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

Optimal. Leaf size=147 \[ -\frac {a^3 \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{c f (c-c \sin (e+f x))^{3/2}}+\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f (c-c \sin (e+f x))^{5/2}} \]

[Out]

1/2*a*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f/(c-c*sin(f*x+e))^(5/2)-a^2*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/c/f/(c-
c*sin(f*x+e))^(3/2)-a^3*cos(f*x+e)*ln(1-sin(f*x+e))/c^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 = 0.29, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2739, 2737, 2667, 31} \[ -\frac {a^3 \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{c f (c-c \sin (e+f x))^{3/2}}+\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f (c-c \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 2737

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(
a*c*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), Int[Cos[e + f*x]/(c + d*Sin[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 2739

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-2*b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)), x] - Dist[(b*(2*m - 1)
)/(d*(2*n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e
, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] && LtQ[n, -1] &&  !(ILtQ[m + n, 0] && G
tQ[2*m + n + 1, 0])

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx &=\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {a \int \frac {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{3/2}} \, dx}{c}\\ &=\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c f (c-c \sin (e+f x))^{3/2}}+\frac {a^2 \int \frac {\sqrt {a+a \sin (e+f x)}}{\sqrt {c-c \sin (e+f x)}} \, dx}{c^2}\\ &=\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c f (c-c \sin (e+f x))^{3/2}}+\frac {\left (a^3 \cos (e+f x)\right ) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)} \, dx}{c \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c f (c-c \sin (e+f x))^{3/2}}-\frac {\left (a^3 \cos (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{c+x} \, dx,x,-c \sin (e+f x)\right )}{c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c f (c-c \sin (e+f x))^{3/2}}-\frac {a^3 \cos (e+f x) \log (1-\sin (e+f x))}{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 = 1.16, size = 190, normalized size = 1.29 \[ \frac {a^2 \sqrt {a (\sin (e+f x)+1)} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos (2 (e+f x)) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-3 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+4 \sin (e+f x) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+1\right )-2\right )}{c^2 f (\sin (e+f x)-1)^2 \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.27, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 2 \, a^{2}\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3} - {\left (c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3}\right )} \sin \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^2)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/
(3*c^3*cos(f*x + e)^2 - 4*c^3 - (c^3*cos(f*x + e)^2 - 4*c^3)*sin(f*x + e)), x)

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

[Out]

Timed out

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maple [B]  time = 0.26, size = 553, normalized size = 3.76 \[ \frac {\left (\sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-2 \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 )+\left (\cos ^{3}\left (f x +e \right )\right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-2 \left (\cos ^{3}\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 )+2 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )-4 \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )+2 \left (\cos ^{3}\left (f x +e \right )\right )-3 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+6 \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 )-4 \sin \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+8 \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 \left (\cos ^{2}\left (f x +e \right )\right )-2 \cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+4 \cos \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 \sin \left (f x +e \right )-2 \cos \left (f x +e \right )+4 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-8 \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+2\right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}}}{f \left (\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-\left (\cos ^{3}\left (f x +e \right )\right )+2 \sin \left (f x +e \right ) \cos \left (f x +e \right )+3 \left (\cos ^{2}\left (f x +e \right )\right )-4 \sin \left (f x +e \right )+2 \cos \left (f x +e \right )-4\right ) \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+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(5/2),x)

[Out]

1/f*(sin(f*x+e)*cos(f*x+e)^2*ln(2/(cos(f*x+e)+1))-2*sin(f*x+e)*cos(f*x+e)^2*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin
(f*x+e))+cos(f*x+e)^3*ln(2/(cos(f*x+e)+1))-2*cos(f*x+e)^3*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+2*cos(f*x
+e)^2*sin(f*x+e)+2*ln(2/(cos(f*x+e)+1))*sin(f*x+e)*cos(f*x+e)-4*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*sin
(f*x+e)*cos(f*x+e)+2*cos(f*x+e)^3-3*cos(f*x+e)^2*ln(2/(cos(f*x+e)+1))+6*cos(f*x+e)^2*ln(-(-1+cos(f*x+e)+sin(f*
x+e))/sin(f*x+e))-4*sin(f*x+e)*ln(2/(cos(f*x+e)+1))+8*sin(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-2*
cos(f*x+e)^2-2*cos(f*x+e)*ln(2/(cos(f*x+e)+1))+4*cos(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-2*sin(f
*x+e)-2*cos(f*x+e)+4*ln(2/(cos(f*x+e)+1))-8*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+2)*(a*(1+sin(f*x+e)))^(
5/2)/(cos(f*x+e)^2*sin(f*x+e)-cos(f*x+e)^3+2*sin(f*x+e)*cos(f*x+e)+3*cos(f*x+e)^2-4*sin(f*x+e)+2*cos(f*x+e)-4)
/(-c*(sin(f*x+e)-1))^(5/2)

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maxima [A]  time = 1.92, size = 184, normalized size = 1.25 \[ -\frac {\frac {8 \, a^{\frac {5}{2}} \sqrt {c} \sin \left (f x + e\right )^{2}}{{\left (c^{3} - \frac {4 \, c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {6 \, c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {4 \, c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {2 \, a^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{\frac {5}{2}}} + \frac {a^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{c^{\frac {5}{2}}}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(8*a^(5/2)*sqrt(c)*sin(f*x + e)^2/((c^3 - 4*c^3*sin(f*x + e)/(cos(f*x + e) + 1) + 6*c^3*sin(f*x + e)^2/(cos(f
*x + e) + 1)^2 - 4*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4)*(cos(f*x
+ e) + 1)^2) - 2*a^(5/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c^(5/2) + a^(5/2)*log(sin(f*x + e)^2/(cos(f
*x + e) + 1)^2 + 1)/c^(5/2))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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