\(\int \frac {\sin ^3(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx\) [205]

Optimal result

Integrand size = 21, antiderivative size = 43 \[ \int \frac {\sin ^3(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=-\frac {2 \sqrt {d \cos (a+b x)}}{b d}+\frac {2 (d \cos (a+b x))^{5/2}}{5 b d^3} \]

[Out]

2/5*(d*cos(b*x+a))^(5/2)/b/d^3-2*(d*cos(b*x+a))^(1/2)/b/d

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 43, 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 = {2645, 14} \[ \int \frac {\sin ^3(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\frac {2 (d \cos (a+b x))^{5/2}}{5 b d^3}-\frac {2 \sqrt {d \cos (a+b x)}}{b d} \]

[In]

Int[Sin[a + b*x]^3/Sqrt[d*Cos[a + b*x]],x]

[Out]

(-2*Sqrt[d*Cos[a + b*x]])/(b*d) + (2*(d*Cos[a + b*x])^(5/2))/(5*b*d^3)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 \frac {1-\frac {x^2}{d^2}}{\sqrt {x}} \, dx,x,d \cos (a+b x)\right )}{b d} \\ & = -\frac {\text {Subst}\left (\int \left (\frac {1}{\sqrt {x}}-\frac {x^{3/2}}{d^2}\right ) \, dx,x,d \cos (a+b x)\right )}{b d} \\ & = -\frac {2 \sqrt {d \cos (a+b x)}}{b d}+\frac {2 (d \cos (a+b x))^{5/2}}{5 b d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.33 \[ \int \frac {\sin ^3(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\frac {\cos (a+b x) (-9+\cos (2 (a+b x)))+8 \cos ^2(a+b x)^{3/4} \sec (a+b x)}{5 b \sqrt {d \cos (a+b x)}} \]

[In]

Integrate[Sin[a + b*x]^3/Sqrt[d*Cos[a + b*x]],x]

[Out]

(Cos[a + b*x]*(-9 + Cos[2*(a + b*x)]) + 8*(Cos[a + b*x]^2)^(3/4)*Sec[a + b*x])/(5*b*Sqrt[d*Cos[a + b*x]])

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {\frac {2 \left (d \cos \left (b x +a \right )\right )^{\frac {5}{2}}}{5}-2 d^{2} \sqrt {d \cos \left (b x +a \right )}}{b \,d^{3}}\)	\(37\)
default	\(\frac {\frac {2 \left (d \cos \left (b x +a \right )\right )^{\frac {5}{2}}}{5}-2 d^{2} \sqrt {d \cos \left (b x +a \right )}}{b \,d^{3}}\)	\(37\)

[In]

int(sin(b*x+a)^3/(d*cos(b*x+a))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/b/d^3*(1/5*(d*cos(b*x+a))^(5/2)-d^2*(d*cos(b*x+a))^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.65 \[ \int \frac {\sin ^3(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\frac {2 \, \sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right )^{2} - 5\right )}}{5 \, b d} \]

[In]

integrate(sin(b*x+a)^3/(d*cos(b*x+a))^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.78 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.65 \[ \int \frac {\sin ^3(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\begin {cases} - \frac {2 \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{b \sqrt {d \cos {\left (a + b x \right )}}} - \frac {8 \cos ^{3}{\left (a + b x \right )}}{5 b \sqrt {d \cos {\left (a + b x \right )}}} & \text {for}\: b \neq 0 \\\frac {x \sin ^{3}{\left (a \right )}}{\sqrt {d \cos {\left (a \right )}}} & \text {otherwise} \end {cases} \]

[In]

integrate(sin(b*x+a)**3/(d*cos(b*x+a))**(1/2),x)

[Out]

Piecewise((-2*sin(a + b*x)**2*cos(a + b*x)/(b*sqrt(d*cos(a + b*x))) - 8*cos(a + b*x)**3/(5*b*sqrt(d*cos(a + b*
x))), Ne(b, 0)), (x*sin(a)**3/sqrt(d*cos(a)), True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84 \[ \int \frac {\sin ^3(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=-\frac {2 \, {\left (5 \, \sqrt {d \cos \left (b x + a\right )} - \frac {\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}}}{d^{2}}\right )}}{5 \, b d} \]

[In]

integrate(sin(b*x+a)^3/(d*cos(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.07 \[ \int \frac {\sin ^3(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\frac {2 \, {\left (\sqrt {d \cos \left (b x + a\right )} d^{2} \cos \left (b x + a\right )^{2} - 5 \, \sqrt {d \cos \left (b x + a\right )} d^{2}\right )}}{5 \, b d^{3}} \]

[In]

integrate(sin(b*x+a)^3/(d*cos(b*x+a))^(1/2),x, algorithm="giac")

[Out]

2/5*(sqrt(d*cos(b*x + a))*d^2*cos(b*x + a)^2 - 5*sqrt(d*cos(b*x + a))*d^2)/(b*d^3)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^3(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\int \frac {{\sin \left (a+b\,x\right )}^3}{\sqrt {d\,\cos \left (a+b\,x\right )}} \,d x \]

[In]

int(sin(a + b*x)^3/(d*cos(a + b*x))^(1/2),x)

[Out]

int(sin(a + b*x)^3/(d*cos(a + b*x))^(1/2), x)