\(\int \cos ^3(c+d x) (a+b \sin (c+d x))^m \, dx\) [632]

Optimal result

Integrand size = 21, antiderivative size = 92 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^m \, dx=-\frac {\left (a^2-b^2\right ) (a+b \sin (c+d x))^{1+m}}{b^3 d (1+m)}+\frac {2 a (a+b \sin (c+d x))^{2+m}}{b^3 d (2+m)}-\frac {(a+b \sin (c+d x))^{3+m}}{b^3 d (3+m)} \]

[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)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2747, 711} \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^m \, dx=-\frac {\left (a^2-b^2\right ) (a+b \sin (c+d x))^{m+1}}{b^3 d (m+1)}+\frac {2 a (a+b \sin (c+d x))^{m+2}}{b^3 d (m+2)}-\frac {(a+b \sin (c+d x))^{m+3}}{b^3 d (m+3)} \]

[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 711

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 2747

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*} \text {integral}& = \frac {\text {Subst}\left (\int (a+x)^m \left (b^2-x^2\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d} \\ & = \frac {\text {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*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.80 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^m \, dx=\frac {(a+b \sin (c+d x))^{1+m} \left (\frac {-a^2+b^2}{1+m}+\frac {2 a (a+b \sin (c+d x))}{2+m}-\frac {(a+b \sin (c+d x))^2}{3+m}\right )}{b^3 d} \]

[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)

Maple [A] (verified)

Time = 1.82 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.33

method	result	size
parallelrisch	\(-\frac {2 \left (-\frac {\left (2+m \right ) \left (1+m \right ) b^{3} \sin \left (3 d x +3 c \right )}{8}-\frac {a \,b^{2} m \left (1+m \right ) \cos \left (2 d x +2 c \right )}{4}-\left (\frac {\left (m +9\right ) \left (2+m \right ) b^{2}}{8}+a^{2} m \right ) b \sin \left (d x +c \right )+\left (\left (-\frac {1}{4} m^{2}-\frac {9}{4} m -3\right ) b^{2}+a^{2}\right ) a \right ) \left (a +b \sin \left (d x +c \right )\right )^{m}}{b^{3} \left (m^{3}+6 m^{2}+11 m +6\right ) d}\)	\(122\)
derivativedivides	\(\frac {\left (b^{2} m^{2}+2 a^{2} m +5 b^{2} m +6 b^{2}\right ) \sin \left (d x +c \right ) {\mathrm e}^{m \ln \left (a +b \sin \left (d x +c \right )\right )}}{b^{2} \left (m^{3}+6 m^{2}+11 m +6\right ) d}-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) {\mathrm e}^{m \ln \left (a +b \sin \left (d x +c \right )\right )}}{d \left (3+m \right )}-\frac {a \left (-b^{2} m^{2}-5 b^{2} m +2 a^{2}-6 b^{2}\right ) {\mathrm e}^{m \ln \left (a +b \sin \left (d x +c \right )\right )}}{b^{3} d \left (m^{3}+6 m^{2}+11 m +6\right )}-\frac {a m \left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{m \ln \left (a +b \sin \left (d x +c \right )\right )}}{b d \left (m^{2}+5 m +6\right )}\)	\(206\)
default	\(\frac {\left (b^{2} m^{2}+2 a^{2} m +5 b^{2} m +6 b^{2}\right ) \sin \left (d x +c \right ) {\mathrm e}^{m \ln \left (a +b \sin \left (d x +c \right )\right )}}{b^{2} \left (m^{3}+6 m^{2}+11 m +6\right ) d}-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) {\mathrm e}^{m \ln \left (a +b \sin \left (d x +c \right )\right )}}{d \left (3+m \right )}-\frac {a \left (-b^{2} m^{2}-5 b^{2} m +2 a^{2}-6 b^{2}\right ) {\mathrm e}^{m \ln \left (a +b \sin \left (d x +c \right )\right )}}{b^{3} d \left (m^{3}+6 m^{2}+11 m +6\right )}-\frac {a m \left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{m \ln \left (a +b \sin \left (d x +c \right )\right )}}{b d \left (m^{2}+5 m +6\right )}\)	\(206\)

[In]

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

[Out]

-2*(-1/8*(2+m)*(1+m)*b^3*sin(3*d*x+3*c)-1/4*a*b^2*m*(1+m)*cos(2*d*x+2*c)-(1/8*(m+9)*(2+m)*b^2+a^2*m)*b*sin(d*x
+c)+((-1/4*m^2-9/4*m-3)*b^2+a^2)*a)*(a+b*sin(d*x+c))^m/b^3/(m^3+6*m^2+11*m+6)/d

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.54 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^m \, dx=\frac {{\left (4 \, a b^{2} m - 2 \, a^{3} + 6 \, a b^{2} + {\left (a b^{2} m^{2} + a b^{2} m\right )} \cos \left (d x + c\right )^{2} + {\left (4 \, b^{3} + {\left (b^{3} m^{2} + 3 \, b^{3} m + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{2} b + b^{3}\right )} m\right )} \sin \left (d x + c\right )\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{m}}{b^{3} d m^{3} + 6 \, b^{3} d m^{2} + 11 \, b^{3} d m + 6 \, b^{3} d} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2837 vs. \(2 (75) = 150\).

Time = 125.39 (sec) , antiderivative size = 2837, normalized size of antiderivative = 30.84 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^m \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((a**m*(2*sin(c + d*x)**3/(3*d) + sin(c + d*x)*cos(c + d*x)**2/d), Eq(b, 0)), (x*(a + b*sin(c))**m*co
s(c)**3, Eq(d, 0)), (-2*a**2*log(a/b + sin(c + d*x))/(2*a**2*b**3*d + 4*a*b**4*d*sin(c + d*x) + 2*b**5*d*sin(c
+ d*x)**2) - 2*a**2/(2*a**2*b**3*d + 4*a*b**4*d*sin(c + d*x) + 2*b**5*d*sin(c + d*x)**2) - 4*a*b*log(a/b + si
n(c + d*x))*sin(c + d*x)/(2*a**2*b**3*d + 4*a*b**4*d*sin(c + d*x) + 2*b**5*d*sin(c + d*x)**2) - 2*a*b*sin(c +
d*x)/(2*a**2*b**3*d + 4*a*b**4*d*sin(c + d*x) + 2*b**5*d*sin(c + d*x)**2) - 2*b**2*log(a/b + sin(c + d*x))*sin
(c + d*x)**2/(2*a**2*b**3*d + 4*a*b**4*d*sin(c + d*x) + 2*b**5*d*sin(c + d*x)**2) - b**2*cos(c + d*x)**2/(2*a*
*2*b**3*d + 4*a*b**4*d*sin(c + d*x) + 2*b**5*d*sin(c + d*x)**2), Eq(m, -3)), (2*a**2*log(a/b + sin(c + d*x))/(
a*b**3*d + b**4*d*sin(c + d*x)) + 2*a**2/(a*b**3*d + b**4*d*sin(c + d*x)) + 2*a*b*log(a/b + sin(c + d*x))*sin(
c + d*x)/(a*b**3*d + b**4*d*sin(c + d*x)) - 2*b**2*sin(c + d*x)**2/(a*b**3*d + b**4*d*sin(c + d*x)) - b**2*cos
(c + d*x)**2/(a*b**3*d + b**4*d*sin(c + d*x)), Eq(m, -2)), (a**2*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)
**4/(b**3*d*tan(c/2 + d*x/2)**4 + 2*b**3*d*tan(c/2 + d*x/2)**2 + b**3*d) + 2*a**2*log(tan(c/2 + d*x/2)**2 + 1)
*tan(c/2 + d*x/2)**2/(b**3*d*tan(c/2 + d*x/2)**4 + 2*b**3*d*tan(c/2 + d*x/2)**2 + b**3*d) + a**2*log(tan(c/2 +
d*x/2)**2 + 1)/(b**3*d*tan(c/2 + d*x/2)**4 + 2*b**3*d*tan(c/2 + d*x/2)**2 + b**3*d) - a**2*log(tan(c/2 + d*x/
2) + b/a - sqrt(-a**2 + b**2)/a)*tan(c/2 + d*x/2)**4/(b**3*d*tan(c/2 + d*x/2)**4 + 2*b**3*d*tan(c/2 + d*x/2)**
2 + b**3*d) - 2*a**2*log(tan(c/2 + d*x/2) + b/a - sqrt(-a**2 + b**2)/a)*tan(c/2 + d*x/2)**2/(b**3*d*tan(c/2 +
d*x/2)**4 + 2*b**3*d*tan(c/2 + d*x/2)**2 + b**3*d) - a**2*log(tan(c/2 + d*x/2) + b/a - sqrt(-a**2 + b**2)/a)/(
b**3*d*tan(c/2 + d*x/2)**4 + 2*b**3*d*tan(c/2 + d*x/2)**2 + b**3*d) - a**2*log(tan(c/2 + d*x/2) + b/a + sqrt(-
a**2 + b**2)/a)*tan(c/2 + d*x/2)**4/(b**3*d*tan(c/2 + d*x/2)**4 + 2*b**3*d*tan(c/2 + d*x/2)**2 + b**3*d) - 2*a
**2*log(tan(c/2 + d*x/2) + b/a + sqrt(-a**2 + b**2)/a)*tan(c/2 + d*x/2)**2/(b**3*d*tan(c/2 + d*x/2)**4 + 2*b**
3*d*tan(c/2 + d*x/2)**2 + b**3*d) - a**2*log(tan(c/2 + d*x/2) + b/a + sqrt(-a**2 + b**2)/a)/(b**3*d*tan(c/2 +
d*x/2)**4 + 2*b**3*d*tan(c/2 + d*x/2)**2 + b**3*d) + 2*a*b*tan(c/2 + d*x/2)**3/(b**3*d*tan(c/2 + d*x/2)**4 + 2
*b**3*d*tan(c/2 + d*x/2)**2 + b**3*d) + 2*a*b*tan(c/2 + d*x/2)/(b**3*d*tan(c/2 + d*x/2)**4 + 2*b**3*d*tan(c/2
+ d*x/2)**2 + b**3*d) - b**2*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**4/(b**3*d*tan(c/2 + d*x/2)**4 + 2*
b**3*d*tan(c/2 + d*x/2)**2 + b**3*d) - 2*b**2*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**2/(b**3*d*tan(c/2
+ d*x/2)**4 + 2*b**3*d*tan(c/2 + d*x/2)**2 + b**3*d) - b**2*log(tan(c/2 + d*x/2)**2 + 1)/(b**3*d*tan(c/2 + d*
x/2)**4 + 2*b**3*d*tan(c/2 + d*x/2)**2 + b**3*d) + b**2*log(tan(c/2 + d*x/2) + b/a - sqrt(-a**2 + b**2)/a)*tan
(c/2 + d*x/2)**4/(b**3*d*tan(c/2 + d*x/2)**4 + 2*b**3*d*tan(c/2 + d*x/2)**2 + b**3*d) + 2*b**2*log(tan(c/2 + d
*x/2) + b/a - sqrt(-a**2 + b**2)/a)*tan(c/2 + d*x/2)**2/(b**3*d*tan(c/2 + d*x/2)**4 + 2*b**3*d*tan(c/2 + d*x/2
)**2 + b**3*d) + b**2*log(tan(c/2 + d*x/2) + b/a - sqrt(-a**2 + b**2)/a)/(b**3*d*tan(c/2 + d*x/2)**4 + 2*b**3*
d*tan(c/2 + d*x/2)**2 + b**3*d) + b**2*log(tan(c/2 + d*x/2) + b/a + sqrt(-a**2 + b**2)/a)*tan(c/2 + d*x/2)**4/
(b**3*d*tan(c/2 + d*x/2)**4 + 2*b**3*d*tan(c/2 + d*x/2)**2 + b**3*d) + 2*b**2*log(tan(c/2 + d*x/2) + b/a + sqr
t(-a**2 + b**2)/a)*tan(c/2 + d*x/2)**2/(b**3*d*tan(c/2 + d*x/2)**4 + 2*b**3*d*tan(c/2 + d*x/2)**2 + b**3*d) +
b**2*log(tan(c/2 + d*x/2) + b/a + sqrt(-a**2 + b**2)/a)/(b**3*d*tan(c/2 + d*x/2)**4 + 2*b**3*d*tan(c/2 + d*x/2
)**2 + b**3*d) - 2*b**2*tan(c/2 + d*x/2)**2/(b**3*d*tan(c/2 + d*x/2)**4 + 2*b**3*d*tan(c/2 + d*x/2)**2 + b**3*
d), Eq(m, -1)), (-2*a**3*(a + b*sin(c + d*x))**m/(b**3*d*m**3 + 6*b**3*d*m**2 + 11*b**3*d*m + 6*b**3*d) + 2*a*
*2*b*m*(a + b*sin(c + d*x))**m*sin(c + d*x)/(b**3*d*m**3 + 6*b**3*d*m**2 + 11*b**3*d*m + 6*b**3*d) + a*b**2*m*
*2*(a + b*sin(c + d*x))**m*cos(c + d*x)**2/(b**3*d*m**3 + 6*b**3*d*m**2 + 11*b**3*d*m + 6*b**3*d) + 4*a*b**2*m
*(a + b*sin(c + d*x))**m*sin(c + d*x)**2/(b**3*d*m**3 + 6*b**3*d*m**2 + 11*b**3*d*m + 6*b**3*d) + 5*a*b**2*m*(
a + b*sin(c + d*x))**m*cos(c + d*x)**2/(b**3*d*m**3 + 6*b**3*d*m**2 + 11*b**3*d*m + 6*b**3*d) + 6*a*b**2*(a +
b*sin(c + d*x))**m*sin(c + d*x)**2/(b**3*d*m**3 + 6*b**3*d*m**2 + 11*b**3*d*m + 6*b**3*d) + 6*a*b**2*(a + b*si
n(c + d*x))**m*cos(c + d*x)**2/(b**3*d*m**3 + 6*b**3*d*m**2 + 11*b**3*d*m + 6*b**3*d) + b**3*m**2*(a + b*sin(c
+ d*x))**m*sin(c + d*x)*cos(c + d*x)**2/(b**3*d*m**3 + 6*b**3*d*m**2 + 11*b**3*d*m + 6*b**3*d) + 2*b**3*m*(a
+ b*sin(c + d*x))**m*sin(c + d*x)**3/(b**3*d*m**3 + 6*b**3*d*m**2 + 11*b**3*d*m + 6*b**3*d) + 5*b**3*m*(a + b*
sin(c + d*x))**m*sin(c + d*x)*cos(c + d*x)**2/(b**3*d*m**3 + 6*b**3*d*m**2 + 11*b**3*d*m + 6*b**3*d) + 4*b**3*
(a + b*sin(c + d*x))**m*sin(c + d*x)**3/(b**3*d*m**3 + 6*b**3*d*m**2 + 11*b**3*d*m + 6*b**3*d) + 6*b**3*(a + b
*sin(c + d*x))**m*sin(c + d*x)*cos(c + d*x)**2/(b**3*d*m**3 + 6*b**3*d*m**2 + 11*b**3*d*m + 6*b**3*d), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.27 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^m \, dx=\frac {\frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{m + 1}}{b {\left (m + 1\right )}} - \frac {{\left ({\left (m^{2} + 3 \, m + 2\right )} b^{3} \sin \left (d x + c\right )^{3} + {\left (m^{2} + m\right )} a b^{2} \sin \left (d x + c\right )^{2} - 2 \, a^{2} b m \sin \left (d x + c\right ) + 2 \, a^{3}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{m}}{{\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )} b^{3}}}{d} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 373 vs. \(2 (92) = 184\).

Time = 0.32 (sec) , antiderivative size = 373, normalized size of antiderivative = 4.05 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^m \, dx=-\frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{3} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} m^{2} - 2 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{2} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} a m^{2} + {\left (b \sin \left (d x + c\right ) + a\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} a^{2} m^{2} - {\left (b \sin \left (d x + c\right ) + a\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} b^{2} m^{2} + 3 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{3} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} m - 8 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{2} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} a m + 5 \, {\left (b \sin \left (d x + c\right ) + a\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} a^{2} m - 5 \, {\left (b \sin \left (d x + c\right ) + a\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} b^{2} m + 2 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{3} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} - 6 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{2} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} a + 6 \, {\left (b \sin \left (d x + c\right ) + a\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} a^{2} - 6 \, {\left (b \sin \left (d x + c\right ) + a\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} b^{2}}{{\left (b^{2} m^{3} + 6 \, b^{2} m^{2} + 11 \, b^{2} m + 6 \, b^{2}\right )} b d} \]

[In]

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

[Out]

-((b*sin(d*x + c) + a)^3*(b*sin(d*x + c) + a)^m*m^2 - 2*(b*sin(d*x + c) + a)^2*(b*sin(d*x + c) + a)^m*a*m^2 +
(b*sin(d*x + c) + a)*(b*sin(d*x + c) + a)^m*a^2*m^2 - (b*sin(d*x + c) + a)*(b*sin(d*x + c) + a)^m*b^2*m^2 + 3*
(b*sin(d*x + c) + a)^3*(b*sin(d*x + c) + a)^m*m - 8*(b*sin(d*x + c) + a)^2*(b*sin(d*x + c) + a)^m*a*m + 5*(b*s
in(d*x + c) + a)*(b*sin(d*x + c) + a)^m*a^2*m - 5*(b*sin(d*x + c) + a)*(b*sin(d*x + c) + a)^m*b^2*m + 2*(b*sin
(d*x + c) + a)^3*(b*sin(d*x + c) + a)^m - 6*(b*sin(d*x + c) + a)^2*(b*sin(d*x + c) + a)^m*a + 6*(b*sin(d*x + c
) + a)*(b*sin(d*x + c) + a)^m*a^2 - 6*(b*sin(d*x + c) + a)*(b*sin(d*x + c) + a)^m*b^2)/((b^2*m^3 + 6*b^2*m^2 +
11*b^2*m + 6*b^2)*b*d)

Mupad [B] (verification not implemented)

Time = 7.12 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.14 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^m \, dx=\frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^m\,\left (24\,a\,b^2+18\,b^3\,\sin \left (c+d\,x\right )-8\,a^3+2\,b^3\,\sin \left (3\,c+3\,d\,x\right )+2\,a\,b^2\,m^2+3\,b^3\,m\,\sin \left (3\,c+3\,d\,x\right )+b^3\,m^2\,\sin \left (c+d\,x\right )+b^3\,m^2\,\sin \left (3\,c+3\,d\,x\right )+18\,a\,b^2\,m+11\,b^3\,m\,\sin \left (c+d\,x\right )+8\,a^2\,b\,m\,\sin \left (c+d\,x\right )-2\,a\,b^2\,m\,\left (2\,{\sin \left (c+d\,x\right )}^2-1\right )-2\,a\,b^2\,m^2\,\left (2\,{\sin \left (c+d\,x\right )}^2-1\right )\right )}{4\,b^3\,d\,\left (m^3+6\,m^2+11\,m+6\right )} \]

[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))