3.179 \(\int \frac{A+B \sin (e+f x)}{\sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=153 \[ \frac{(A-B) \cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{4 c^2 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{(A-B) \cos (e+f x)}{4 c f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}+\frac{(A+B) \cos (e+f x)}{4 f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}} \]

[Out]

((A + B)*Cos[e + f*x])/(4*f*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(5/2)) + ((A - B)*Cos[e + f*x])/(4*c
*f*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(3/2)) + ((A - B)*ArcTanh[Sin[e + f*x]]*Cos[e + f*x])/(4*c^2*
f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]])

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Rubi [A]  time = 0.355057, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {2972, 2743, 2741, 3770} \[ \frac{(A-B) \cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{4 c^2 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{(A-B) \cos (e+f x)}{4 c f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}+\frac{(A+B) \cos (e+f x)}{4 f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sin[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(5/2)),x]

[Out]

((A + B)*Cos[e + f*x])/(4*f*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(5/2)) + ((A - B)*Cos[e + f*x])/(4*c
*f*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(3/2)) + ((A - B)*ArcTanh[Sin[e + f*x]]*Cos[e + f*x])/(4*c^2*
f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]])

Rule 2972

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

Rule 2743

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

Rule 2741

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Di
st[Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), Int[1/Cos[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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.618092, size = 222, normalized size = 1.45 \[ \frac{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left ((A-B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2+(B-A) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+(A-B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+A+B\right )}{4 f \sqrt{a (\sin (e+f x)+1)} (c-c \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sin[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(5/2)),x]

[Out]

((A + B + (A - B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2 + (-A + B)*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]*
(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4 + (A - B)*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]*(Cos[(e + f*x)/2] -
Sin[(e + f*x)/2])^4)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))/(4*f*Sqrt[a
*(1 + Sin[e + f*x])]*(c - c*Sin[e + f*x])^(5/2))

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Maple [B]  time = 0.344, size = 465, normalized size = 3. \begin{align*}{\frac{\cos \left ( fx+e \right ) }{4\,f} \left ( A \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 ) -A\ln \left ({\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \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 ({\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,A\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 ) -2\,A\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 ) +2\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\,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 ) +2\,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 ) +3\,A\sin \left ( fx+e \right ) -2\,A\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +2\,A\ln \left ({\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +B\sin \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 ) -2\,B\ln \left ({\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -2\,A \right ){\frac{1}{\sqrt{a \left ( 1+\sin \left ( fx+e \right ) \right ) }}} \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(1/2),x)

[Out]

1/4/f*(A*cos(f*x+e)^2*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-A*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*co
s(f*x+e)^2-B*cos(f*x+e)^2*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+B*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e)
)*cos(f*x+e)^2+2*A*sin(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-2*A*sin(f*x+e)*ln((1-cos(f*x+e)+sin(f
*x+e))/sin(f*x+e))+2*A*cos(f*x+e)^2-2*B*sin(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+2*B*sin(f*x+e)*l
n((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+3*A*sin(f*x+e)-2*A*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+2*A*ln((
1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+B*sin(f*x+e)+2*B*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-2*B*ln((1-cos
(f*x+e)+sin(f*x+e))/sin(f*x+e))-2*A)*cos(f*x+e)/(a*(1+sin(f*x+e)))^(1/2)/(-c*(-1+sin(f*x+e)))^(5/2)

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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 (-c \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2)/(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)*(-c*sin(f*x + e) + c)^(5/2)), x)

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Fricas [A]  time = 2.1617, size = 1083, normalized size = 7.08 \begin{align*} \left [-\frac{{\left ({\left (A - B\right )} \cos \left (f x + e\right )^{3} + 2 \,{\left (A - B\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \,{\left (A - B\right )} \cos \left (f x + e\right )\right )} \sqrt{a c} \log \left (-\frac{a c \cos \left (f x + e\right )^{3} - 2 \, a c \cos \left (f x + e\right ) + 2 \, \sqrt{a c} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{3}}\right ) - 2 \,{\left ({\left (A - B\right )} \sin \left (f x + e\right ) - 2 \, A\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{8 \,{\left (a c^{3} f \cos \left (f x + e\right )^{3} + 2 \, a c^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a c^{3} f \cos \left (f x + e\right )\right )}}, -\frac{{\left ({\left (A - B\right )} \cos \left (f x + e\right )^{3} + 2 \,{\left (A - B\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \,{\left (A - B\right )} \cos \left (f x + e\right )\right )} \sqrt{-a c} \arctan \left (\frac{\sqrt{-a c} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{a c \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) -{\left ({\left (A - B\right )} \sin \left (f x + e\right ) - 2 \, A\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{4 \,{\left (a c^{3} f \cos \left (f x + e\right )^{3} + 2 \, a c^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a c^{3} f \cos \left (f x + e\right )\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[-1/8*(((A - B)*cos(f*x + e)^3 + 2*(A - B)*cos(f*x + e)*sin(f*x + e) - 2*(A - B)*cos(f*x + e))*sqrt(a*c)*log(-
(a*c*cos(f*x + e)^3 - 2*a*c*cos(f*x + e) + 2*sqrt(a*c)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*sin(
f*x + e))/cos(f*x + e)^3) - 2*((A - B)*sin(f*x + e) - 2*A)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c))
/(a*c^3*f*cos(f*x + e)^3 + 2*a*c^3*f*cos(f*x + e)*sin(f*x + e) - 2*a*c^3*f*cos(f*x + e)), -1/4*(((A - B)*cos(f
*x + e)^3 + 2*(A - B)*cos(f*x + e)*sin(f*x + e) - 2*(A - B)*cos(f*x + e))*sqrt(-a*c)*arctan(sqrt(-a*c)*sqrt(a*
sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(a*c*cos(f*x + e)*sin(f*x + e))) - ((A - B)*sin(f*x + e) - 2*A)*sq
rt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c))/(a*c^3*f*cos(f*x + e)^3 + 2*a*c^3*f*cos(f*x + e)*sin(f*x + e
) - 2*a*c^3*f*cos(f*x + e))]

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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-c*sin(f*x+e))**(5/2)/(a+a*sin(f*x+e))**(1/2),x)

[Out]

Timed out

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Giac [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 (-c \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)/(sqrt(a*sin(f*x + e) + a)*(-c*sin(f*x + e) + c)^(5/2)), x)