3.11.22 \(\int \cos ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx\) [1022]
Optimal. Leaf size=93 \[ \frac {2 (A-B) (a+a \sin (e+f x))^{2+m}}{a^2 f (2+m)}-\frac {(A-3 B) (a+a \sin (e+f x))^{3+m}}{a^3 f (3+m)}-\frac {B (a+a \sin (e+f x))^{4+m}}{a^4 f (4+m)} \]
[Out]
2*(A-B)*(a+a*sin(f*x+e))^(2+m)/a^2/f/(2+m)-(A-3*B)*(a+a*sin(f*x+e))^(3+m)/a^3/f/(3+m)-B*(a+a*sin(f*x+e))^(4+m)
/a^4/f/(4+m)
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Rubi [A]
time = 0.08, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {2915, 78}
\begin {gather*} -\frac {B (a \sin (e+f x)+a)^{m+4}}{a^4 f (m+4)}-\frac {(A-3 B) (a \sin (e+f x)+a)^{m+3}}{a^3 f (m+3)}+\frac {2 (A-B) (a \sin (e+f x)+a)^{m+2}}{a^2 f (m+2)} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Cos[e + f*x]^3*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]
[Out]
(2*(A - B)*(a + a*Sin[e + f*x])^(2 + m))/(a^2*f*(2 + m)) - ((A - 3*B)*(a + a*Sin[e + f*x])^(3 + m))/(a^3*f*(3
+ m)) - (B*(a + a*Sin[e + f*x])^(4 + m))/(a^4*f*(4 + m))
Rule 78
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Rule 2915
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]
Rubi steps
\begin {align*} \int \cos ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx &=\frac {\text {Subst}\left (\int (a-x) (a+x)^{1+m} \left (A+\frac {B x}{a}\right ) \, dx,x,a \sin (e+f x)\right )}{a^3 f}\\ &=\frac {\text {Subst}\left (\int \left (2 a (A-B) (a+x)^{1+m}+(-A+3 B) (a+x)^{2+m}-\frac {B (a+x)^{3+m}}{a}\right ) \, dx,x,a \sin (e+f x)\right )}{a^3 f}\\ &=\frac {2 (A-B) (a+a \sin (e+f x))^{2+m}}{a^2 f (2+m)}-\frac {(A-3 B) (a+a \sin (e+f x))^{3+m}}{a^3 f (3+m)}-\frac {B (a+a \sin (e+f x))^{4+m}}{a^4 f (4+m)}\\ \end {align*}
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Mathematica [A]
time = 0.22, size = 93, normalized size = 1.00 \begin {gather*} \frac {2 (A-B) (a+a \sin (e+f x))^{2+m}}{a^2 f (2+m)}-\frac {(A-3 B) (a+a \sin (e+f x))^{3+m}}{a^3 f (3+m)}-\frac {B (a+a \sin (e+f x))^{4+m}}{a^4 f (4+m)} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Cos[e + f*x]^3*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]
[Out]
(2*(A - B)*(a + a*Sin[e + f*x])^(2 + m))/(a^2*f*(2 + m)) - ((A - 3*B)*(a + a*Sin[e + f*x])^(3 + m))/(a^3*f*(3
+ m)) - (B*(a + a*Sin[e + f*x])^(4 + m))/(a^4*f*(4 + m))
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Maple [F]
time = 0.33, size = 0, normalized size = 0.00 \[\int \left (\cos ^{3}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(f*x+e)^3*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)
[Out]
int(cos(f*x+e)^3*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 298 vs.
\(2 (96) = 192\).
time = 0.31, size = 298, normalized size = 3.20 \begin {gather*} -\frac {\frac {{\left ({\left (m^{2} + 3 \, m + 2\right )} a^{m} \sin \left (f x + e\right )^{3} + {\left (m^{2} + m\right )} a^{m} \sin \left (f x + e\right )^{2} - 2 \, a^{m} m \sin \left (f x + e\right ) + 2 \, a^{m}\right )} A {\left (\sin \left (f x + e\right ) + 1\right )}^{m}}{m^{3} + 6 \, m^{2} + 11 \, m + 6} + \frac {{\left ({\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )} a^{m} \sin \left (f x + e\right )^{4} + {\left (m^{3} + 3 \, m^{2} + 2 \, m\right )} a^{m} \sin \left (f x + e\right )^{3} - 3 \, {\left (m^{2} + m\right )} a^{m} \sin \left (f x + e\right )^{2} + 6 \, a^{m} m \sin \left (f x + e\right ) - 6 \, a^{m}\right )} B {\left (\sin \left (f x + e\right ) + 1\right )}^{m}}{m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24} - \frac {{\left (a^{m} {\left (m + 1\right )} \sin \left (f x + e\right )^{2} + a^{m} m \sin \left (f x + e\right ) - a^{m}\right )} B {\left (\sin \left (f x + e\right ) + 1\right )}^{m}}{m^{2} + 3 \, m + 2} - \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m + 1} A}{a {\left (m + 1\right )}}}{f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(f*x+e)^3*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="maxima")
[Out]
-(((m^2 + 3*m + 2)*a^m*sin(f*x + e)^3 + (m^2 + m)*a^m*sin(f*x + e)^2 - 2*a^m*m*sin(f*x + e) + 2*a^m)*A*(sin(f*
x + e) + 1)^m/(m^3 + 6*m^2 + 11*m + 6) + ((m^3 + 6*m^2 + 11*m + 6)*a^m*sin(f*x + e)^4 + (m^3 + 3*m^2 + 2*m)*a^
m*sin(f*x + e)^3 - 3*(m^2 + m)*a^m*sin(f*x + e)^2 + 6*a^m*m*sin(f*x + e) - 6*a^m)*B*(sin(f*x + e) + 1)^m/(m^4
+ 10*m^3 + 35*m^2 + 50*m + 24) - (a^m*(m + 1)*sin(f*x + e)^2 + a^m*m*sin(f*x + e) - a^m)*B*(sin(f*x + e) + 1)^
m/(m^2 + 3*m + 2) - (a*sin(f*x + e) + a)^(m + 1)*A/(a*(m + 1)))/f
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Fricas [A]
time = 0.40, size = 140, normalized size = 1.51 \begin {gather*} -\frac {{\left ({\left (B m^{2} + 5 \, B m + 6 \, B\right )} \cos \left (f x + e\right )^{4} - {\left ({\left (A + B\right )} m^{2} + 4 \, A m\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left (A + B\right )} m - {\left ({\left ({\left (A + B\right )} m^{2} + 2 \, {\left (3 \, A + B\right )} m + 8 \, A\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left (A + B\right )} m + 16 \, A\right )} \sin \left (f x + e\right ) - 16 \, A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{f m^{3} + 9 \, f m^{2} + 26 \, f m + 24 \, f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(f*x+e)^3*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="fricas")
[Out]
-((B*m^2 + 5*B*m + 6*B)*cos(f*x + e)^4 - ((A + B)*m^2 + 4*A*m)*cos(f*x + e)^2 - 4*(A + B)*m - (((A + B)*m^2 +
2*(3*A + B)*m + 8*A)*cos(f*x + e)^2 + 4*(A + B)*m + 16*A)*sin(f*x + e) - 16*A)*(a*sin(f*x + e) + a)^m/(f*m^3 +
9*f*m^2 + 26*f*m + 24*f)
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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 5243 vs.
\(2 (78) = 156\).
time = 17.09, size = 5243, normalized size = 56.38 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(f*x+e)**3*(a+a*sin(f*x+e))**m*(A+B*sin(f*x+e)),x)
[Out]
Piecewise((x*(A + B*sin(e))*(a*sin(e) + a)**m*cos(e)**3, Eq(f, 0)), (4*A*sin(e + f*x)**2/(6*a**4*f*sin(e + f*x
)**3 + 18*a**4*f*sin(e + f*x)**2 + 18*a**4*f*sin(e + f*x) + 6*a**4*f) + 6*A*sin(e + f*x)/(6*a**4*f*sin(e + f*x
)**3 + 18*a**4*f*sin(e + f*x)**2 + 18*a**4*f*sin(e + f*x) + 6*a**4*f) - 2*A*cos(e + f*x)**2/(6*a**4*f*sin(e +
f*x)**3 + 18*a**4*f*sin(e + f*x)**2 + 18*a**4*f*sin(e + f*x) + 6*a**4*f) + 2*A/(6*a**4*f*sin(e + f*x)**3 + 18*
a**4*f*sin(e + f*x)**2 + 18*a**4*f*sin(e + f*x) + 6*a**4*f) - 6*B*log(sin(e + f*x) + 1)*sin(e + f*x)**3/(6*a**
4*f*sin(e + f*x)**3 + 18*a**4*f*sin(e + f*x)**2 + 18*a**4*f*sin(e + f*x) + 6*a**4*f) - 18*B*log(sin(e + f*x) +
1)*sin(e + f*x)**2/(6*a**4*f*sin(e + f*x)**3 + 18*a**4*f*sin(e + f*x)**2 + 18*a**4*f*sin(e + f*x) + 6*a**4*f)
- 18*B*log(sin(e + f*x) + 1)*sin(e + f*x)/(6*a**4*f*sin(e + f*x)**3 + 18*a**4*f*sin(e + f*x)**2 + 18*a**4*f*s
in(e + f*x) + 6*a**4*f) - 6*B*log(sin(e + f*x) + 1)/(6*a**4*f*sin(e + f*x)**3 + 18*a**4*f*sin(e + f*x)**2 + 18
*a**4*f*sin(e + f*x) + 6*a**4*f) - 10*B*sin(e + f*x)**2/(6*a**4*f*sin(e + f*x)**3 + 18*a**4*f*sin(e + f*x)**2
+ 18*a**4*f*sin(e + f*x) + 6*a**4*f) - 3*B*sin(e + f*x)*cos(e + f*x)**2/(6*a**4*f*sin(e + f*x)**3 + 18*a**4*f*
sin(e + f*x)**2 + 18*a**4*f*sin(e + f*x) + 6*a**4*f) - 18*B*sin(e + f*x)/(6*a**4*f*sin(e + f*x)**3 + 18*a**4*f
*sin(e + f*x)**2 + 18*a**4*f*sin(e + f*x) + 6*a**4*f) - B*cos(e + f*x)**2/(6*a**4*f*sin(e + f*x)**3 + 18*a**4*
f*sin(e + f*x)**2 + 18*a**4*f*sin(e + f*x) + 6*a**4*f) - 8*B/(6*a**4*f*sin(e + f*x)**3 + 18*a**4*f*sin(e + f*x
)**2 + 18*a**4*f*sin(e + f*x) + 6*a**4*f), Eq(m, -4)), (-2*A*log(sin(e + f*x) + 1)*sin(e + f*x)**2/(2*a**3*f*s
in(e + f*x)**2 + 4*a**3*f*sin(e + f*x) + 2*a**3*f) - 4*A*log(sin(e + f*x) + 1)*sin(e + f*x)/(2*a**3*f*sin(e +
f*x)**2 + 4*a**3*f*sin(e + f*x) + 2*a**3*f) - 2*A*log(sin(e + f*x) + 1)/(2*a**3*f*sin(e + f*x)**2 + 4*a**3*f*s
in(e + f*x) + 2*a**3*f) - 2*A*sin(e + f*x)/(2*a**3*f*sin(e + f*x)**2 + 4*a**3*f*sin(e + f*x) + 2*a**3*f) - A*c
os(e + f*x)**2/(2*a**3*f*sin(e + f*x)**2 + 4*a**3*f*sin(e + f*x) + 2*a**3*f) - 2*A/(2*a**3*f*sin(e + f*x)**2 +
4*a**3*f*sin(e + f*x) + 2*a**3*f) + 6*B*log(sin(e + f*x) + 1)*sin(e + f*x)**2/(2*a**3*f*sin(e + f*x)**2 + 4*a
**3*f*sin(e + f*x) + 2*a**3*f) + 12*B*log(sin(e + f*x) + 1)*sin(e + f*x)/(2*a**3*f*sin(e + f*x)**2 + 4*a**3*f*
sin(e + f*x) + 2*a**3*f) + 6*B*log(sin(e + f*x) + 1)/(2*a**3*f*sin(e + f*x)**2 + 4*a**3*f*sin(e + f*x) + 2*a**
3*f) - 4*B*sin(e + f*x)**3/(2*a**3*f*sin(e + f*x)**2 + 4*a**3*f*sin(e + f*x) + 2*a**3*f) - 2*B*sin(e + f*x)*co
s(e + f*x)**2/(2*a**3*f*sin(e + f*x)**2 + 4*a**3*f*sin(e + f*x) + 2*a**3*f) + 14*B*sin(e + f*x)/(2*a**3*f*sin(
e + f*x)**2 + 4*a**3*f*sin(e + f*x) + 2*a**3*f) - B*cos(e + f*x)**2/(2*a**3*f*sin(e + f*x)**2 + 4*a**3*f*sin(e
+ f*x) + 2*a**3*f) + 10*B/(2*a**3*f*sin(e + f*x)**2 + 4*a**3*f*sin(e + f*x) + 2*a**3*f), Eq(m, -3)), (4*A*log
(tan(e/2 + f*x/2) + 1)*tan(e/2 + f*x/2)**4/(a**2*f*tan(e/2 + f*x/2)**4 + 2*a**2*f*tan(e/2 + f*x/2)**2 + a**2*f
) + 8*A*log(tan(e/2 + f*x/2) + 1)*tan(e/2 + f*x/2)**2/(a**2*f*tan(e/2 + f*x/2)**4 + 2*a**2*f*tan(e/2 + f*x/2)*
*2 + a**2*f) + 4*A*log(tan(e/2 + f*x/2) + 1)/(a**2*f*tan(e/2 + f*x/2)**4 + 2*a**2*f*tan(e/2 + f*x/2)**2 + a**2
*f) - 2*A*log(tan(e/2 + f*x/2)**2 + 1)*tan(e/2 + f*x/2)**4/(a**2*f*tan(e/2 + f*x/2)**4 + 2*a**2*f*tan(e/2 + f*
x/2)**2 + a**2*f) - 4*A*log(tan(e/2 + f*x/2)**2 + 1)*tan(e/2 + f*x/2)**2/(a**2*f*tan(e/2 + f*x/2)**4 + 2*a**2*
f*tan(e/2 + f*x/2)**2 + a**2*f) - 2*A*log(tan(e/2 + f*x/2)**2 + 1)/(a**2*f*tan(e/2 + f*x/2)**4 + 2*a**2*f*tan(
e/2 + f*x/2)**2 + a**2*f) - 2*A*tan(e/2 + f*x/2)**3/(a**2*f*tan(e/2 + f*x/2)**4 + 2*a**2*f*tan(e/2 + f*x/2)**2
+ a**2*f) - 2*A*tan(e/2 + f*x/2)/(a**2*f*tan(e/2 + f*x/2)**4 + 2*a**2*f*tan(e/2 + f*x/2)**2 + a**2*f) - 4*B*l
og(tan(e/2 + f*x/2) + 1)*tan(e/2 + f*x/2)**4/(a**2*f*tan(e/2 + f*x/2)**4 + 2*a**2*f*tan(e/2 + f*x/2)**2 + a**2
*f) - 8*B*log(tan(e/2 + f*x/2) + 1)*tan(e/2 + f*x/2)**2/(a**2*f*tan(e/2 + f*x/2)**4 + 2*a**2*f*tan(e/2 + f*x/2
)**2 + a**2*f) - 4*B*log(tan(e/2 + f*x/2) + 1)/(a**2*f*tan(e/2 + f*x/2)**4 + 2*a**2*f*tan(e/2 + f*x/2)**2 + a*
*2*f) + 2*B*log(tan(e/2 + f*x/2)**2 + 1)*tan(e/2 + f*x/2)**4/(a**2*f*tan(e/2 + f*x/2)**4 + 2*a**2*f*tan(e/2 +
f*x/2)**2 + a**2*f) + 4*B*log(tan(e/2 + f*x/2)**2 + 1)*tan(e/2 + f*x/2)**2/(a**2*f*tan(e/2 + f*x/2)**4 + 2*a**
2*f*tan(e/2 + f*x/2)**2 + a**2*f) + 2*B*log(tan(e/2 + f*x/2)**2 + 1)/(a**2*f*tan(e/2 + f*x/2)**4 + 2*a**2*f*ta
n(e/2 + f*x/2)**2 + a**2*f) + 4*B*tan(e/2 + f*x/2)**3/(a**2*f*tan(e/2 + f*x/2)**4 + 2*a**2*f*tan(e/2 + f*x/2)*
*2 + a**2*f) - 2*B*tan(e/2 + f*x/2)**2/(a**2*f*tan(e/2 + f*x/2)**4 + 2*a**2*f*tan(e/2 + f*x/2)**2 + a**2*f) +
4*B*tan(e/2 + f*x/2)/(a**2*f*tan(e/2 + f*x/2)**4 + 2*a**2*f*tan(e/2 + f*x/2)**2 + a**2*f), Eq(m, -2)), (6*A*ta
n(e/2 + f*x/2)**5/(3*a*f*tan(e/2 + f*x/2)**6 + 9*a*f*tan(e/2 + f*x/2)**4 + 9*a*f*tan(e/2 + f*x/2)**2 + 3*a*f)
- 6*A*tan(e/2 + f*x/2)**4/(3*a*f*tan(e/2 + f*x/2)**6 + 9*a*f*tan(e/2 + f*x/2)**4 + 9*a*f*tan(e/2 + f*x/2)**2 +
3*a*f) + 12*A*tan(e/2 + f*x/2)**3/(3*a*f*tan(e...
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 458 vs.
\(2 (96) = 192\).
time = 0.46, size = 458, normalized size = 4.92 \begin {gather*} -\frac {\frac {{\left ({\left (a \sin \left (f x + e\right ) + a\right )}^{3} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} m - 2 \, {\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} a m + 2 \, {\left (a \sin \left (f x + e\right ) + a\right )}^{3} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} - 6 \, {\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} a\right )} A}{a^{2} m^{2} + 5 \, a^{2} m + 6 \, a^{2}} + \frac {{\left ({\left (a \sin \left (f x + e\right ) + a\right )}^{4} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} m^{2} - 3 \, {\left (a \sin \left (f x + e\right ) + a\right )}^{3} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} a m^{2} + 2 \, {\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} a^{2} m^{2} + 5 \, {\left (a \sin \left (f x + e\right ) + a\right )}^{4} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} m - 18 \, {\left (a \sin \left (f x + e\right ) + a\right )}^{3} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} a m + 14 \, {\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} a^{2} m + 6 \, {\left (a \sin \left (f x + e\right ) + a\right )}^{4} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} - 24 \, {\left (a \sin \left (f x + e\right ) + a\right )}^{3} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} a + 24 \, {\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} a^{2}\right )} B}{{\left (a^{2} m^{3} + 9 \, a^{2} m^{2} + 26 \, a^{2} m + 24 \, a^{2}\right )} a}}{a f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(f*x+e)^3*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="giac")
[Out]
-(((a*sin(f*x + e) + a)^3*(a*sin(f*x + e) + a)^m*m - 2*(a*sin(f*x + e) + a)^2*(a*sin(f*x + e) + a)^m*a*m + 2*(
a*sin(f*x + e) + a)^3*(a*sin(f*x + e) + a)^m - 6*(a*sin(f*x + e) + a)^2*(a*sin(f*x + e) + a)^m*a)*A/(a^2*m^2 +
5*a^2*m + 6*a^2) + ((a*sin(f*x + e) + a)^4*(a*sin(f*x + e) + a)^m*m^2 - 3*(a*sin(f*x + e) + a)^3*(a*sin(f*x +
e) + a)^m*a*m^2 + 2*(a*sin(f*x + e) + a)^2*(a*sin(f*x + e) + a)^m*a^2*m^2 + 5*(a*sin(f*x + e) + a)^4*(a*sin(f
*x + e) + a)^m*m - 18*(a*sin(f*x + e) + a)^3*(a*sin(f*x + e) + a)^m*a*m + 14*(a*sin(f*x + e) + a)^2*(a*sin(f*x
+ e) + a)^m*a^2*m + 6*(a*sin(f*x + e) + a)^4*(a*sin(f*x + e) + a)^m - 24*(a*sin(f*x + e) + a)^3*(a*sin(f*x +
e) + a)^m*a + 24*(a*sin(f*x + e) + a)^2*(a*sin(f*x + e) + a)^m*a^2)*B/((a^2*m^3 + 9*a^2*m^2 + 26*a^2*m + 24*a^
2)*a))/(a*f)
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Mupad [B]
time = 11.74, size = 272, normalized size = 2.92 \begin {gather*} \frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\left (128\,A-18\,B+48\,A\,m+17\,B\,m+144\,A\,\sin \left (e+f\,x\right )-24\,B\,\cos \left (2\,e+2\,f\,x\right )-6\,B\,\cos \left (4\,e+4\,f\,x\right )+4\,A\,m^2+B\,m^2+16\,A\,\sin \left (3\,e+3\,f\,x\right )+4\,A\,m^2\,\cos \left (2\,e+2\,f\,x\right )-B\,m^2\,\cos \left (4\,e+4\,f\,x\right )+2\,A\,m^2\,\sin \left (3\,e+3\,f\,x\right )+2\,B\,m^2\,\sin \left (3\,e+3\,f\,x\right )+44\,A\,m\,\sin \left (e+f\,x\right )+36\,B\,m\,\sin \left (e+f\,x\right )+16\,A\,m\,\cos \left (2\,e+2\,f\,x\right )-20\,B\,m\,\cos \left (2\,e+2\,f\,x\right )-5\,B\,m\,\cos \left (4\,e+4\,f\,x\right )+12\,A\,m\,\sin \left (3\,e+3\,f\,x\right )+2\,A\,m^2\,\sin \left (e+f\,x\right )+4\,B\,m\,\sin \left (3\,e+3\,f\,x\right )+2\,B\,m^2\,\sin \left (e+f\,x\right )\right )}{8\,f\,\left (m^3+9\,m^2+26\,m+24\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(e + f*x)^3*(A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m,x)
[Out]
((a*(sin(e + f*x) + 1))^m*(128*A - 18*B + 48*A*m + 17*B*m + 144*A*sin(e + f*x) - 24*B*cos(2*e + 2*f*x) - 6*B*c
os(4*e + 4*f*x) + 4*A*m^2 + B*m^2 + 16*A*sin(3*e + 3*f*x) + 4*A*m^2*cos(2*e + 2*f*x) - B*m^2*cos(4*e + 4*f*x)
+ 2*A*m^2*sin(3*e + 3*f*x) + 2*B*m^2*sin(3*e + 3*f*x) + 44*A*m*sin(e + f*x) + 36*B*m*sin(e + f*x) + 16*A*m*cos
(2*e + 2*f*x) - 20*B*m*cos(2*e + 2*f*x) - 5*B*m*cos(4*e + 4*f*x) + 12*A*m*sin(3*e + 3*f*x) + 2*A*m^2*sin(e + f
*x) + 4*B*m*sin(3*e + 3*f*x) + 2*B*m^2*sin(e + f*x)))/(8*f*(26*m + 9*m^2 + m^3 + 24))
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