3.4.75 \(\int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{3/2}} \, dx\) [375]

Optimal. Leaf size=192 \[ \frac {a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{f (c-c \sin (e+f x))^{3/2}}+\frac {12 a^4 \cos (e+f x) \log (1-\sin (e+f x))}{c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {6 a^3 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c f \sqrt {c-c \sin (e+f x)}}+\frac {3 a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.27, antiderivative size = 192, 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 a^4 \cos (e+f x) \log (1-\sin (e+f x))}{c f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {6 a^3 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{c f \sqrt {c-c \sin (e+f x)}}+\frac {3 a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 c f \sqrt {c-c \sin (e+f x)}}+\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{f (c-c \sin (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 1.02, size = 179, normalized size = 0.93 \begin {gather*} -\frac {a^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\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 c f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(489\) vs. \(2(174)=348\).
time = 17.42, size = 490, normalized size = 2.55

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 ) \sin \left (f x +e \right ) \cos \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 \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 )-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 (a \left (1+\sin \left (f x +e \right )\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 (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}}}\)	\(490\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^(7/2)/(c-c*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))*sin(f*x+e)*cos(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*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(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)*(a*(1+sin(f*x+e)))^(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)/(-c*(sin(f*x+e)-1))^(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((a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

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

[Out]

integral((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))*sqrt(a*sin(f*x + e) + a)*s
qrt(-c*sin(f*x + e) + c)/(c^2*cos(f*x + e)^2 + 2*c^2*sin(f*x + e) - 2*c^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((a+a*sin(f*x+e))**(7/2)/(c-c*sin(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [A]
time = 0.52, size = 179, normalized size = 0.93 \begin {gather*} -\frac {2 \, a^{\frac {7}{2}} \sqrt {c} {\left (\frac {6 \, \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {c^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 4 \, c^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{c^{4}} - \frac {2}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \mathrm {sgn}\left (\cos \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((a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

-2*a^(7/2)*sqrt(c)*(6*log(-cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 1)/(c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + (
c^2*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 4*c^2*cos(-1/4*pi + 1/2*f*x + 1/2*e
)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))/c^4 - 2/((cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1)*c^2*sgn(sin(-1/4*pi +
1/2*f*x + 1/2*e))))*sgn(cos(-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 (a+a\,\sin \left (e+f\,x\right )\right )}^{7/2}}{{\left (c-c\,\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((a + a*sin(e + f*x))^(7/2)/(c - c*sin(e + f*x))^(3/2),x)

[Out]

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

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