3.4.98 \(\int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx\) [398]

Optimal. Leaf size=143 \[ \frac {c^3 \cos (e+f x) \log (1+\sin (e+f x))}{a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f (a+a \sin (e+f x))^{5/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.21, antiderivative size = 143, 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 = {2818, 2816, 2746, 31} \begin {gather*} \frac {c^3 \cos (e+f x) \log (\sin (e+f x)+1)}{a^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f (a \sin (e+f x)+a)^{3/2}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f (a \sin (e+f x)+a)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 2746

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 2816

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 2818

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 {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx &=-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f (a+a \sin (e+f x))^{5/2}}-\frac {c \int \frac {(c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2}} \, dx}{a}\\ &=\frac {c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f (a+a \sin (e+f x))^{5/2}}+\frac {c^2 \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a^2}\\ &=\frac {c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f (a+a \sin (e+f x))^{5/2}}+\frac {\left (c^3 \cos (e+f x)\right ) \int \frac {\cos (e+f x)}{a+a \sin (e+f x)} \, dx}{a \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f (a+a \sin (e+f x))^{5/2}}+\frac {\left (c^3 \cos (e+f x)\right ) \text {Subst}\left (\int \frac {1}{a+x} \, dx,x,a \sin (e+f x)\right )}{a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {c^3 \cos (e+f x) \log (1+\sin (e+f x))}{a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f (a+a \sin (e+f x))^{5/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.75, size = 172, normalized size = 1.20 \begin {gather*} \frac {c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)} \left (2+3 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )-\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 )+4 \left (1+\log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)\right )}{f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(565\) vs. \(2(129)=258\).
time = 18.65, size = 566, normalized size = 3.96

method	result	size
default	\(\frac {\left (\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 )-\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+2 \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 ) \sin \left (f x +e \right )+2 \left (\cos ^{3}\left (f x +e \right )\right )-2 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )-3 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+6 \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 )-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 ) \cos \left (f x +e \right ) \sin \left (f x +e \right )-2 \left (\cos ^{2}\left (f x +e \right )\right )-2 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right ) \cos \left (f x +e \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 )+4 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right ) \sin \left (f x +e \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 ) \sin \left (f x +e \right )-2 \cos \left (f x +e \right )+2 \sin \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 (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {5}{2}}}{f \left (\cos ^{3}\left (f x +e \right )+\sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )-3 \left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right ) \sin \left (f x +e \right )-2 \cos \left (f x +e \right )-4 \sin \left (f x +e \right )+4\right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}}}\)	\(566\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.51, size = 197, normalized size = 1.38 \begin {gather*} \frac {\frac {8 \, \sqrt {a} c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (a^{3} + \frac {4 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {6 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {4 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {a^{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 \, c^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{\frac {5}{2}}} + \frac {c^{\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 )}{a^{\frac {5}{2}}}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3004 deep

________________________________________________________________________________________

Giac [A]
time = 0.47, size = 135, normalized size = 0.94 \begin {gather*} -\frac {\sqrt {2} \sqrt {a} c^{\frac {5}{2}} {\left (\frac {2 \, \sqrt {2} \log \left (-2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {\sqrt {2} {\left (4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 3\right )}}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{4 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________