3.402 \(\int \frac {1}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=188 \[ \frac {3 \cos (e+f x)}{8 a^2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}+\frac {3 \cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{8 a^2 c f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {3 \cos (e+f x)}{8 a f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{3/2}}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.38, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2743, 2741, 3770} \[ \frac {3 \cos (e+f x)}{8 a^2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}+\frac {3 \cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{8 a^2 c f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {3 \cos (e+f x)}{8 a f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{3/2}}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(3/2)),x]

[Out]

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

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Mathematica [A]  time = 0.75, size = 287, normalized size = 1.53 \[ \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (-2 \cos ^2(e+f x)+\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4-\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )}{8 f (a (\sin (e+f x)+1))^{5/2} (c-c \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(3/2)),x]

[Out]

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

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fricas [A]  time = 0.50, size = 371, normalized size = 1.97 \[ \left [\frac {3 \, {\left (\cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + \cos \left (f x + e\right )^{3}\right )} \sqrt {a c} \log \left (-\frac {a c \cos \left (f x + e\right )^{3} - 2 \, a c \cos \left (f x + e\right ) - 2 \, \sqrt {a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{3}}\right ) - 2 \, {\left (3 \, \cos \left (f x + e\right )^{2} - 3 \, \sin \left (f x + e\right ) - 1\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{16 \, {\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + a^{3} c^{2} f \cos \left (f x + e\right )^{3}\right )}}, -\frac {3 \, {\left (\cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + \cos \left (f x + e\right )^{3}\right )} \sqrt {-a c} \arctan \left (\frac {\sqrt {-a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{a c \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + {\left (3 \, \cos \left (f x + e\right )^{2} - 3 \, \sin \left (f x + e\right ) - 1\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{8 \, {\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + a^{3} c^{2} f \cos \left (f x + e\right )^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

[1/16*(3*(cos(f*x + e)^3*sin(f*x + e) + 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*cos(f*
x + e)^2 - 3*sin(f*x + e) - 1)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c))/(a^3*c^2*f*cos(f*x + e)^3*s
in(f*x + e) + a^3*c^2*f*cos(f*x + e)^3), -1/8*(3*(cos(f*x + e)^3*sin(f*x + e) + cos(f*x + e)^3)*sqrt(-a*c)*arc
tan(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*cos(f*
x + e)^2 - 3*sin(f*x + e) - 1)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c))/(a^3*c^2*f*cos(f*x + e)^3*s
in(f*x + e) + a^3*c^2*f*cos(f*x + e)^3)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate(1/((a*sin(f*x + e) + a)^(5/2)*(-c*sin(f*x + e) + c)^(3/2)), x)

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maple [A]  time = 0.27, size = 229, normalized size = 1.22 \[ \frac {\left (3 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-3 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+2 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+3 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-3 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-\left (\cos ^{2}\left (f x +e \right )\right )+3 \sin \left (f x +e \right )+1\right ) \cos \left (f x +e \right )}{8 f \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(3/2),x)

[Out]

1/8/f*(3*sin(f*x+e)*cos(f*x+e)^2*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-3*sin(f*x+e)*cos(f*x+e)^2*ln(-(-1+
cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+2*cos(f*x+e)^2*sin(f*x+e)+3*cos(f*x+e)^2*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin
(f*x+e))-3*cos(f*x+e)^2*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-cos(f*x+e)^2+3*sin(f*x+e)+1)*cos(f*x+e)/(a*
(1+sin(f*x+e)))^(5/2)/(-c*(sin(f*x+e)-1))^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((a*sin(f*x + e) + a)^(5/2)*(-c*sin(f*x + e) + c)^(3/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + a*sin(e + f*x))^(5/2)*(c - c*sin(e + f*x))^(3/2)),x)

[Out]

int(1/((a + a*sin(e + f*x))^(5/2)*(c - c*sin(e + f*x))^(3/2)), 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(1/(a+a*sin(f*x+e))**(5/2)/(c-c*sin(f*x+e))**(3/2),x)

[Out]

Timed out

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