3.4.67 \(\int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx\) [367]

Optimal. Leaf size=133 \[ \frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{40 c f (c-c \sin (e+f x))^{9/2}}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{240 c^2 f (c-c \sin (e+f x))^{7/2}} \]

[Out]

1/10*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/f/(c-c*sin(f*x+e))^(11/2)+1/40*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/c/f/(c
-c*sin(f*x+e))^(9/2)+1/240*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/c^2/f/(c-c*sin(f*x+e))^(7/2)

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Rubi [A]
time = 0.20, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2822, 2821} \begin {gather*} \frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{240 c^2 f (c-c \sin (e+f x))^{7/2}}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{40 c f (c-c \sin (e+f x))^{9/2}}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(5/2)/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

(Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(10*f*(c - c*Sin[e + f*x])^(11/2)) + (Cos[e + f*x]*(a + a*Sin[e + f*
x])^(5/2))/(40*c*f*(c - c*Sin[e + f*x])^(9/2)) + (Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(240*c^2*f*(c - c*S
in[e + f*x])^(7/2))

Rule 2821

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] /; FreeQ[{a, b, c, d, e, f
, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] && NeQ[m, -2^(-1)]

Rule 2822

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])

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac {\int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{9/2}} \, dx}{5 c}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{40 c f (c-c \sin (e+f x))^{9/2}}+\frac {\int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{7/2}} \, dx}{40 c^2}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{40 c f (c-c \sin (e+f x))^{9/2}}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{240 c^2 f (c-c \sin (e+f x))^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 2.12, size = 118, normalized size = 0.89 \begin {gather*} -\frac {a^2 \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))} (9-5 \cos (2 (e+f x))+10 \sin (e+f x))}{30 c^5 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))^5 \sqrt {c-c \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^(5/2)/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

-1/30*(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(9 - 5*Cos[2*(e + f*x)] + 10*Sin[e
+ f*x]))/(c^5*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^5*Sqrt[c - c*Sin[e + f*x]])

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Maple [A]
time = 17.88, size = 225, normalized size = 1.69

method	result	size
default	\(\frac {\sin \left (f x +e \right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \left (2 \left (\cos ^{5}\left (f x +e \right )\right )+2 \sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right )-12 \left (\cos ^{4}\left (f x +e \right )\right )+10 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )-24 \left (\cos ^{3}\left (f x +e \right )\right )-34 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )+59 \left (\cos ^{2}\left (f x +e \right )\right )-25 \cos \left (f x +e \right ) \sin \left (f x +e \right )+37 \cos \left (f x +e \right )+62 \sin \left (f x +e \right )-62\right )}{15 f \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {11}{2}} \left (\cos ^{3}\left (f x +e \right )-\sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )-3 \left (\cos ^{2}\left (f x +e \right )\right )-2 \cos \left (f x +e \right ) \sin \left (f x +e \right )-2 \cos \left (f x +e \right )+4 \sin \left (f x +e \right )+4\right )}\)	\(225\)

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))^(11/2),x,method=_RETURNVERBOSE)

[Out]

1/15/f*sin(f*x+e)*(a*(1+sin(f*x+e)))^(5/2)*(2*cos(f*x+e)^5+2*sin(f*x+e)*cos(f*x+e)^4-12*cos(f*x+e)^4+10*cos(f*
x+e)^3*sin(f*x+e)-24*cos(f*x+e)^3-34*sin(f*x+e)*cos(f*x+e)^2+59*cos(f*x+e)^2-25*cos(f*x+e)*sin(f*x+e)+37*cos(f
*x+e)+62*sin(f*x+e)-62)/(-c*(sin(f*x+e)-1))^(11/2)/(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)

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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))^(5/2)/(c-c*sin(f*x+e))^(11/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(5/2)/(-c*sin(f*x + e) + c)^(11/2), x)

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Fricas [A]
time = 0.35, size = 159, normalized size = 1.20 \begin {gather*} -\frac {{\left (5 \, a^{2} \cos \left (f x + e\right )^{2} - 5 \, a^{2} \sin \left (f x + e\right ) - 7 \, a^{2}\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{15 \, {\left (5 \, c^{6} f \cos \left (f x + e\right )^{5} - 20 \, c^{6} f \cos \left (f x + e\right )^{3} + 16 \, c^{6} f \cos \left (f x + e\right ) - {\left (c^{6} f \cos \left (f x + e\right )^{5} - 12 \, c^{6} f \cos \left (f x + e\right )^{3} + 16 \, c^{6} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \end {gather*}

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))^(11/2),x, algorithm="fricas")

[Out]

-1/15*(5*a^2*cos(f*x + e)^2 - 5*a^2*sin(f*x + e) - 7*a^2)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(
5*c^6*f*cos(f*x + e)^5 - 20*c^6*f*cos(f*x + e)^3 + 16*c^6*f*cos(f*x + e) - (c^6*f*cos(f*x + e)^5 - 12*c^6*f*co
s(f*x + e)^3 + 16*c^6*f*cos(f*x + e))*sin(f*x + e))

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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))**(5/2)/(c-c*sin(f*x+e))**(11/2),x)

[Out]

Timed out

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Giac [A]
time = 0.49, size = 139, normalized size = 1.05 \begin {gather*} -\frac {{\left (10 \, a^{2} \sqrt {c} \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} - 15 \, a^{2} \sqrt {c} \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} + 6 \, a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a}}{240 \, c^{6} f \mathrm {sgn}\left (\sin \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 )^{10}} \end {gather*}

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))^(11/2),x, algorithm="giac")

[Out]

-1/240*(10*a^2*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^4 - 15*a^2*sqrt(c)*s
gn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 6*a^2*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x
+ 1/2*e)))*sqrt(a)/(c^6*f*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^10)

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Mupad [B]
time = 12.08, size = 273, normalized size = 2.05 \begin {gather*} \frac {\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,96{}\mathrm {i}}{5\,c^6\,f}+\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,64{}\mathrm {i}}{3\,c^6\,f}-\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,32{}\mathrm {i}}{3\,c^6\,f}\right )}{\cos \left (e+f\,x\right )\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,264{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (3\,e+3\,f\,x\right )\,220{}\mathrm {i}+{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (5\,e+5\,f\,x\right )\,20{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (2\,e+2\,f\,x\right )\,330{}\mathrm {i}+{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (4\,e+4\,f\,x\right )\,88{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (6\,e+6\,f\,x\right )\,2{}\mathrm {i}} \end {gather*}

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))^(11/2),x)

[Out]

((c - c*sin(e + f*x))^(1/2)*((a^2*exp(e*6i + f*x*6i)*(a + a*sin(e + f*x))^(1/2)*96i)/(5*c^6*f) + (a^2*exp(e*6i
+ f*x*6i)*sin(e + f*x)*(a + a*sin(e + f*x))^(1/2)*64i)/(3*c^6*f) - (a^2*exp(e*6i + f*x*6i)*cos(2*e + 2*f*x)*(
a + a*sin(e + f*x))^(1/2)*32i)/(3*c^6*f)))/(cos(e + f*x)*exp(e*6i + f*x*6i)*264i - exp(e*6i + f*x*6i)*cos(3*e
+ 3*f*x)*220i + exp(e*6i + f*x*6i)*cos(5*e + 5*f*x)*20i - exp(e*6i + f*x*6i)*sin(2*e + 2*f*x)*330i + exp(e*6i
+ f*x*6i)*sin(4*e + 4*f*x)*88i - exp(e*6i + f*x*6i)*sin(6*e + 6*f*x)*2i)

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