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

Optimal. Leaf size=151 \[ \frac {(A+B) \cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{4 a^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {(A+B) \cos (e+f x)}{4 a f (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.35, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2972, 2743, 2741, 3770} \[ \frac {(A+B) \cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{4 a^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {(A+B) \cos (e+f x)}{4 a f (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((A - B)*Cos[e + f*x])/(4*f*(a + a*Sin[e + f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]]) - ((A + B)*Cos[e + f*x])/(4*
a*f*(a + a*Sin[e + f*x])^(3/2)*Sqrt[c - c*Sin[e + f*x]]) + ((A + B)*ArcTanh[Sin[e + f*x]]*Cos[e + f*x])/(4*a^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} \sqrt {c-c \sin (e+f x)}} \, dx &=-\frac {(A-B) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}+\frac {(A+B) \int \frac {1}{(a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}} \, dx}{2 a}\\ &=-\frac {(A-B) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}-\frac {(A+B) \cos (e+f x)}{4 a f (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}+\frac {(A+B) \int \frac {1}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \, dx}{4 a^2}\\ &=-\frac {(A-B) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}-\frac {(A+B) \cos (e+f x)}{4 a f (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}+\frac {((A+B) \cos (e+f x)) \int \sec (e+f x) \, dx}{4 a^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} \sqrt {c-c \sin (e+f x)}}-\frac {(A+B) \cos (e+f x)}{4 a f (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}+\frac {(A+B) \tanh ^{-1}(\sin (e+f x)) \cos (e+f x)}{4 a^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 = 0.67, size = 214, normalized size = 1.42 \[ \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (-(A+B) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2-\left ((A+B) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )+(A+B) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )-A+B\right )}{4 f (a (\sin (e+f x)+1))^{5/2} \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.53, size = 408, normalized size = 2.70 \[ \left [\frac {{\left ({\left (A + B\right )} \cos \left (f x + e\right )^{3} - 2 \, {\left (A + B\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, {\left (A + B\right )} \cos \left (f x + e\right )\right )} \sqrt {a c} \log \left (-\frac {a c \cos \left (f x + e\right )^{3} - 2 \, a c \cos \left (f x + e\right ) - 2 \, \sqrt {a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{3}}\right ) + 2 \, {\left ({\left (A + B\right )} \sin \left (f x + e\right ) + 2 \, A\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{8 \, {\left (a^{3} c f \cos \left (f x + e\right )^{3} - 2 \, a^{3} c f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} c f \cos \left (f x + e\right )\right )}}, -\frac {{\left ({\left (A + B\right )} \cos \left (f x + e\right )^{3} - 2 \, {\left (A + B\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, {\left (A + B\right )} \cos \left (f x + e\right )\right )} \sqrt {-a c} \arctan \left (\frac {\sqrt {-a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{a c \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) - {\left ({\left (A + B\right )} \sin \left (f x + e\right ) + 2 \, A\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{4 \, {\left (a^{3} c f \cos \left (f x + e\right )^{3} - 2 \, a^{3} c f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} c f \cos \left (f x + e\right )\right )}}\right ] \]

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))^(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^3*c*f*cos(f*x + e)^3 - 2*a^3*c*f*cos(f*x + e)*sin(f*x + e) - 2*a^3*c*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*s
in(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)*sqr
t(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c))/(a^3*c*f*cos(f*x + e)^3 - 2*a^3*c*f*cos(f*x + e)*sin(f*x + e)
- 2*a^3*c*f*cos(f*x + e))]

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

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))^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.63, size = 466, normalized size = 3.09 \[ \frac {\left (-A \left (\cos ^{2}\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 )+A \left (\cos ^{2}\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 )-B \left (\cos ^{2}\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 )+B \left (\cos ^{2}\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 A \sin \left (f x +e \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 A \sin \left (f x +e \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 A \left (\cos ^{2}\left (f x +e \right )\right )+2 B \sin \left (f x +e \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 \sin \left (f x +e \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 A \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 \sin \left (f x +e \right )-2 A \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 \ln \left (\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-B \sin \left (f x +e \right )-2 B \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+2 A \right ) \cos \left (f x +e \right )}{4 f \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}} \]

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

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

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))^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+B\,\sin \left (e+f\,x\right )}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \]

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))^(1/2)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B \sin {\left (e + f x \right )}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}} \sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )}}\, dx \]

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))**(1/2),x)

[Out]

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

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