3.186 \(\int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=217 \[ \frac{(3 A-B) \cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{8 a c^2 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{(3 A-B) \cos (e+f x)}{8 a c f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}+\frac{(3 A-B) \cos (e+f x)}{8 a f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}-\frac{(A-B) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}} \]

[Out]

-((A - B)*Cos[e + f*x])/(2*f*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(5/2)) + ((3*A - B)*Cos[e + f*x])
/(8*a*f*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(5/2)) + ((3*A - B)*Cos[e + f*x])/(8*a*c*f*Sqrt[a + a*Si
n[e + f*x]]*(c - c*Sin[e + f*x])^(3/2)) + ((3*A - B)*ArcTanh[Sin[e + f*x]]*Cos[e + f*x])/(8*a*c^2*f*Sqrt[a + a
*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]])

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Rubi [A]  time = 0.479328, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {2972, 2743, 2741, 3770} \[ \frac{(3 A-B) \cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{8 a c^2 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{(3 A-B) \cos (e+f x)}{8 a c f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}+\frac{(3 A-B) \cos (e+f x)}{8 a f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}-\frac{(A-B) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(5/2)),x]

[Out]

-((A - B)*Cos[e + f*x])/(2*f*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(5/2)) + ((3*A - B)*Cos[e + f*x])
/(8*a*f*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(5/2)) + ((3*A - B)*Cos[e + f*x])/(8*a*c*f*Sqrt[a + a*Si
n[e + f*x]]*(c - c*Sin[e + f*x])^(3/2)) + ((3*A - B)*ArcTanh[Sin[e + f*x]]*Cos[e + f*x])/(8*a*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)}{(a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}} \, dx &=-\frac{(A-B) \cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}+\frac{(3 A-B) \int \frac{1}{\sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}} \, dx}{2 a}\\ &=-\frac{(A-B) \cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}+\frac{(3 A-B) \cos (e+f x)}{8 a f \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac{(3 A-B) \int \frac{1}{\sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx}{4 a c}\\ &=-\frac{(A-B) \cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}+\frac{(3 A-B) \cos (e+f x)}{8 a f \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac{(3 A-B) \cos (e+f x)}{8 a c f \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac{(3 A-B) \int \frac{1}{\sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}} \, dx}{8 a c^2}\\ &=-\frac{(A-B) \cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}+\frac{(3 A-B) \cos (e+f x)}{8 a f \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac{(3 A-B) \cos (e+f x)}{8 a c f \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac{((3 A-B) \cos (e+f x)) \int \sec (e+f x) \, dx}{8 a c^2 \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=-\frac{(A-B) \cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}+\frac{(3 A-B) \cos (e+f x)}{8 a f \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac{(3 A-B) \cos (e+f x)}{8 a c f \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac{(3 A-B) \tanh ^{-1}(\sin (e+f x)) \cos (e+f x)}{8 a c^2 f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.948476, size = 306, normalized size = 1.41 \[ \frac{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left ((B-A) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4+(A+B) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2+(B-3 A) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+(3 A-B) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+2 A \cos ^2(e+f x)\right )}{8 f (a (\sin (e+f x)+1))^{3/2} (c-c \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(5/2)),x]

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(2*A*Cos[e + f*x]^2 + (-A + B)*(C
os[(e + f*x)/2] - Sin[(e + f*x)/2])^4 + (A + B)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + (-3*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])^
2 + (3*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])^2))/(8*f*(a*(1 + Sin[e + f*x]))^(3/2)*(c - c*Sin[e + f*x])^(5/2))

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Maple [B]  time = 0.272, size = 431, normalized size = 2. \begin{align*}{\frac{\cos \left ( fx+e \right ) }{8\,f} \left ( 3\,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}\sin \left ( fx+e \right ) -3\,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}\sin \left ( fx+e \right ) -B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\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 ) +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}\sin \left ( fx+e \right ) +2\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -3\,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 ) +3\,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}+2\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +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}+A \left ( \cos \left ( fx+e \right ) \right ) ^{2}-3\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}+3\,A\sin \left ( fx+e \right ) -B\sin \left ( fx+e \right ) -A+3\,B \right ) \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{-{\frac{3}{2}}} \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))/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/2),x)

[Out]

1/8/f*(3*A*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2*sin(f*x+e)-3*A*ln((1-cos(f*x+e)+sin(f*x+e))
/sin(f*x+e))*cos(f*x+e)^2*sin(f*x+e)-B*cos(f*x+e)^2*sin(f*x+e)*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*sin(f*x+e)+2*A*cos(f*x+e)^2*sin(f*x+e)-3*A*cos(f*x+e)^2*ln
(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+3*A*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2+2*B*cos(f*x
+e)^2*sin(f*x+e)+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+A*cos(f*x+e)^2-3*B*cos(f*x+e)^2+3*A*sin(f*x+e)-B*sin(f*x+e)-A+3*B)*cos(f*x+e)/(a*(1+sin(f
*x+e)))^(3/2)/(-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}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}}{\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))/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)/((a*sin(f*x + e) + a)^(3/2)*(-c*sin(f*x + e) + c)^(5/2)), x)

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Fricas [A]  time = 2.62385, size = 1060, normalized size = 4.88 \begin{align*} \left [-\frac{{\left ({\left (3 \, A - B\right )} \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) -{\left (3 \, A - B\right )} \cos \left (f x + e\right )^{3}\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 (3 \, A - B\right )} \cos \left (f x + e\right )^{2} +{\left (3 \, A - B\right )} \sin \left (f x + e\right ) - A + 3 \, B\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{16 \,{\left (a^{2} c^{3} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - a^{2} c^{3} f \cos \left (f x + e\right )^{3}\right )}}, -\frac{{\left ({\left (3 \, A - B\right )} \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) -{\left (3 \, A - B\right )} \cos \left (f x + e\right )^{3}\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 (3 \, A - B\right )} \cos \left (f x + e\right )^{2} +{\left (3 \, A - B\right )} \sin \left (f x + e\right ) - A + 3 \, B\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{8 \,{\left (a^{2} c^{3} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - a^{2} c^{3} f \cos \left (f x + e\right )^{3}\right )}}\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))^(3/2)/(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

[-1/16*(((3*A - B)*cos(f*x + e)^3*sin(f*x + e) - (3*A - B)*cos(f*x + e)^3)*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*((3*A - B)*cos(f*x + e)^2 + (3*A - B)*sin(f*x + e) - A + 3*B)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f
*x + e) + c))/(a^2*c^3*f*cos(f*x + e)^3*sin(f*x + e) - a^2*c^3*f*cos(f*x + e)^3), -1/8*(((3*A - B)*cos(f*x + e
)^3*sin(f*x + e) - (3*A - B)*cos(f*x + e)^3)*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))) + ((3*A - B)*cos(f*x + e)^2 + (3*A - B)*sin(f*x + e) - A + 3*B
)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c))/(a^2*c^3*f*cos(f*x + e)^3*sin(f*x + e) - a^2*c^3*f*cos(f
*x + e)^3)]

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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))/(a+a*sin(f*x+e))**(3/2)/(c-c*sin(f*x+e))**(5/2),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}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}}{\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))/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)/((a*sin(f*x + e) + a)^(3/2)*(-c*sin(f*x + e) + c)^(5/2)), x)