3.34 \(\int \frac{(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx\)
Optimal. Leaf size=75 \[ \frac{a^2 c^2 (A+B) \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^6}+\frac{a^2 c (A-6 B) \cos ^5(e+f x)}{35 f (c-c \sin (e+f x))^5} \]
[Out]
(a^2*(A + B)*c^2*Cos[e + f*x]^5)/(7*f*(c - c*Sin[e + f*x])^6) + (a^2*(A - 6*B)*c*Cos[e + f*x]^5)/(35*f*(c - c*
Sin[e + f*x])^5)
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Rubi [A] time = 0.228854, antiderivative size = 75, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 3, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used =
{2967, 2859, 2671} \[ \frac{a^2 c^2 (A+B) \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^6}+\frac{a^2 c (A-6 B) \cos ^5(e+f x)}{35 f (c-c \sin (e+f x))^5} \]
Antiderivative was successfully verified.
[In]
Int[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^4,x]
[Out]
(a^2*(A + B)*c^2*Cos[e + f*x]^5)/(7*f*(c - c*Sin[e + f*x])^6) + (a^2*(A - 6*B)*c*Cos[e + f*x]^5)/(35*f*(c - c*
Sin[e + f*x])^5)
Rule 2967
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] && !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
Rule 2859
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(2*m +
p + 1)), x] + Dist[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^
(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[
m + p], 0]) && NeQ[2*m + p + 1, 0]
Rule 2671
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*m), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && EqQ[Simplify[m + p + 1], 0] && !ILtQ[p, 0]
Rubi steps
\begin{align*} \int \frac{(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx &=\left (a^2 c^2\right ) \int \frac{\cos ^4(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^6} \, dx\\ &=\frac{a^2 (A+B) c^2 \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^6}+\frac{1}{7} \left (a^2 (A-6 B) c\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^5} \, dx\\ &=\frac{a^2 (A+B) c^2 \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^6}+\frac{a^2 (A-6 B) c \cos ^5(e+f x)}{35 f (c-c \sin (e+f x))^5}\\ \end{align*}
Mathematica [B] time = 0.912054, size = 191, normalized size = 2.55 \[ -\frac{a^2 \left (-35 (A+4 B) \cos \left (\frac{1}{2} (e+f x)\right )+7 (2 A+13 B) \cos \left (\frac{3}{2} (e+f x)\right )-70 A \sin \left (\frac{1}{2} (e+f x)\right )-35 A \sin \left (\frac{3}{2} (e+f x)\right )+7 A \sin \left (\frac{5}{2} (e+f x)\right )+A \cos \left (\frac{7}{2} (e+f x)\right )+70 B \sin \left (\frac{1}{2} (e+f x)\right )+35 B \sin \left (\frac{3}{2} (e+f x)\right )-7 B \sin \left (\frac{5}{2} (e+f x)\right )+35 B \cos \left (\frac{5}{2} (e+f x)\right )-6 B \cos \left (\frac{7}{2} (e+f x)\right )\right )}{140 c^4 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^7} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^4,x]
[Out]
-(a^2*(-35*(A + 4*B)*Cos[(e + f*x)/2] + 7*(2*A + 13*B)*Cos[(3*(e + f*x))/2] + 35*B*Cos[(5*(e + f*x))/2] + A*Co
s[(7*(e + f*x))/2] - 6*B*Cos[(7*(e + f*x))/2] - 70*A*Sin[(e + f*x)/2] + 70*B*Sin[(e + f*x)/2] - 35*A*Sin[(3*(e
+ f*x))/2] + 35*B*Sin[(3*(e + f*x))/2] + 7*A*Sin[(5*(e + f*x))/2] - 7*B*Sin[(5*(e + f*x))/2]))/(140*c^4*f*(Co
s[(e + f*x)/2] - Sin[(e + f*x)/2])^7)
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Maple [B] time = 0.131, size = 161, normalized size = 2.2 \begin{align*} 2\,{\frac{{a}^{2}}{f{c}^{4}} \left ( -1/6\,{\frac{96\,A+96\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{6}}}-1/3\,{\frac{42\,A+18\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{3}}}-1/4\,{\frac{96\,A+64\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{4}}}-1/7\,{\frac{32\,A+32\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{7}}}-1/2\,{\frac{10\,A+2\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}-{\frac{A}{\tan \left ( 1/2\,fx+e/2 \right ) -1}}-1/5\,{\frac{128\,A+112\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{5}}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^4,x)
[Out]
2/f*a^2/c^4*(-1/6*(96*A+96*B)/(tan(1/2*f*x+1/2*e)-1)^6-1/3*(42*A+18*B)/(tan(1/2*f*x+1/2*e)-1)^3-1/4*(96*A+64*B
)/(tan(1/2*f*x+1/2*e)-1)^4-1/7*(32*A+32*B)/(tan(1/2*f*x+1/2*e)-1)^7-1/2*(10*A+2*B)/(tan(1/2*f*x+1/2*e)-1)^2-A/
(tan(1/2*f*x+1/2*e)-1)-1/5*(128*A+112*B)/(tan(1/2*f*x+1/2*e)-1)^5)
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Maxima [B] time = 1.1733, size = 2121, normalized size = 28.28 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorithm="maxima")
[Out]
2/105*(2*A*a^2*(91*sin(f*x + e)/(cos(f*x + e) + 1) - 168*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 280*sin(f*x + e
)^3/(cos(f*x + e) + 1)^3 - 175*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 105*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 -
13)/(c^4 - 7*c^4*sin(f*x + e)/(cos(f*x + e) + 1) + 21*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 35*c^4*sin(f*
x + e)^3/(cos(f*x + e) + 1)^3 + 35*c^4*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 21*c^4*sin(f*x + e)^5/(cos(f*x +
e) + 1)^5 + 7*c^4*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7) + B*a^2*(91*s
in(f*x + e)/(cos(f*x + e) + 1) - 168*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 280*sin(f*x + e)^3/(cos(f*x + e) +
1)^3 - 175*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 105*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 13)/(c^4 - 7*c^4*si
n(f*x + e)/(cos(f*x + e) + 1) + 21*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 35*c^4*sin(f*x + e)^3/(cos(f*x +
e) + 1)^3 + 35*c^4*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 21*c^4*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 7*c^4*si
n(f*x + e)^6/(cos(f*x + e) + 1)^6 - c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7) - 3*A*a^2*(49*sin(f*x + e)/(cos(f
*x + e) + 1) - 147*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 210*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 210*sin(f*x
+ e)^4/(cos(f*x + e) + 1)^4 + 105*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 35*sin(f*x + e)^6/(cos(f*x + e) + 1)^
6 - 12)/(c^4 - 7*c^4*sin(f*x + e)/(cos(f*x + e) + 1) + 21*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 35*c^4*sin
(f*x + e)^3/(cos(f*x + e) + 1)^3 + 35*c^4*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 21*c^4*sin(f*x + e)^5/(cos(f*x
+ e) + 1)^5 + 7*c^4*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7) - 4*A*a^2*
(14*sin(f*x + e)/(cos(f*x + e) + 1) - 42*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 35*sin(f*x + e)^3/(cos(f*x + e)
+ 1)^3 - 35*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 2)/(c^4 - 7*c^4*sin(f*x + e)/(cos(f*x + e) + 1) + 21*c^4*si
n(f*x + e)^2/(cos(f*x + e) + 1)^2 - 35*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 35*c^4*sin(f*x + e)^4/(cos(f*
x + e) + 1)^4 - 21*c^4*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 7*c^4*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - c^4*s
in(f*x + e)^7/(cos(f*x + e) + 1)^7) - 8*B*a^2*(14*sin(f*x + e)/(cos(f*x + e) + 1) - 42*sin(f*x + e)^2/(cos(f*x
+ e) + 1)^2 + 35*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 35*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 2)/(c^4 - 7*c
^4*sin(f*x + e)/(cos(f*x + e) + 1) + 21*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 35*c^4*sin(f*x + e)^3/(cos(f
*x + e) + 1)^3 + 35*c^4*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 21*c^4*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 7*c
^4*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7) + 6*B*a^2*(7*sin(f*x + e)/(c
os(f*x + e) + 1) - 21*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 35*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 1)/(c^4 -
7*c^4*sin(f*x + e)/(cos(f*x + e) + 1) + 21*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 35*c^4*sin(f*x + e)^3/(c
os(f*x + e) + 1)^3 + 35*c^4*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 21*c^4*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 +
7*c^4*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7))/f
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Fricas [B] time = 1.41181, size = 652, normalized size = 8.69 \begin{align*} -\frac{{\left (A - 6 \, B\right )} a^{2} \cos \left (f x + e\right )^{4} +{\left (4 \, A + 11 \, B\right )} a^{2} \cos \left (f x + e\right )^{3} +{\left (13 \, A + 27 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 10 \,{\left (A + B\right )} a^{2} \cos \left (f x + e\right ) - 20 \,{\left (A + B\right )} a^{2} -{\left ({\left (A - 6 \, B\right )} a^{2} \cos \left (f x + e\right )^{3} -{\left (3 \, A + 17 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} + 10 \,{\left (A + B\right )} a^{2} \cos \left (f x + e\right ) + 20 \,{\left (A + B\right )} a^{2}\right )} \sin \left (f x + e\right )}{35 \,{\left (c^{4} f \cos \left (f x + e\right )^{4} - 3 \, c^{4} f \cos \left (f x + e\right )^{3} - 8 \, c^{4} f \cos \left (f x + e\right )^{2} + 4 \, c^{4} f \cos \left (f x + e\right ) + 8 \, c^{4} f +{\left (c^{4} f \cos \left (f x + e\right )^{3} + 4 \, c^{4} f \cos \left (f x + e\right )^{2} - 4 \, c^{4} f \cos \left (f x + e\right ) - 8 \, c^{4} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorithm="fricas")
[Out]
-1/35*((A - 6*B)*a^2*cos(f*x + e)^4 + (4*A + 11*B)*a^2*cos(f*x + e)^3 + (13*A + 27*B)*a^2*cos(f*x + e)^2 - 10*
(A + B)*a^2*cos(f*x + e) - 20*(A + B)*a^2 - ((A - 6*B)*a^2*cos(f*x + e)^3 - (3*A + 17*B)*a^2*cos(f*x + e)^2 +
10*(A + B)*a^2*cos(f*x + e) + 20*(A + B)*a^2)*sin(f*x + e))/(c^4*f*cos(f*x + e)^4 - 3*c^4*f*cos(f*x + e)^3 - 8
*c^4*f*cos(f*x + e)^2 + 4*c^4*f*cos(f*x + e) + 8*c^4*f + (c^4*f*cos(f*x + e)^3 + 4*c^4*f*cos(f*x + e)^2 - 4*c^
4*f*cos(f*x + e) - 8*c^4*f)*sin(f*x + e))
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))**2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**4,x)
[Out]
Timed out
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Giac [B] time = 1.24011, size = 309, normalized size = 4.12 \begin{align*} -\frac{2 \,{\left (35 \, A a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 35 \, A a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 35 \, B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 140 \, A a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 35 \, B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 70 \, A a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 70 \, B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 91 \, A a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 14 \, B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 7 \, A a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 7 \, B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 6 \, A a^{2} - B a^{2}\right )}}{35 \, c^{4} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{7}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorithm="giac")
[Out]
-2/35*(35*A*a^2*tan(1/2*f*x + 1/2*e)^6 - 35*A*a^2*tan(1/2*f*x + 1/2*e)^5 + 35*B*a^2*tan(1/2*f*x + 1/2*e)^5 + 1
40*A*a^2*tan(1/2*f*x + 1/2*e)^4 + 35*B*a^2*tan(1/2*f*x + 1/2*e)^4 - 70*A*a^2*tan(1/2*f*x + 1/2*e)^3 + 70*B*a^2
*tan(1/2*f*x + 1/2*e)^3 + 91*A*a^2*tan(1/2*f*x + 1/2*e)^2 + 14*B*a^2*tan(1/2*f*x + 1/2*e)^2 - 7*A*a^2*tan(1/2*
f*x + 1/2*e) + 7*B*a^2*tan(1/2*f*x + 1/2*e) + 6*A*a^2 - B*a^2)/(c^4*f*(tan(1/2*f*x + 1/2*e) - 1)^7)