3.4.89 \(\int \frac {(c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^{3/2}} \, dx\) [389]

Optimal. Leaf size=191 \[ -\frac {12 c^4 \cos (e+f x) \log (1+\sin (e+f x))}{a f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {6 c^3 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a+a \sin (e+f x)}}-\frac {3 c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a+a \sin (e+f x))^{3/2}} \]

[Out]

-c*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)/f/(a+a*sin(f*x+e))^(3/2)-3/2*c^2*cos(f*x+e)*(c-c*sin(f*x+e))^(3/2)/a/f/(a
+a*sin(f*x+e))^(1/2)-12*c^4*cos(f*x+e)*ln(1+sin(f*x+e))/a/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)-6*c^
3*cos(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a/f/(a+a*sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.26, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2818, 2819, 2816, 2746, 31} \begin {gather*} -\frac {12 c^4 \cos (e+f x) \log (\sin (e+f x)+1)}{a f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {6 c^3 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a \sin (e+f x)+a}}-\frac {3 c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a \sin (e+f x)+a}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a \sin (e+f x)+a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c - c*Sin[e + f*x])^(7/2)/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

(-12*c^4*Cos[e + f*x]*Log[1 + Sin[e + f*x]])/(a*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]) - (6*c^3*
Cos[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(a*f*Sqrt[a + a*Sin[e + f*x]]) - (3*c^2*Cos[e + f*x]*(c - c*Sin[e + f*x
])^(3/2))/(2*a*f*Sqrt[a + a*Sin[e + f*x]]) - (c*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(f*(a + a*Sin[e + f*x
])^(3/2))

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])

Rule 2819

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 - 1)*((c + d*Sin[e + f*x])^n/(f*(m + n))), x] + Dist[a*((2*m - 1)/(
m + n)), 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, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m
]) &&  !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])

Rubi steps

\begin {align*} \int \frac {(c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^{3/2}} \, dx &=-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a+a \sin (e+f x))^{3/2}}-\frac {(3 c) \int \frac {(c-c \sin (e+f x))^{5/2}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a}\\ &=-\frac {3 c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a+a \sin (e+f x))^{3/2}}-\frac {\left (6 c^2\right ) \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a}\\ &=-\frac {6 c^3 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a+a \sin (e+f x)}}-\frac {3 c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a+a \sin (e+f x))^{3/2}}-\frac {\left (12 c^3\right ) \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a}\\ &=-\frac {6 c^3 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a+a \sin (e+f x)}}-\frac {3 c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a+a \sin (e+f x))^{3/2}}-\frac {\left (12 c^4 \cos (e+f x)\right ) \int \frac {\cos (e+f x)}{a+a \sin (e+f x)} \, dx}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=-\frac {6 c^3 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a+a \sin (e+f x)}}-\frac {3 c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a+a \sin (e+f x))^{3/2}}-\frac {\left (12 c^4 \cos (e+f x)\right ) \text {Subst}\left (\int \frac {1}{a+x} \, dx,x,a \sin (e+f x)\right )}{a f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=-\frac {12 c^4 \cos (e+f x) \log (1+\sin (e+f x))}{a f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {6 c^3 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a+a \sin (e+f x)}}-\frac {3 c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a+a \sin (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 1.02, size = 162, normalized size = 0.85 \begin {gather*} \frac {c^3 \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 (-44-18 \cos (2 (e+f x))-192 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+\left (39-192 \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)+\sin (3 (e+f x))\right )}{8 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)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c - c*Sin[e + f*x])^(7/2)/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

(c^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(-44 - 18*Cos[2*(e + f*x)] - 192*Log[Cos[(
e + f*x)/2] + Sin[(e + f*x)/2]] + (39 - 192*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]])*Sin[e + f*x] + Sin[3*(e
+ f*x)]))/(8*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(499\) vs. \(2(173)=346\).
time = 17.26, size = 500, normalized size = 2.62

method	result	size
default	\(\frac {\left (\cos ^{4}\left (f x +e \right )+\left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )+24 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+24 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )-48 \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 )-48 \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 )-9 \left (\cos ^{3}\left (f x +e \right )\right )+8 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )+24 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right ) \cos \left (f x +e \right )-48 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right ) \sin \left (f x +e \right )-48 \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 )+96 \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 )+33 \left (\cos ^{2}\left (f x +e \right )\right )+25 \cos \left (f x +e \right ) \sin \left (f x +e \right )-48 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+96 \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+9 \cos \left (f x +e \right )-34 \sin \left (f x +e \right )-34\right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {7}{2}}}{2 f \left (\cos ^{4}\left (f x +e \right )-\left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )+3 \left (\cos ^{3}\left (f x +e \right )\right )+4 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )-8 \left (\cos ^{2}\left (f x +e \right )\right )+4 \cos \left (f x +e \right ) \sin \left (f x +e \right )-4 \cos \left (f x +e \right )-8 \sin \left (f x +e \right )+8\right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}}}\)	\(500\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/2/f*(cos(f*x+e)^4+cos(f*x+e)^3*sin(f*x+e)+24*ln(2/(cos(f*x+e)+1))*cos(f*x+e)^2+24*ln(2/(cos(f*x+e)+1))*sin(f
*x+e)*cos(f*x+e)-48*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2-48*ln(-(-1+cos(f*x+e)-sin(f*x+e))/
sin(f*x+e))*cos(f*x+e)*sin(f*x+e)-9*cos(f*x+e)^3+8*sin(f*x+e)*cos(f*x+e)^2+24*ln(2/(cos(f*x+e)+1))*cos(f*x+e)-
48*ln(2/(cos(f*x+e)+1))*sin(f*x+e)-48*cos(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))+96*ln(-(-1+cos(f*x
+e)-sin(f*x+e))/sin(f*x+e))*sin(f*x+e)+33*cos(f*x+e)^2+25*cos(f*x+e)*sin(f*x+e)-48*ln(2/(cos(f*x+e)+1))+96*ln(
-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))+9*cos(f*x+e)-34*sin(f*x+e)-34)*(-c*(sin(f*x+e)-1))^(7/2)/(cos(f*x+e)^4
-cos(f*x+e)^3*sin(f*x+e)+3*cos(f*x+e)^3+4*sin(f*x+e)*cos(f*x+e)^2-8*cos(f*x+e)^2+4*cos(f*x+e)*sin(f*x+e)-4*cos
(f*x+e)-8*sin(f*x+e)+8)/(a*(1+sin(f*x+e)))^(3/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((-c*sin(f*x + e) + c)^(7/2)/(a*sin(f*x + e) + a)^(3/2), x)

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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))^(7/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))**(7/2)/(a+a*sin(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [A]
time = 0.50, size = 179, normalized size = 0.94 \begin {gather*} \frac {2 \, \sqrt {a} c^{\frac {7}{2}} {\left (\frac {6 \, \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 4 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{4}} - \frac {2}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} a^{2} \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 )}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

2*sqrt(a)*c^(7/2)*(6*log(-sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 1)/(a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) + (a
^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^4 + 4*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/
2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2)/a^4 - 2/((sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1)*a^2*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

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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 )}^{7/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c*sin(e + f*x))^(7/2)/(a + a*sin(e + f*x))^(3/2),x)

[Out]

int((c - c*sin(e + f*x))^(7/2)/(a + a*sin(e + f*x))^(3/2), x)

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