3.340 \(\int \cos ^5(a+b x) (c \sin (a+b x))^m \, dx\)
Optimal. Leaf size=74 \[ \frac {(c \sin (a+b x))^{m+5}}{b c^5 (m+5)}-\frac {2 (c \sin (a+b x))^{m+3}}{b c^3 (m+3)}+\frac {(c \sin (a+b x))^{m+1}}{b c (m+1)} \]
[Out]
(c*sin(b*x+a))^(1+m)/b/c/(1+m)-2*(c*sin(b*x+a))^(3+m)/b/c^3/(3+m)+(c*sin(b*x+a))^(5+m)/b/c^5/(5+m)
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Rubi [A] time = 0.07, antiderivative size = 74, normalized size of antiderivative = 1.00,
number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used =
{2564, 270} \[ -\frac {2 (c \sin (a+b x))^{m+3}}{b c^3 (m+3)}+\frac {(c \sin (a+b x))^{m+5}}{b c^5 (m+5)}+\frac {(c \sin (a+b x))^{m+1}}{b c (m+1)} \]
Antiderivative was successfully verified.
[In]
Int[Cos[a + b*x]^5*(c*Sin[a + b*x])^m,x]
[Out]
(c*Sin[a + b*x])^(1 + m)/(b*c*(1 + m)) - (2*(c*Sin[a + b*x])^(3 + m))/(b*c^3*(3 + m)) + (c*Sin[a + b*x])^(5 +
m)/(b*c^5*(5 + m))
Rule 270
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]
Rule 2564
Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])
Rubi steps
\begin {align*} \int \cos ^5(a+b x) (c \sin (a+b x))^m \, dx &=\frac {\operatorname {Subst}\left (\int x^m \left (1-\frac {x^2}{c^2}\right )^2 \, dx,x,c \sin (a+b x)\right )}{b c}\\ &=\frac {\operatorname {Subst}\left (\int \left (x^m-\frac {2 x^{2+m}}{c^2}+\frac {x^{4+m}}{c^4}\right ) \, dx,x,c \sin (a+b x)\right )}{b c}\\ &=\frac {(c \sin (a+b x))^{1+m}}{b c (1+m)}-\frac {2 (c \sin (a+b x))^{3+m}}{b c^3 (3+m)}+\frac {(c \sin (a+b x))^{5+m}}{b c^5 (5+m)}\\ \end {align*}
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Mathematica [A] time = 0.31, size = 55, normalized size = 0.74 \[ \frac {\sin (a+b x) \left (\frac {\sin ^4(a+b x)}{m+5}-\frac {2 \sin ^2(a+b x)}{m+3}+\frac {1}{m+1}\right ) (c \sin (a+b x))^m}{b} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[a + b*x]^5*(c*Sin[a + b*x])^m,x]
[Out]
(Sin[a + b*x]*(c*Sin[a + b*x])^m*((1 + m)^(-1) - (2*Sin[a + b*x]^2)/(3 + m) + Sin[a + b*x]^4/(5 + m)))/b
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fricas [A] time = 0.63, size = 70, normalized size = 0.95 \[ \frac {{\left ({\left (m^{2} + 4 \, m + 3\right )} \cos \left (b x + a\right )^{4} + 4 \, {\left (m + 1\right )} \cos \left (b x + a\right )^{2} + 8\right )} \left (c \sin \left (b x + a\right )\right )^{m} \sin \left (b x + a\right )}{b m^{3} + 9 \, b m^{2} + 23 \, b m + 15 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(b*x+a)^5*(c*sin(b*x+a))^m,x, algorithm="fricas")
[Out]
((m^2 + 4*m + 3)*cos(b*x + a)^4 + 4*(m + 1)*cos(b*x + a)^2 + 8)*(c*sin(b*x + a))^m*sin(b*x + a)/(b*m^3 + 9*b*m
^2 + 23*b*m + 15*b)
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(b*x+a)^5*(c*sin(b*x+a))^m,x, algorithm="giac")
[Out]
Timed out
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maple [F] time = 1.78, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{5}\left (b x +a \right )\right ) \left (c \sin \left (b x +a \right )\right )^{m}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(b*x+a)^5*(c*sin(b*x+a))^m,x)
[Out]
int(cos(b*x+a)^5*(c*sin(b*x+a))^m,x)
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maxima [A] time = 0.56, size = 77, normalized size = 1.04 \[ \frac {\frac {c^{m} \sin \left (b x + a\right )^{m} \sin \left (b x + a\right )^{5}}{m + 5} - \frac {2 \, c^{m} \sin \left (b x + a\right )^{m} \sin \left (b x + a\right )^{3}}{m + 3} + \frac {\left (c \sin \left (b x + a\right )\right )^{m + 1}}{c {\left (m + 1\right )}}}{b} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(b*x+a)^5*(c*sin(b*x+a))^m,x, algorithm="maxima")
[Out]
(c^m*sin(b*x + a)^m*sin(b*x + a)^5/(m + 5) - 2*c^m*sin(b*x + a)^m*sin(b*x + a)^3/(m + 3) + (c*sin(b*x + a))^(m
+ 1)/(c*(m + 1)))/b
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mupad [B] time = 1.61, size = 132, normalized size = 1.78 \[ \frac {{\left (c\,\sin \left (a+b\,x\right )\right )}^m\,\left (150\,\sin \left (a+b\,x\right )+25\,\sin \left (3\,a+3\,b\,x\right )+3\,\sin \left (5\,a+5\,b\,x\right )+24\,m\,\sin \left (a+b\,x\right )+28\,m\,\sin \left (3\,a+3\,b\,x\right )+4\,m\,\sin \left (5\,a+5\,b\,x\right )+2\,m^2\,\sin \left (a+b\,x\right )+3\,m^2\,\sin \left (3\,a+3\,b\,x\right )+m^2\,\sin \left (5\,a+5\,b\,x\right )\right )}{16\,b\,\left (m^3+9\,m^2+23\,m+15\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(a + b*x)^5*(c*sin(a + b*x))^m,x)
[Out]
((c*sin(a + b*x))^m*(150*sin(a + b*x) + 25*sin(3*a + 3*b*x) + 3*sin(5*a + 5*b*x) + 24*m*sin(a + b*x) + 28*m*si
n(3*a + 3*b*x) + 4*m*sin(5*a + 5*b*x) + 2*m^2*sin(a + b*x) + 3*m^2*sin(3*a + 3*b*x) + m^2*sin(5*a + 5*b*x)))/(
16*b*(23*m + 9*m^2 + m^3 + 15))
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sympy [A] time = 66.31, size = 2050, normalized size = 27.70 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(b*x+a)**5*(c*sin(b*x+a))**m,x)
[Out]
Piecewise((x*(c*sin(a))**m*cos(a)**5, Eq(b, 0)), ((log(sin(a + b*x))/b + cos(a + b*x)**2/(2*b*sin(a + b*x)**2)
- cos(a + b*x)**4/(4*b*sin(a + b*x)**4))/c**5, Eq(m, -5)), ((16*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)
**6/(8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) + 32*log(tan(a/2 + b*x/2)**
2 + 1)*tan(a/2 + b*x/2)**4/(8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) + 16
*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**2/(8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a/2 + b*x/2)**4 + 8*b*ta
n(a/2 + b*x/2)**2) - 16*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**6/(8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a/2 + b*
x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) - 32*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**4/(8*b*tan(a/2 + b*x/2)**6 + 1
6*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) - 16*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**2/(8*b*tan(a/2
+ b*x/2)**6 + 16*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) - tan(a/2 + b*x/2)**8/(8*b*tan(a/2 + b*x/2)
**6 + 16*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) + 18*tan(a/2 + b*x/2)**4/(8*b*tan(a/2 + b*x/2)**6 +
16*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) - 1/(8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a/2 + b*x/2)**4 +
8*b*tan(a/2 + b*x/2)**2))/c**3, Eq(m, -3)), ((-log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**8/(b*tan(a/2 + b
*x/2)**8 + 4*b*tan(a/2 + b*x/2)**6 + 6*b*tan(a/2 + b*x/2)**4 + 4*b*tan(a/2 + b*x/2)**2 + b) - 4*log(tan(a/2 +
b*x/2)**2 + 1)*tan(a/2 + b*x/2)**6/(b*tan(a/2 + b*x/2)**8 + 4*b*tan(a/2 + b*x/2)**6 + 6*b*tan(a/2 + b*x/2)**4
+ 4*b*tan(a/2 + b*x/2)**2 + b) - 6*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**4/(b*tan(a/2 + b*x/2)**8 + 4
*b*tan(a/2 + b*x/2)**6 + 6*b*tan(a/2 + b*x/2)**4 + 4*b*tan(a/2 + b*x/2)**2 + b) - 4*log(tan(a/2 + b*x/2)**2 +
1)*tan(a/2 + b*x/2)**2/(b*tan(a/2 + b*x/2)**8 + 4*b*tan(a/2 + b*x/2)**6 + 6*b*tan(a/2 + b*x/2)**4 + 4*b*tan(a/
2 + b*x/2)**2 + b) - log(tan(a/2 + b*x/2)**2 + 1)/(b*tan(a/2 + b*x/2)**8 + 4*b*tan(a/2 + b*x/2)**6 + 6*b*tan(a
/2 + b*x/2)**4 + 4*b*tan(a/2 + b*x/2)**2 + b) + log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**8/(b*tan(a/2 + b*x/2)*
*8 + 4*b*tan(a/2 + b*x/2)**6 + 6*b*tan(a/2 + b*x/2)**4 + 4*b*tan(a/2 + b*x/2)**2 + b) + 4*log(tan(a/2 + b*x/2)
)*tan(a/2 + b*x/2)**6/(b*tan(a/2 + b*x/2)**8 + 4*b*tan(a/2 + b*x/2)**6 + 6*b*tan(a/2 + b*x/2)**4 + 4*b*tan(a/2
+ b*x/2)**2 + b) + 6*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**4/(b*tan(a/2 + b*x/2)**8 + 4*b*tan(a/2 + b*x/2)*
*6 + 6*b*tan(a/2 + b*x/2)**4 + 4*b*tan(a/2 + b*x/2)**2 + b) + 4*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**2/(b*t
an(a/2 + b*x/2)**8 + 4*b*tan(a/2 + b*x/2)**6 + 6*b*tan(a/2 + b*x/2)**4 + 4*b*tan(a/2 + b*x/2)**2 + b) + log(ta
n(a/2 + b*x/2))/(b*tan(a/2 + b*x/2)**8 + 4*b*tan(a/2 + b*x/2)**6 + 6*b*tan(a/2 + b*x/2)**4 + 4*b*tan(a/2 + b*x
/2)**2 + b) - 4*tan(a/2 + b*x/2)**6/(b*tan(a/2 + b*x/2)**8 + 4*b*tan(a/2 + b*x/2)**6 + 6*b*tan(a/2 + b*x/2)**4
+ 4*b*tan(a/2 + b*x/2)**2 + b) - 4*tan(a/2 + b*x/2)**4/(b*tan(a/2 + b*x/2)**8 + 4*b*tan(a/2 + b*x/2)**6 + 6*b
*tan(a/2 + b*x/2)**4 + 4*b*tan(a/2 + b*x/2)**2 + b) - 4*tan(a/2 + b*x/2)**2/(b*tan(a/2 + b*x/2)**8 + 4*b*tan(a
/2 + b*x/2)**6 + 6*b*tan(a/2 + b*x/2)**4 + 4*b*tan(a/2 + b*x/2)**2 + b))/c, Eq(m, -1)), (c**m*m**2*sin(a + b*x
)*sin(a + b*x)**m*cos(a + b*x)**4/(b*m**3 + 9*b*m**2 + 23*b*m + 15*b) + 4*c**m*m*sin(a + b*x)**3*sin(a + b*x)*
*m*cos(a + b*x)**2/(b*m**3 + 9*b*m**2 + 23*b*m + 15*b) + 8*c**m*m*sin(a + b*x)*sin(a + b*x)**m*cos(a + b*x)**4
/(b*m**3 + 9*b*m**2 + 23*b*m + 15*b) + 8*c**m*sin(a + b*x)**5*sin(a + b*x)**m/(b*m**3 + 9*b*m**2 + 23*b*m + 15
*b) + 20*c**m*sin(a + b*x)**3*sin(a + b*x)**m*cos(a + b*x)**2/(b*m**3 + 9*b*m**2 + 23*b*m + 15*b) + 15*c**m*si
n(a + b*x)*sin(a + b*x)**m*cos(a + b*x)**4/(b*m**3 + 9*b*m**2 + 23*b*m + 15*b), True))
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