1/2)*a^(1/2)*c+2*B*(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

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 667 vs. \(2 (164) = 328\).
time = 2.07, size = 1633, normalized size = 8.73 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(sqrt(2)*(((A + 3*B)*c - (5*A - B)*d)*cos(f*x + e)^2 - 2*(A + 3*B)*c + 2*(5*A - B)*d - ((A + 3*B)*c - (5*
A - B)*d)*cos(f*x + e) - (2*(A + 3*B)*c - 2*(5*A - B)*d + ((A + 3*B)*c - (5*A - B)*d)*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*(2*B*a*c - 2*A*a*d - (B*a*c - A*a*d)*cos(f*x + e)^2 + (B*a*c - A*a*d)*cos(f*x
+ e) + (2*B*a*c - 2*A*a*d + (B*a*c - A*a*d)*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*((A - B)*c - (A - B)*d + ((A - B)*c - (A - B)*d)*cos(f*x + e) - ((A - B)*c - (A - B)*d)*sin
(f*x + e))*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)*(((A + 3*B)*c - (5*A - B)*d)*cos(f*x +
e)^2 - 2*(A + 3*B)*c + 2*(5*A - B)*d - ((A + 3*B)*c - (5*A - B)*d)*cos(f*x + e) - (2*(A + 3*B)*c - 2*(5*A - B)
*d + ((A + 3*B)*c - (5*A - B)*d)*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)) - 8*(2*B*a*c - 2*A*a*d -
(B*a*c - A*a*d)*cos(f*x + e)^2 + (B*a*c - A*a*d)*cos(f*x + e) + (2*B*a*c - 2*A*a*d + (B*a*c - A*a*d)*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*sin(f*x + e) - c - 2*d)*sqrt(-
d/(a*c + a*d))/(d*cos(f*x + e))) + 4*((A - B)*c - (A - B)*d + ((A - B)*c - (A - B)*d)*cos(f*x + e) - ((A - B)*
c - (A - B)*d)*sin(f*x + e))*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))]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 488 vs. \(2 (164) = 328\).
time = 0.60, size = 488, normalized size = 2.61 \begin {gather*} \frac {\frac {8 \, \sqrt {2} {\left (B c d - A d^{2}\right )} \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 (\sqrt {2} a^{\frac {3}{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^{\frac {3}{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^{\frac {3}{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 (A \sqrt {a} c + 3 \, B \sqrt {a} c - 5 \, A \sqrt {a} d + B \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 (A \sqrt {a} c + 3 \, B \sqrt {a} c - 5 \, A \sqrt {a} d + B \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 \, {\left (A \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - B \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\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((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e)),x, algorithm="giac")

[Out]

1/4*(8*sqrt(2)*(B*c*d - A*d^2)*arctan(sqrt(2)*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)/sqrt(-c*d - d^2))/((sqrt(2)*a^(
3/2)*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 2*sqrt(2)*a^(3/2)*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + sqr
t(2)*a^(3/2)*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(-c*d - d^2)) + (A*sqrt(a)*c + 3*B*sqrt(a)*c - 5*A*s
qrt(a)*d + B*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))) - (A*sqrt(a)*c + 3*B*sqrt(a)*c - 5*A*sqrt(a)*d + B*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))) - 2*(A*sqrt(a)*sin(-1/4*pi + 1/2*f*x + 1/2*e) - B*sqrt(
a)*sin(-1/4*pi + 1/2*f*x + 1/2*e))/((sqrt(2)*a^2*c*sgn(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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+B\,\sin \left (e+f\,x\right )}{{\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((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^(3/2)*(c + d*sin(e + f*x))),x)

[Out]

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

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