3.632 \(\int \cos ^3(c+d x) (a+b \sin (c+d x))^m \, dx\)

Optimal. Leaf size=92 \[ -\frac {\left (a^2-b^2\right ) (a+b \sin (c+d x))^{m+1}}{b^3 d (m+1)}+\frac {2 a (a+b \sin (c+d x))^{m+2}}{b^3 d (m+2)}-\frac {(a+b \sin (c+d x))^{m+3}}{b^3 d (m+3)} \]

[Out]

-(a^2-b^2)*(a+b*sin(d*x+c))^(1+m)/b^3/d/(1+m)+2*a*(a+b*sin(d*x+c))^(2+m)/b^3/d/(2+m)-(a+b*sin(d*x+c))^(3+m)/b^
3/d/(3+m)

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Rubi [A]  time = 0.07, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2668, 697} \[ -\frac {\left (a^2-b^2\right ) (a+b \sin (c+d x))^{m+1}}{b^3 d (m+1)}+\frac {2 a (a+b \sin (c+d x))^{m+2}}{b^3 d (m+2)}-\frac {(a+b \sin (c+d x))^{m+3}}{b^3 d (m+3)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + b*Sin[c + d*x])^m,x]

[Out]

-(((a^2 - b^2)*(a + b*Sin[c + d*x])^(1 + m))/(b^3*d*(1 + m))) + (2*a*(a + b*Sin[c + d*x])^(2 + m))/(b^3*d*(2 +
m)) - (a + b*Sin[c + d*x])^(3 + m)/(b^3*d*(3 + m))

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \cos ^3(c+d x) (a+b \sin (c+d x))^m \, dx &=\frac {\operatorname {Subst}\left (\int (a+x)^m \left (b^2-x^2\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (\left (-a^2+b^2\right ) (a+x)^m+2 a (a+x)^{1+m}-(a+x)^{2+m}\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=-\frac {\left (a^2-b^2\right ) (a+b \sin (c+d x))^{1+m}}{b^3 d (1+m)}+\frac {2 a (a+b \sin (c+d x))^{2+m}}{b^3 d (2+m)}-\frac {(a+b \sin (c+d x))^{3+m}}{b^3 d (3+m)}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 74, normalized size = 0.80 \[ \frac {(a+b \sin (c+d x))^{m+1} \left (\frac {b^2-a^2}{m+1}-\frac {(a+b \sin (c+d x))^2}{m+3}+\frac {2 a (a+b \sin (c+d x))}{m+2}\right )}{b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + b*Sin[c + d*x])^m,x]

[Out]

((a + b*Sin[c + d*x])^(1 + m)*((-a^2 + b^2)/(1 + m) + (2*a*(a + b*Sin[c + d*x]))/(2 + m) - (a + b*Sin[c + d*x]
)^2/(3 + m)))/(b^3*d)

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fricas [A]  time = 0.77, size = 142, normalized size = 1.54 \[ \frac {{\left (4 \, a b^{2} m - 2 \, a^{3} + 6 \, a b^{2} + {\left (a b^{2} m^{2} + a b^{2} m\right )} \cos \left (d x + c\right )^{2} + {\left (4 \, b^{3} + {\left (b^{3} m^{2} + 3 \, b^{3} m + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{2} b + b^{3}\right )} m\right )} \sin \left (d x + c\right )\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{m}}{b^{3} d m^{3} + 6 \, b^{3} d m^{2} + 11 \, b^{3} d m + 6 \, b^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

(4*a*b^2*m - 2*a^3 + 6*a*b^2 + (a*b^2*m^2 + a*b^2*m)*cos(d*x + c)^2 + (4*b^3 + (b^3*m^2 + 3*b^3*m + 2*b^3)*cos
(d*x + c)^2 + 2*(a^2*b + b^3)*m)*sin(d*x + c))*(b*sin(d*x + c) + a)^m/(b^3*d*m^3 + 6*b^3*d*m^2 + 11*b^3*d*m +
6*b^3*d)

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giac [B]  time = 0.36, size = 340, normalized size = 3.70 \[ -\frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{m} b^{3} m^{2} \sin \left (d x + c\right )^{3} + {\left (b \sin \left (d x + c\right ) + a\right )}^{m} a b^{2} m^{2} \sin \left (d x + c\right )^{2} + 3 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{m} b^{3} m \sin \left (d x + c\right )^{3} - {\left (b \sin \left (d x + c\right ) + a\right )}^{m} b^{3} m^{2} \sin \left (d x + c\right ) + {\left (b \sin \left (d x + c\right ) + a\right )}^{m} a b^{2} m \sin \left (d x + c\right )^{2} + 2 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{m} b^{3} \sin \left (d x + c\right )^{3} - {\left (b \sin \left (d x + c\right ) + a\right )}^{m} a b^{2} m^{2} - 2 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{m} a^{2} b m \sin \left (d x + c\right ) - 5 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{m} b^{3} m \sin \left (d x + c\right ) - 5 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{m} a b^{2} m - 6 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{m} b^{3} \sin \left (d x + c\right ) + 2 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{m} a^{3} - 6 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{m} a b^{2}}{{\left (b^{3} m^{3} + 6 \, b^{3} m^{2} + 11 \, b^{3} m + 6 \, b^{3}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sin(d*x+c))^m,x, algorithm="giac")

[Out]

-((b*sin(d*x + c) + a)^m*b^3*m^2*sin(d*x + c)^3 + (b*sin(d*x + c) + a)^m*a*b^2*m^2*sin(d*x + c)^2 + 3*(b*sin(d
*x + c) + a)^m*b^3*m*sin(d*x + c)^3 - (b*sin(d*x + c) + a)^m*b^3*m^2*sin(d*x + c) + (b*sin(d*x + c) + a)^m*a*b
^2*m*sin(d*x + c)^2 + 2*(b*sin(d*x + c) + a)^m*b^3*sin(d*x + c)^3 - (b*sin(d*x + c) + a)^m*a*b^2*m^2 - 2*(b*si
n(d*x + c) + a)^m*a^2*b*m*sin(d*x + c) - 5*(b*sin(d*x + c) + a)^m*b^3*m*sin(d*x + c) - 5*(b*sin(d*x + c) + a)^
m*a*b^2*m - 6*(b*sin(d*x + c) + a)^m*b^3*sin(d*x + c) + 2*(b*sin(d*x + c) + a)^m*a^3 - 6*(b*sin(d*x + c) + a)^
m*a*b^2)/((b^3*m^3 + 6*b^3*m^2 + 11*b^3*m + 6*b^3)*d)

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maple [F]  time = 0.49, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{3}\left (d x +c \right )\right ) \left (a +b \sin \left (d x +c \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+b*sin(d*x+c))^m,x)

[Out]

int(cos(d*x+c)^3*(a+b*sin(d*x+c))^m,x)

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maxima [A]  time = 0.78, size = 117, normalized size = 1.27 \[ \frac {\frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{m + 1}}{b {\left (m + 1\right )}} - \frac {{\left ({\left (m^{2} + 3 \, m + 2\right )} b^{3} \sin \left (d x + c\right )^{3} + {\left (m^{2} + m\right )} a b^{2} \sin \left (d x + c\right )^{2} - 2 \, a^{2} b m \sin \left (d x + c\right ) + 2 \, a^{3}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{m}}{{\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )} b^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

((b*sin(d*x + c) + a)^(m + 1)/(b*(m + 1)) - ((m^2 + 3*m + 2)*b^3*sin(d*x + c)^3 + (m^2 + m)*a*b^2*sin(d*x + c)
^2 - 2*a^2*b*m*sin(d*x + c) + 2*a^3)*(b*sin(d*x + c) + a)^m/((m^3 + 6*m^2 + 11*m + 6)*b^3))/d

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mupad [B]  time = 7.51, size = 197, normalized size = 2.14 \[ \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^m\,\left (24\,a\,b^2+18\,b^3\,\sin \left (c+d\,x\right )-8\,a^3+2\,b^3\,\sin \left (3\,c+3\,d\,x\right )+2\,a\,b^2\,m^2+3\,b^3\,m\,\sin \left (3\,c+3\,d\,x\right )+b^3\,m^2\,\sin \left (c+d\,x\right )+b^3\,m^2\,\sin \left (3\,c+3\,d\,x\right )+18\,a\,b^2\,m+11\,b^3\,m\,\sin \left (c+d\,x\right )+8\,a^2\,b\,m\,\sin \left (c+d\,x\right )-2\,a\,b^2\,m\,\left (2\,{\sin \left (c+d\,x\right )}^2-1\right )-2\,a\,b^2\,m^2\,\left (2\,{\sin \left (c+d\,x\right )}^2-1\right )\right )}{4\,b^3\,d\,\left (m^3+6\,m^2+11\,m+6\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(a + b*sin(c + d*x))^m,x)

[Out]

((a + b*sin(c + d*x))^m*(24*a*b^2 + 18*b^3*sin(c + d*x) - 8*a^3 + 2*b^3*sin(3*c + 3*d*x) + 2*a*b^2*m^2 + 3*b^3
*m*sin(3*c + 3*d*x) + b^3*m^2*sin(c + d*x) + b^3*m^2*sin(3*c + 3*d*x) + 18*a*b^2*m + 11*b^3*m*sin(c + d*x) + 8
*a^2*b*m*sin(c + d*x) - 2*a*b^2*m*(2*sin(c + d*x)^2 - 1) - 2*a*b^2*m^2*(2*sin(c + d*x)^2 - 1)))/(4*b^3*d*(11*m
+ 6*m^2 + m^3 + 6))

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sympy [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(d*x+c)**3*(a+b*sin(d*x+c))**m,x)

[Out]

Timed out

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