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

Optimal. Leaf size=245 \[ \frac{3 A \cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{8 a^2 c^2 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{3 A \cos (e+f x)}{8 a^2 c f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}+\frac{3 A \cos (e+f x)}{8 a^2 f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}-\frac{(A-B) \cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}-\frac{A \cos (e+f x)}{2 a f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.569916, antiderivative size = 245, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {2972, 2743, 2741, 3770} \[ \frac{3 A \cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{8 a^2 c^2 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{3 A \cos (e+f x)}{8 a^2 c f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}+\frac{3 A \cos (e+f x)}{8 a^2 f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}-\frac{(A-B) \cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}-\frac{A \cos (e+f x)}{2 a f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.939692, size = 246, normalized size = 1. \[ \frac{\sec ^3(e+f x) \left (22 A \sin (e+f x)+6 A \sin (3 (e+f x))-9 A \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-12 A \cos (2 (e+f x)) \left (\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )-3 A \cos (4 (e+f x)) \left (\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )+9 A \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+16 B\right )}{64 a^2 c^2 f \sqrt{a (\sin (e+f x)+1)} \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]^3*(16*B - 9*A*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - 12*A*Cos[2*(e + f*x)]*(Log[Cos[(e + f*x
)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]) - 3*A*Cos[4*(e + f*x)]*(Log[Cos[(e + f*x)
/2] - Sin[(e + f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]) + 9*A*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)
/2]] + 22*A*Sin[e + f*x] + 6*A*Sin[3*(e + f*x)]))/(64*a^2*c^2*f*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e +
f*x]])

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Maple [A]  time = 0.297, size = 151, normalized size = 0.6 \begin{align*} -{\frac{\cos \left ( fx+e \right ) }{8\,f} \left ( 3\,A \left ( \cos \left ( fx+e \right ) \right ) ^{4}\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -3\,A \left ( \cos \left ( fx+e \right ) \right ) ^{4}\ln \left ({\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +2\,B \left ( \cos \left ( fx+e \right ) \right ) ^{4}-3\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -2\,A\sin \left ( fx+e \right ) -2\,B \right ) \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{-{\frac{5}{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))^(5/2)/(c-c*sin(f*x+e))^(5/2),x)

[Out]

-1/8/f*(3*A*cos(f*x+e)^4*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-3*A*cos(f*x+e)^4*ln((1-cos(f*x+e)+sin(f*x+
e))/sin(f*x+e))+2*B*cos(f*x+e)^4-3*A*cos(f*x+e)^2*sin(f*x+e)-2*A*sin(f*x+e)-2*B)*cos(f*x+e)/(a*(1+sin(f*x+e)))
^(5/2)/(-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{5}{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))^(5/2)/(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)^(5/2)*(-c*sin(f*x + e) + c)^(5/2)), x)

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Fricas [A]  time = 2.59602, size = 783, normalized size = 3.2 \begin{align*} \left [\frac{3 \, \sqrt{a c} A \cos \left (f x + e\right )^{5} \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 \cos \left (f x + e\right )^{2} + 2 \, A\right )} \sin \left (f x + e\right ) + 2 \, B\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{16 \, a^{3} c^{3} f \cos \left (f x + e\right )^{5}}, -\frac{3 \, \sqrt{-a c} A \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 ) \cos \left (f x + e\right )^{5} -{\left ({\left (3 \, A \cos \left (f x + e\right )^{2} + 2 \, A\right )} \sin \left (f x + e\right ) + 2 \, B\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{8 \, a^{3} c^{3} f \cos \left (f x + e\right )^{5}}\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))^(5/2)/(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

[1/16*(3*sqrt(a*c)*A*cos(f*x + e)^5*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*cos(f*x + e)^2 + 2*A)*sin(f*x + e
) + 2*B)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c))/(a^3*c^3*f*cos(f*x + e)^5), -1/8*(3*sqrt(-a*c)*A*
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)))*cos(f*x
+ e)^5 - ((3*A*cos(f*x + e)^2 + 2*A)*sin(f*x + e) + 2*B)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c))/(
a^3*c^3*f*cos(f*x + e)^5)]

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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))**(5/2)/(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{5}{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))^(5/2)/(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)^(5/2)*(-c*sin(f*x + e) + c)^(5/2)), x)