3.311 \(\int \frac{A+B \sin (e+f x)}{\sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx\)

Optimal. Leaf size=136 \[ -\frac{\sqrt{2} (A-B) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{\sqrt{a} f (c-d)}-\frac{2 (B c-A d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{\sqrt{a} \sqrt{d} f (c-d) \sqrt{c+d}} \]

[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)*f)) -
(2*(B*c - A*d)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[a]*(c - d
)*Sqrt[d]*Sqrt[c + d]*f)

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Rubi [A]  time = 0.282783, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.135, Rules used = {2985, 2649, 206, 2773, 208} \[ -\frac{\sqrt{2} (A-B) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{\sqrt{a} f (c-d)}-\frac{2 (B c-A d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{\sqrt{a} \sqrt{d} f (c-d) \sqrt{c+d}} \]

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])),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)*f)) -
(2*(B*c - A*d)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[a]*(c - d
)*Sqrt[d]*Sqrt[c + d]*f)

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]

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 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 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 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]

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))} \, dx &=\frac{(A-B) \int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx}{c-d}+\frac{(B c-A d) \int \frac{\sqrt{a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{a (c-d)}\\ &=-\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) f}-\frac{(2 (B c-A d)) \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) 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) f}-\frac{2 (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 )}{\sqrt{a} (c-d) \sqrt{d} \sqrt{c+d} f}\\ \end{align*}

Mathematica [C]  time = 3.0525, size = 238, normalized size = 1.75 \[ \frac{(-1)^{3/4} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sqrt [4]{-1} (B c-A d) \left (\log \left (\sec ^2\left (\frac{1}{4} (e+f x)\right ) \left (\sqrt{c+d}-\sqrt{d} \sin \left (\frac{1}{2} (e+f x)\right )+\sqrt{d} \cos \left (\frac{1}{2} (e+f x)\right )\right )\right )-\log \left (\sec ^2\left (\frac{1}{4} (e+f x)\right ) \left (\sqrt{c+d}+\sqrt{d} \sin \left (\frac{1}{2} (e+f x)\right )-\sqrt{d} \cos \left (\frac{1}{2} (e+f x)\right )\right )\right )\right )+(2+2 i) \sqrt{d} (A-B) \sqrt{c+d} \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )\right )}{\sqrt{d} f (c-d) \sqrt{c+d} \sqrt{a (\sin (e+f x)+1)}} \]

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])),x]

[Out]

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

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Maple [A]  time = 1.713, size = 199, normalized size = 1.5 \begin{align*} -{\frac{1+\sin \left ( fx+e \right ) }{ \left ( c-d \right ) \cos \left ( fx+e \right ) f}\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) } \left ( \sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }{\frac{1}{\sqrt{a}}}} \right ) \sqrt{a \left ( c+d \right ) d}A-2\,A\sqrt{a}{\it Artanh} \left ({\frac{\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }d}{\sqrt{a \left ( c+d \right ) d}}} \right ) d-\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }{\frac{1}{\sqrt{a}}}} \right ) \sqrt{a \left ( c+d \right ) d}B+2\,B\sqrt{a}{\it Artanh} \left ({\frac{\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }d}{\sqrt{a \left ( c+d \right ) d}}} \right ) c \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{a \left ( c+d \right ) d}}}{\frac{1}{\sqrt{a+a\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(1+sin(f*x+e))*(-a*(-1+sin(f*x+e)))^(1/2)*(2^(1/2)*arctanh(1/2*(-a*(-1+sin(f*x+e)))^(1/2)*2^(1/2)/a^(1/2))*(a
*(c+d)*d)^(1/2)*A-2*A*a^(1/2)*arctanh((-a*(-1+sin(f*x+e)))^(1/2)*d/(a*(c+d)*d)^(1/2))*d-2^(1/2)*arctanh(1/2*(-
a*(-1+sin(f*x+e)))^(1/2)*2^(1/2)/a^(1/2))*(a*(c+d)*d)^(1/2)*B+2*B*a^(1/2)*arctanh((-a*(-1+sin(f*x+e)))^(1/2)*d
/(a*(c+d)*d)^(1/2))*c)/(c-d)/a^(1/2)/(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., size = 0, normalized size = 0. \begin{align*} \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 )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [B]  time = 10.0327, size = 1831, normalized size = 13.46 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(a*c*d + a*d^2)*(B*c - A*d)*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))) + sqrt(2)*((A - B)*a*c*d + (A - B)*a*d^2)*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))/((a*c^2*
d - a*d^3)*f), -1/2*(2*sqrt(-a*c*d - a*d^2)*(B*c - A*d)*arctan(1/2*sqrt(-a*c*d - a*d^2)*sqrt(a*sin(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*d + (A - B)*a*d^2)*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))/((a*c^2*d - a*d^3)*f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError