3.312 \(\int \frac {A+B \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2} \, dx\)
Optimal. Leaf size=207 \[ -\frac {(B c-A d) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}+\frac {\left (A d (3 c+d)-B \left (c^2+c d+2 d^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} \sqrt {d} f (c-d)^2 (c+d)^{3/2}}-\frac {\sqrt {2} (A-B) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} f (c-d)^2} \]
[Out]
-(A-B)*arctanh(1/2*cos(f*x+e)*a^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))*2^(1/2)/(c-d)^2/f/a^(1/2)+(A*d*(3*c+d)-B
*(c^2+c*d+2*d^2))*arctanh(cos(f*x+e)*a^(1/2)*d^(1/2)/(c+d)^(1/2)/(a+a*sin(f*x+e))^(1/2))/(c-d)^2/(c+d)^(3/2)/f
/a^(1/2)/d^(1/2)-(-A*d+B*c)*cos(f*x+e)/(c^2-d^2)/f/(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2)
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Rubi [A] time = 0.62, antiderivative size = 207, 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 =
{2984, 2985, 2649, 206, 2773, 208} \[ -\frac {(B c-A d) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}+\frac {\left (A d (3 c+d)-B \left (c^2+c d+2 d^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} \sqrt {d} f (c-d)^2 (c+d)^{3/2}}-\frac {\sqrt {2} (A-B) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} f (c-d)^2} \]
Antiderivative was successfully verified.
[In]
Int[(A + B*Sin[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^2),x]
[Out]
-((Sqrt[2]*(A - B)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[a]*(c - d)^2*f))
+ ((A*d*(3*c + d) - B*(c^2 + c*d + 2*d^2))*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[
e + f*x]])])/(Sqrt[a]*(c - d)^2*Sqrt[d]*(c + d)^(3/2)*f) - ((B*c - A*d)*Cos[e + f*x])/((c^2 - d^2)*f*Sqrt[a +
a*Sin[e + f*x]]*(c + d*Sin[e + f*x]))
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rule 2649
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 2773
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 2984
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 2985
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)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2} \, dx &=-\frac {(B c-A d) \cos (e+f x)}{\left (c^2-d^2\right ) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))}-\frac {\int \frac {-\frac {1}{2} a (A (2 c+d)-B (c+2 d))-\frac {1}{2} a (B c-A d) \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx}{a \left (c^2-d^2\right )}\\ &=-\frac {(B c-A d) \cos (e+f x)}{\left (c^2-d^2\right ) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))}+\frac {(A-B) \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx}{(c-d)^2}-\frac {\left (A d (3 c+d)-B \left (c^2+c d+2 d^2\right )\right ) \int \frac {\sqrt {a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{2 a (c-d)^2 (c+d)}\\ &=-\frac {(B c-A d) \cos (e+f x)}{\left (c^2-d^2\right ) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))}-\frac {(2 (A-B)) \operatorname {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 )}{(c-d)^2 f}+\frac {\left (A d (3 c+d)-B \left (c^2+c d+2 d^2\right )\right ) \operatorname {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 )}{(c-d)^2 (c+d) f}\\ &=-\frac {\sqrt {2} (A-B) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} (c-d)^2 f}+\frac {\left (A d (3 c+d)-B \left (c^2+c d+2 d^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} (c-d)^2 \sqrt {d} (c+d)^{3/2} f}-\frac {(B c-A d) \cos (e+f x)}{\left (c^2-d^2\right ) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))}\\ \end {align*}
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Mathematica [C] time = 6.97, size = 374, normalized size = 1.81 \[ \frac {\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (-\frac {\left (B \left (c^2+c d+2 d^2\right )-A d (3 c+d)\right ) \left (2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right ) \left (\sqrt {c+d}-\sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right )+\sqrt {d} \cos \left (\frac {1}{2} (e+f x)\right )\right )\right )-2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )+e+f x\right )}{\sqrt {d} (c+d)^{3/2}}+\frac {\left (B \left (c^2+c d+2 d^2\right )-A d (3 c+d)\right ) \left (2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right ) \left (\sqrt {c+d}+\sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right )-\sqrt {d} \cos \left (\frac {1}{2} (e+f x)\right )\right )\right )-2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )+e+f x\right )}{\sqrt {d} (c+d)^{3/2}}-\frac {4 (c-d) (B c-A d) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )}{(c+d) (c+d \sin (e+f x))}+(8+8 i) (-1)^{3/4} (A-B) \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (e+f x)\right )-1\right )\right )\right )}{4 f (c-d)^2 \sqrt {a (\sin (e+f x)+1)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(A + B*Sin[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^2),x]
[Out]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*((8 + 8*I)*(-1)^(3/4)*(A - B)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[
(e + f*x)/4])] - ((-(A*d*(3*c + d)) + B*(c^2 + c*d + 2*d^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])]))/(Sqrt[d]*(c + d)^(3/2))
+ ((-(A*d*(3*c + d)) + B*(c^2 + c*d + 2*d^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])]))/(Sqrt[d]*(c + d)^(3/2)) - (4*(c - d)*(B
*c - A*d)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))/((c + d)*(c + d*Sin[e + f*x]))))/(4*(c - d)^2*f*Sqrt[a*(1 + S
in[e + f*x])])
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fricas [B] time = 3.77, size = 2159, normalized size = 10.43 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))/(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
[-1/4*((B*c^3 - (3*A - 2*B)*c^2*d - (4*A - 3*B)*c*d^2 - (A - 2*B)*d^3 - (B*c^2*d - (3*A - B)*c*d^2 - (A - 2*B)
*d^3)*cos(f*x + e)^2 + (B*c^3 - (3*A - B)*c^2*d - (A - 2*B)*c*d^2)*cos(f*x + e) + (B*c^3 - (3*A - 2*B)*c^2*d -
(4*A - 3*B)*c*d^2 - (A - 2*B)*d^3 + (B*c^2*d - (3*A - B)*c*d^2 - (A - 2*B)*d^3)*cos(f*x + e))*sin(f*x + e))*s
qrt(a*c*d + a*d^2)*log((a*d^2*cos(f*x + e)^3 - a*c^2 - 2*a*c*d - a*d^2 - (6*a*c*d + 7*a*d^2)*cos(f*x + e)^2 -
4*sqrt(a*c*d + a*d^2)*(d*cos(f*x + e)^2 - (c + 2*d)*cos(f*x + e) + (d*cos(f*x + e) + c + 3*d)*sin(f*x + e) - c
- 3*d)*sqrt(a*sin(f*x + e) + a) - (a*c^2 + 8*a*c*d + 9*a*d^2)*cos(f*x + e) + (a*d^2*cos(f*x + e)^2 - a*c^2 -
2*a*c*d - a*d^2 + 2*(3*a*c*d + 4*a*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))) - 2*sqrt(2)*((A - B)*a*c^3*d + 3*(A - B)*a*c^2*d^2 + 3*(A - B)*a*c*d^3 + (A - B)*a*d^
4 - ((A - B)*a*c^2*d^2 + 2*(A - B)*a*c*d^3 + (A - B)*a*d^4)*cos(f*x + e)^2 + ((A - B)*a*c^3*d + 2*(A - B)*a*c^
2*d^2 + (A - B)*a*c*d^3)*cos(f*x + e) + ((A - B)*a*c^3*d + 3*(A - B)*a*c^2*d^2 + 3*(A - B)*a*c*d^3 + (A - B)*a
*d^4 + ((A - B)*a*c^2*d^2 + 2*(A - B)*a*c*d^3 + (A - B)*a*d^4)*cos(f*x + e))*sin(f*x + e))*log(-(cos(f*x + e)^
2 - (cos(f*x + e) - 2)*sin(f*x + e) + 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*(cos(f*x + e) - sin(f*x + e) + 1)/sqr
t(a) + 3*cos(f*x + e) + 2)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2))/sqrt(a) - 4*
(B*c^3*d - A*c^2*d^2 - B*c*d^3 + A*d^4 + (B*c^3*d - A*c^2*d^2 - B*c*d^3 + A*d^4)*cos(f*x + e) - (B*c^3*d - A*c
^2*d^2 - B*c*d^3 + A*d^4)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((a*c^4*d^2 - 2*a*c^2*d^4 + a*d^6)*f*cos(f*x
+ e)^2 - (a*c^5*d - 2*a*c^3*d^3 + a*c*d^5)*f*cos(f*x + e) - (a*c^5*d + a*c^4*d^2 - 2*a*c^3*d^3 - 2*a*c^2*d^4
+ a*c*d^5 + a*d^6)*f - ((a*c^4*d^2 - 2*a*c^2*d^4 + a*d^6)*f*cos(f*x + e) + (a*c^5*d + a*c^4*d^2 - 2*a*c^3*d^3
- 2*a*c^2*d^4 + a*c*d^5 + a*d^6)*f)*sin(f*x + e)), 1/2*((B*c^3 - (3*A - 2*B)*c^2*d - (4*A - 3*B)*c*d^2 - (A -
2*B)*d^3 - (B*c^2*d - (3*A - B)*c*d^2 - (A - 2*B)*d^3)*cos(f*x + e)^2 + (B*c^3 - (3*A - B)*c^2*d - (A - 2*B)*c
*d^2)*cos(f*x + e) + (B*c^3 - (3*A - 2*B)*c^2*d - (4*A - 3*B)*c*d^2 - (A - 2*B)*d^3 + (B*c^2*d - (3*A - B)*c*d
^2 - (A - 2*B)*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(-a*c*d - a*d^2)*arctan(1/2*sqrt(-a*c*d - a*d^2)*sqrt(a*si
n(f*x + e) + a)*(d*sin(f*x + e) - c - 2*d)/((a*c*d + a*d^2)*cos(f*x + e))) + sqrt(2)*((A - B)*a*c^3*d + 3*(A -
B)*a*c^2*d^2 + 3*(A - B)*a*c*d^3 + (A - B)*a*d^4 - ((A - B)*a*c^2*d^2 + 2*(A - B)*a*c*d^3 + (A - B)*a*d^4)*co
s(f*x + e)^2 + ((A - B)*a*c^3*d + 2*(A - B)*a*c^2*d^2 + (A - B)*a*c*d^3)*cos(f*x + e) + ((A - B)*a*c^3*d + 3*(
A - B)*a*c^2*d^2 + 3*(A - B)*a*c*d^3 + (A - B)*a*d^4 + ((A - B)*a*c^2*d^2 + 2*(A - B)*a*c*d^3 + (A - B)*a*d^4)
*cos(f*x + e))*sin(f*x + e))*log(-(cos(f*x + e)^2 - (cos(f*x + e) - 2)*sin(f*x + e) + 2*sqrt(2)*sqrt(a*sin(f*x
+ e) + a)*(cos(f*x + e) - sin(f*x + e) + 1)/sqrt(a) + 3*cos(f*x + e) + 2)/(cos(f*x + e)^2 - (cos(f*x + e) + 2
)*sin(f*x + e) - cos(f*x + e) - 2))/sqrt(a) + 2*(B*c^3*d - A*c^2*d^2 - B*c*d^3 + A*d^4 + (B*c^3*d - A*c^2*d^2
- B*c*d^3 + A*d^4)*cos(f*x + e) - (B*c^3*d - A*c^2*d^2 - B*c*d^3 + A*d^4)*sin(f*x + e))*sqrt(a*sin(f*x + e) +
a))/((a*c^4*d^2 - 2*a*c^2*d^4 + a*d^6)*f*cos(f*x + e)^2 - (a*c^5*d - 2*a*c^3*d^3 + a*c*d^5)*f*cos(f*x + e) - (
a*c^5*d + a*c^4*d^2 - 2*a*c^3*d^3 - 2*a*c^2*d^4 + a*c*d^5 + a*d^6)*f - ((a*c^4*d^2 - 2*a*c^2*d^4 + a*d^6)*f*co
s(f*x + e) + (a*c^5*d + a*c^4*d^2 - 2*a*c^3*d^3 - 2*a*c^2*d^4 + a*c*d^5 + a*d^6)*f)*sin(f*x + e))]
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))/(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")
[Out]
Timed out
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maple [B] time = 2.72, size = 899, normalized size = 4.34 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*sin(f*x+e))/(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(1/2),x)
[Out]
(1+sin(f*x+e))*(-a*(sin(f*x+e)-1))^(1/2)/a^(5/2)*(sin(f*x+e)*d*(3*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*
d^2)^(1/2))*a^(5/2)*c*d+A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(5/2)*d^2-B*arctanh((a-a*sin
(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(5/2)*c^2-B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(5
/2)*c*d-2*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(5/2)*d^2-A*(a*(c+d)*d)^(1/2)*2^(1/2)*arct
anh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c-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^2*d+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^2*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^2*d)+3*A*arctanh((a-a*
sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(5/2)*c^2*d+A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))
*a^(5/2)*c*d^2-B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(5/2)*c^3-B*arctanh((a-a*sin(f*x+e))^
(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(5/2)*c^2*d-2*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(5/2)*c
*d^2+A*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a^(3/2)*c*d-A*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a^(3/2)
*d^2-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^2*c^2-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^2*c*d-B*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1
/2)*a^(3/2)*c^2+B*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a^(3/2)*c*d+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^2*c^2+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^2*c*d)/(c-d)^2/(c+d)/(c+d*sin(f*x+e))/(a*(c+d)*d)^(1/2)/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 \[ \int \frac {B \sin \left (f x + e\right ) + A}{\sqrt {a \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) + c\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))/(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
integrate((B*sin(f*x + e) + A)/(sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,\sin \left (e+f\,x\right )}{\sqrt {a+a\,\sin \left (e+f\,x\right )}\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^(1/2)*(c + d*sin(e + f*x))^2),x)
[Out]
int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^(1/2)*(c + d*sin(e + f*x))^2), x)
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))/(c+d*sin(f*x+e))**2/(a+a*sin(f*x+e))**(1/2),x)
[Out]
Timed out
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