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]

-1/4*(A-B)*cos(f*x+e)/f/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(5/2)-1/2*A*cos(f*x+e)/a/f/(a+a*sin(f*x+e))^(3
/2)/(c-c*sin(f*x+e))^(5/2)+3/8*A*cos(f*x+e)/a^2/f/(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(1/2)+3/8*A*cos(f*x+
e)/a^2/c/f/(c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2)+3/8*A*arctanh(sin(f*x+e))*cos(f*x+e)/a^2/c^2/f/(a+a*s
in(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.57, antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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 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 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 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 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*}

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Mathematica [A]  time = 0.94, size = 246, normalized size = 1.00 \[ \frac {\sec ^3(e+f x) \left (22 A \sin (e+f x)+6 A \sin (3 (e+f x))-9 A \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-12 A \cos (2 (e+f x)) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )-3 A \cos (4 (e+f x)) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )+9 A \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )+16 B\right )}{64 a^2 c^2 f \sqrt {a (\sin (e+f x)+1)} \sqrt {c-c \sin (e+f x)}} \]

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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fricas [A]  time = 0.56, size = 306, normalized size = 1.25 \[ \left [\frac {3 \, \sqrt {a c} A \cos \left (f x + e\right )^{5} \log \left (-\frac {a c \cos \left (f x + e\right )^{3} - 2 \, a c \cos \left (f x + e\right ) - 2 \, \sqrt {a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{3}}\right ) + 2 \, {\left ({\left (3 \, A \cos \left (f x + e\right )^{2} + 2 \, A\right )} \sin \left (f x + e\right ) + 2 \, B\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{16 \, a^{3} c^{3} f \cos \left (f x + e\right )^{5}}, -\frac {3 \, \sqrt {-a c} A \arctan \left (\frac {\sqrt {-a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{a c \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{5} - {\left ({\left (3 \, A \cos \left (f x + e\right )^{2} + 2 \, A\right )} \sin \left (f x + e\right ) + 2 \, B\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{8 \, a^{3} c^{3} f \cos \left (f x + e\right )^{5}}\right ] \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \sin \left (f x + e\right ) + A}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]

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)

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maple [A]  time = 0.60, size = 151, normalized size = 0.62 \[ \frac {\left (3 A \left (\cos ^{4}\left (f x +e \right )\right ) \ln \left (\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-3 A \left (\cos ^{4}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-2 B \left (\cos ^{4}\left (f x +e \right )\right )+3 A \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+2 A \sin \left (f x +e \right )+2 B \right ) \cos \left (f x +e \right )}{8 f \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {5}{2}}} \]

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*(sin(f*x+e)-1))^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \sin \left (f x + e\right ) + A}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,\sin \left (e+f\,x\right )}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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