\(\int (d \cos (a+b x))^{3/2} \sin (a+b x) \, dx\) [187]

Optimal result

Integrand size = 19, antiderivative size = 22 \[ \int (d \cos (a+b x))^{3/2} \sin (a+b x) \, dx=-\frac {2 (d \cos (a+b x))^{5/2}}{5 b d} \]

[Out]

-2/5*(d*cos(b*x+a))^(5/2)/b/d

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2645, 30} \[ \int (d \cos (a+b x))^{3/2} \sin (a+b x) \, dx=-\frac {2 (d \cos (a+b x))^{5/2}}{5 b d} \]

[In]

Int[(d*Cos[a + b*x])^(3/2)*Sin[a + b*x],x]

[Out]

(-2*(d*Cos[a + b*x])^(5/2))/(5*b*d)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int x^{3/2} \, dx,x,d \cos (a+b x)\right )}{b d} \\ & = -\frac {2 (d \cos (a+b x))^{5/2}}{5 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int (d \cos (a+b x))^{3/2} \sin (a+b x) \, dx=-\frac {2 (d \cos (a+b x))^{5/2}}{5 b d} \]

[In]

Integrate[(d*Cos[a + b*x])^(3/2)*Sin[a + b*x],x]

[Out]

(-2*(d*Cos[a + b*x])^(5/2))/(5*b*d)

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(-\frac {2 \left (d \cos \left (b x +a \right )\right )^{\frac {5}{2}}}{5 b d}\)	\(19\)
default	\(-\frac {2 \left (d \cos \left (b x +a \right )\right )^{\frac {5}{2}}}{5 b d}\)	\(19\)

[In]

int((d*cos(b*x+a))^(3/2)*sin(b*x+a),x,method=_RETURNVERBOSE)

[Out]

-2/5*(d*cos(b*x+a))^(5/2)/b/d

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int (d \cos (a+b x))^{3/2} \sin (a+b x) \, dx=-\frac {2 \, \sqrt {d \cos \left (b x + a\right )} d \cos \left (b x + a\right )^{2}}{5 \, b} \]

[In]

integrate((d*cos(b*x+a))^(3/2)*sin(b*x+a),x, algorithm="fricas")

[Out]

-2/5*sqrt(d*cos(b*x + a))*d*cos(b*x + a)^2/b

Sympy [A] (verification not implemented)

Time = 2.89 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int (d \cos (a+b x))^{3/2} \sin (a+b x) \, dx=\begin {cases} - \frac {2 \left (d \cos {\left (a + b x \right )}\right )^{\frac {3}{2}} \cos {\left (a + b x \right )}}{5 b} & \text {for}\: b \neq 0 \\x \left (d \cos {\left (a \right )}\right )^{\frac {3}{2}} \sin {\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

integrate((d*cos(b*x+a))**(3/2)*sin(b*x+a),x)

[Out]

Piecewise((-2*(d*cos(a + b*x))**(3/2)*cos(a + b*x)/(5*b), Ne(b, 0)), (x*(d*cos(a))**(3/2)*sin(a), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int (d \cos (a+b x))^{3/2} \sin (a+b x) \, dx=-\frac {2 \, \left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}}}{5 \, b d} \]

[In]

integrate((d*cos(b*x+a))^(3/2)*sin(b*x+a),x, algorithm="maxima")

[Out]

-2/5*(d*cos(b*x + a))^(5/2)/(b*d)

Giac [F]

\[ \int (d \cos (a+b x))^{3/2} \sin (a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}} \sin \left (b x + a\right ) \,d x } \]

[In]

integrate((d*cos(b*x+a))^(3/2)*sin(b*x+a),x, algorithm="giac")

[Out]

integrate((d*cos(b*x + a))^(3/2)*sin(b*x + a), x)

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int (d \cos (a+b x))^{3/2} \sin (a+b x) \, dx=-\frac {2\,{\left (d\,\cos \left (a+b\,x\right )\right )}^{5/2}}{5\,b\,d} \]

[In]

int(sin(a + b*x)*(d*cos(a + b*x))^(3/2),x)

[Out]

-(2*(d*cos(a + b*x))^(5/2))/(5*b*d)