3.47 \(\int \frac{(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx\)
Optimal. Leaf size=151 \[ \frac{a^3 c^3 (A+B) \cos ^7(e+f x)}{7 f (c-c \sin (e+f x))^7}+\frac{2 a^3 B c^2 \cos ^3(e+f x)}{3 f \left (c^2-c^2 \sin (e+f x)\right )^3}-\frac{2 a^3 B \cos (e+f x)}{f \left (c^4-c^4 \sin (e+f x)\right )}+\frac{a^3 B x}{c^4}-\frac{2 a^3 B c \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5} \]
[Out]
(a^3*B*x)/c^4 + (a^3*(A + B)*c^3*Cos[e + f*x]^7)/(7*f*(c - c*Sin[e + f*x])^7) - (2*a^3*B*c*Cos[e + f*x]^5)/(5*
f*(c - c*Sin[e + f*x])^5) + (2*a^3*B*c^2*Cos[e + f*x]^3)/(3*f*(c^2 - c^2*Sin[e + f*x])^3) - (2*a^3*B*Cos[e + f
*x])/(f*(c^4 - c^4*Sin[e + f*x]))
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Rubi [A] time = 0.329975, antiderivative size = 151, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 4, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used =
{2967, 2859, 2680, 8} \[ \frac{a^3 c^3 (A+B) \cos ^7(e+f x)}{7 f (c-c \sin (e+f x))^7}+\frac{2 a^3 B c^2 \cos ^3(e+f x)}{3 f \left (c^2-c^2 \sin (e+f x)\right )^3}-\frac{2 a^3 B \cos (e+f x)}{f \left (c^4-c^4 \sin (e+f x)\right )}+\frac{a^3 B x}{c^4}-\frac{2 a^3 B c \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5} \]
Antiderivative was successfully verified.
[In]
Int[((a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^4,x]
[Out]
(a^3*B*x)/c^4 + (a^3*(A + B)*c^3*Cos[e + f*x]^7)/(7*f*(c - c*Sin[e + f*x])^7) - (2*a^3*B*c*Cos[e + f*x]^5)/(5*
f*(c - c*Sin[e + f*x])^5) + (2*a^3*B*c^2*Cos[e + f*x]^3)/(3*f*(c^2 - c^2*Sin[e + f*x])^3) - (2*a^3*B*Cos[e + f
*x])/(f*(c^4 - c^4*Sin[e + f*x]))
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 2680
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] && !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]
Rule 8
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
Rubi steps
\begin{align*} \int \frac{(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx &=\left (a^3 c^3\right ) \int \frac{\cos ^6(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^7} \, dx\\ &=\frac{a^3 (A+B) c^3 \cos ^7(e+f x)}{7 f (c-c \sin (e+f x))^7}-\left (a^3 B c^2\right ) \int \frac{\cos ^6(e+f x)}{(c-c \sin (e+f x))^6} \, dx\\ &=\frac{a^3 (A+B) c^3 \cos ^7(e+f x)}{7 f (c-c \sin (e+f x))^7}-\frac{2 a^3 B c \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}+\left (a^3 B\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^4} \, dx\\ &=\frac{a^3 (A+B) c^3 \cos ^7(e+f x)}{7 f (c-c \sin (e+f x))^7}-\frac{2 a^3 B c \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}+\frac{2 a^3 B \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^3}-\frac{\left (a^3 B\right ) \int \frac{\cos ^2(e+f x)}{(c-c \sin (e+f x))^2} \, dx}{c^2}\\ &=\frac{a^3 (A+B) c^3 \cos ^7(e+f x)}{7 f (c-c \sin (e+f x))^7}-\frac{2 a^3 B c \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}+\frac{2 a^3 B \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^3}-\frac{2 a^3 B \cos (e+f x)}{f \left (c^4-c^4 \sin (e+f x)\right )}+\frac{\left (a^3 B\right ) \int 1 \, dx}{c^4}\\ &=\frac{a^3 B x}{c^4}+\frac{a^3 (A+B) c^3 \cos ^7(e+f x)}{7 f (c-c \sin (e+f x))^7}-\frac{2 a^3 B c \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}+\frac{2 a^3 B \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^3}-\frac{2 a^3 B \cos (e+f x)}{f \left (c^4-c^4 \sin (e+f x)\right )}\\ \end{align*}
Mathematica [B] time = 1.15233, size = 356, normalized size = 2.36 \[ \frac{a^3 (\sin (e+f x)+1)^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (240 (A+B) \sin \left (\frac{1}{2} (e+f x)\right )-2 (15 A+337 B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^6+2 (45 A+199 B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5+4 (45 A+199 B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4-12 (15 A+29 B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3-24 (15 A+29 B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2+120 (A+B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+105 B (e+f x) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^7\right )}{105 f (c-c \sin (e+f x))^4 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^6} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^4,x]
[Out]
(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(120*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - 12*(15*A + 29*
B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 + 2*(45*A + 199*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5 + 105*B*
(e + f*x)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7 + 240*(A + B)*Sin[(e + f*x)/2] - 24*(15*A + 29*B)*(Cos[(e +
f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] + 4*(45*A + 199*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sin[
(e + f*x)/2] - 2*(15*A + 337*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^6*Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^3
)/(105*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*(c - c*Sin[e + f*x])^4)
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Maple [B] time = 0.142, size = 374, normalized size = 2.5 \begin{align*} -2\,{\frac{A{a}^{3}}{f{c}^{4} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }}+2\,{\frac{B{a}^{3}}{f{c}^{4} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }}-12\,{\frac{A{a}^{3}}{f{c}^{4} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}-4\,{\frac{B{a}^{3}}{f{c}^{4} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}-40\,{\frac{A{a}^{3}}{f{c}^{4} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{3}}}-{\frac{40\,B{a}^{3}}{3\,f{c}^{4}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-3}}-{\frac{128\,A{a}^{3}}{7\,f{c}^{4}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-7}}-{\frac{128\,B{a}^{3}}{7\,f{c}^{4}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-7}}-80\,{\frac{A{a}^{3}}{f{c}^{4} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{4}}}-48\,{\frac{B{a}^{3}}{f{c}^{4} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{4}}}-64\,{\frac{A{a}^{3}}{f{c}^{4} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{6}}}-64\,{\frac{B{a}^{3}}{f{c}^{4} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{6}}}-96\,{\frac{A{a}^{3}}{f{c}^{4} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{5}}}-{\frac{416\,B{a}^{3}}{5\,f{c}^{4}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-5}}+2\,{\frac{B{a}^{3}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{c}^{4}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^4,x)
[Out]
-2/f*a^3/c^4/(tan(1/2*f*x+1/2*e)-1)*A+2/f*a^3/c^4/(tan(1/2*f*x+1/2*e)-1)*B-12/f*a^3/c^4/(tan(1/2*f*x+1/2*e)-1)
^2*A-4/f*a^3/c^4/(tan(1/2*f*x+1/2*e)-1)^2*B-40/f*a^3/c^4/(tan(1/2*f*x+1/2*e)-1)^3*A-40/3/f*a^3/c^4/(tan(1/2*f*
x+1/2*e)-1)^3*B-128/7/f*a^3/c^4/(tan(1/2*f*x+1/2*e)-1)^7*A-128/7/f*a^3/c^4/(tan(1/2*f*x+1/2*e)-1)^7*B-80/f*a^3
/c^4/(tan(1/2*f*x+1/2*e)-1)^4*A-48/f*a^3/c^4/(tan(1/2*f*x+1/2*e)-1)^4*B-64/f*a^3/c^4/(tan(1/2*f*x+1/2*e)-1)^6*
A-64/f*a^3/c^4/(tan(1/2*f*x+1/2*e)-1)^6*B-96/f*a^3/c^4/(tan(1/2*f*x+1/2*e)-1)^5*A-416/5/f*a^3/c^4/(tan(1/2*f*x
+1/2*e)-1)^5*B+2/f*a^3/c^4*B*arctan(tan(1/2*f*x+1/2*e))
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Maxima [B] time = 1.71883, size = 2859, normalized size = 18.93 \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))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorithm="maxima")
[Out]
2/105*(5*B*a^3*((203*sin(f*x + e)/(cos(f*x + e) + 1) - 525*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 686*sin(f*x +
e)^3/(cos(f*x + e) + 1)^3 - 434*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 147*sin(f*x + e)^5/(cos(f*x + e) + 1)^5
- 21*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 32)/(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) + 21*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^4) + 3*A*a^3*(91*sin(f*x + e)/(co
s(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/(c
os(f*x + e) + 1)^6 - c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7) + B*a^3*(91*sin(f*x + e)/(cos(f*x + e) + 1) - 16
8*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) - 3*A*a^3*(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) - 12*A*a^3*(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
) - 12*B*a^3*(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*A*a^3*(7*sin(f*x + e)/(cos(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/(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) + 18*B*a^3*(7*sin(f*x + e)/(cos(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/(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))/f
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Fricas [B] time = 1.56404, size = 891, normalized size = 5.9 \begin{align*} \frac{840 \, B a^{3} f x +{\left (105 \, B a^{3} f x +{\left (15 \, A + 337 \, B\right )} a^{3}\right )} \cos \left (f x + e\right )^{4} + 120 \,{\left (A + B\right )} a^{3} -{\left (315 \, B a^{3} f x +{\left (45 \, A - 613 \, B\right )} a^{3}\right )} \cos \left (f x + e\right )^{3} - 24 \,{\left (35 \, B a^{3} f x +{\left (5 \, A + 26 \, B\right )} a^{3}\right )} \cos \left (f x + e\right )^{2} + 60 \,{\left (7 \, B a^{3} f x +{\left (A - 13 \, B\right )} a^{3}\right )} \cos \left (f x + e\right ) -{\left (840 \, B a^{3} f x - 120 \,{\left (A + B\right )} a^{3} -{\left (105 \, B a^{3} f x -{\left (15 \, A + 337 \, B\right )} a^{3}\right )} \cos \left (f x + e\right )^{3} - 12 \,{\left (35 \, B a^{3} f x -{\left (5 \, A - 23 \, B\right )} a^{3}\right )} \cos \left (f x + e\right )^{2} + 60 \,{\left (7 \, B a^{3} f x -{\left (A + 15 \, B\right )} a^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{105 \,{\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))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorithm="fricas")
[Out]
1/105*(840*B*a^3*f*x + (105*B*a^3*f*x + (15*A + 337*B)*a^3)*cos(f*x + e)^4 + 120*(A + B)*a^3 - (315*B*a^3*f*x
+ (45*A - 613*B)*a^3)*cos(f*x + e)^3 - 24*(35*B*a^3*f*x + (5*A + 26*B)*a^3)*cos(f*x + e)^2 + 60*(7*B*a^3*f*x +
(A - 13*B)*a^3)*cos(f*x + e) - (840*B*a^3*f*x - 120*(A + B)*a^3 - (105*B*a^3*f*x - (15*A + 337*B)*a^3)*cos(f*
x + e)^3 - 12*(35*B*a^3*f*x - (5*A - 23*B)*a^3)*cos(f*x + e)^2 + 60*(7*B*a^3*f*x - (A + 15*B)*a^3)*cos(f*x + e
))*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))**3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**4,x)
[Out]
Timed out
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Giac [A] time = 1.20714, size = 288, normalized size = 1.91 \begin{align*} \frac{\frac{105 \,{\left (f x + e\right )} B a^{3}}{c^{4}} - \frac{2 \,{\left (105 \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 105 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} + 840 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 525 \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1925 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 3920 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 315 \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 2667 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1064 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 15 \, A a^{3} - 167 \, B a^{3}\right )}}{c^{4}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{7}}}{105 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorithm="giac")
[Out]
1/105*(105*(f*x + e)*B*a^3/c^4 - 2*(105*A*a^3*tan(1/2*f*x + 1/2*e)^6 - 105*B*a^3*tan(1/2*f*x + 1/2*e)^6 + 840*
B*a^3*tan(1/2*f*x + 1/2*e)^5 + 525*A*a^3*tan(1/2*f*x + 1/2*e)^4 - 1925*B*a^3*tan(1/2*f*x + 1/2*e)^4 + 3920*B*a
^3*tan(1/2*f*x + 1/2*e)^3 + 315*A*a^3*tan(1/2*f*x + 1/2*e)^2 - 2667*B*a^3*tan(1/2*f*x + 1/2*e)^2 + 1064*B*a^3*
tan(1/2*f*x + 1/2*e) + 15*A*a^3 - 167*B*a^3)/(c^4*(tan(1/2*f*x + 1/2*e) - 1)^7))/f