3.136 \(\int \frac{\sqrt{a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=99 \[ \frac{a (A+B) \cos (e+f x)}{f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}+\frac{a B \cos (e+f x) \log (1-\sin (e+f x))}{c f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}} \]

[Out]

(a*(A + B)*Cos[e + f*x])/(f*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(3/2)) + (a*B*Cos[e + f*x]*Log[1 - S
in[e + f*x]])/(c*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]])

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Rubi [A]  time = 0.357862, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2971, 2737, 2667, 31, 2738} \[ \frac{a (A+B) \cos (e+f x)}{f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}+\frac{a B \cos (e+f x) \log (1-\sin (e+f x))}{c f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(A + B)*Cos[e + f*x])/(f*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(3/2)) + (a*B*Cos[e + f*x]*Log[1 - S
in[e + f*x]])/(c*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]])

Rule 2971

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

Rule 2737

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(
a*c*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), Int[Cos[e + f*x]/(c + d*Sin[e + f*x]),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2738

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

Rubi steps

\begin{align*} \int \frac{\sqrt{a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx &=(A+B) \int \frac{\sqrt{a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{3/2}} \, dx-\frac{B \int \frac{\sqrt{a+a \sin (e+f x)}}{\sqrt{c-c \sin (e+f x)}} \, dx}{c}\\ &=\frac{a (A+B) \cos (e+f x)}{f \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}-\frac{(a B \cos (e+f x)) \int \frac{\cos (e+f x)}{c-c \sin (e+f x)} \, dx}{\sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{a (A+B) \cos (e+f x)}{f \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac{(a B \cos (e+f x)) \operatorname{Subst}\left (\int \frac{1}{c+x} \, dx,x,-c \sin (e+f x)\right )}{c f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{a (A+B) \cos (e+f x)}{f \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac{a B \cos (e+f x) \log (1-\sin (e+f x))}{c f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 1.16847, size = 147, normalized size = 1.48 \[ \frac{\sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (A+2 B \log \left (i-e^{i (e+f x)}\right )+i B \left (f x+2 i \log \left (i-e^{i (e+f x)}\right )\right ) \sin (e+f x)-i B f x+B\right )}{f (c-c \sin (e+f x))^{3/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + a*Sin[e + f*x]]*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(3/2),x]

[Out]

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

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Maple [B]  time = 0.351, size = 403, normalized size = 4.1 \begin{align*}{\frac{1}{f \left ( -1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) \right ) } \left ( B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) -2\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -B\ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) \sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +2\,B\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +A \left ( \cos \left ( fx+e \right ) \right ) ^{2}-A\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +B \left ( \cos \left ( fx+e \right ) \right ) ^{2}-B\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +B\cos \left ( fx+e \right ) \ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) -2\,B\cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +2\,B\sin \left ( fx+e \right ) \ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) -4\,B\sin \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +A\sin \left ( fx+e \right ) +B\sin \left ( fx+e \right ) -2\,B\ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) +4\,B\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -A-B \right ) \sqrt{a \left ( 1+\sin \left ( fx+e \right ) \right ) } \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sin \left (f x + e\right ) + A\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (B \sin \left (f x + e\right ) + A\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c^{2} \sin \left (f x + e\right ) - 2 \, c^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(B*sin(f*x + e) + A)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(c^2*cos(f*x + e)^2 + 2*c^2*
sin(f*x + e) - 2*c^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a \left (\sin{\left (e + f x \right )} + 1\right )} \left (A + B \sin{\left (e + f x \right )}\right )}{\left (- c \left (\sin{\left (e + f x \right )} - 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*(sin(e + f*x) + 1))*(A + B*sin(e + f*x))/(-c*(sin(e + f*x) - 1))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sin \left (f x + e\right ) + A\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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