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

Optimal. Leaf size=43 \[ \frac {2 (d \cos (a+b x))^{5/2}}{5 b d^3}-\frac {2 \sqrt {d \cos (a+b x)}}{b d} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 43, 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 = {2565, 14} \[ \frac {2 (d \cos (a+b x))^{5/2}}{5 b d^3}-\frac {2 \sqrt {d \cos (a+b x)}}{b d} \]

Antiderivative was successfully verified.

[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 2565

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

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Mathematica [A]  time = 0.18, size = 57, normalized size = 1.33 \[ \frac {\cos (a+b x) (\cos (2 (a+b x))-9)+8 \cos ^2(a+b x)^{3/4} \sec (a+b x)}{5 b \sqrt {d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[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]])

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fricas [A]  time = 0.44, size = 28, normalized size = 0.65 \[ \frac {2 \, \sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right )^{2} - 5\right )}}{5 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 1.03, size = 40, normalized size = 0.93 \[ \frac {2 \, {\left (\sqrt {d \cos \left (b x + a\right )} \cos \left (b x + a\right )^{2} - 5 \, \sqrt {d \cos \left (b x + a\right )}\right )}}{5 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

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

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maple [B]  time = 0.05, size = 92, normalized size = 2.14 \[ \frac {8 \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}\, \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-8 \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}\, \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-8 \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}}{5 d b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(8*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*sin(1/2*b*x+1/2*a)^4-8*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*sin(1/2*
b*x+1/2*a)^2-8*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2))/d/b

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maxima [A]  time = 0.41, size = 36, normalized size = 0.84 \[ -\frac {2 \, {\left (5 \, \sqrt {d \cos \left (b x + a\right )} - \frac {\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}}}{d^{2}}\right )}}{5 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\sin \left (a+b\,x\right )}^3}{\sqrt {d\,\cos \left (a+b\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 6.17, size = 63, normalized size = 1.47 \[ \begin {cases} - \frac {2 \sin ^{2}{\left (a + b x \right )} \sqrt {\cos {\left (a + b x \right )}}}{b \sqrt {d}} - \frac {8 \cos ^{\frac {5}{2}}{\left (a + b x \right )}}{5 b \sqrt {d}} & \text {for}\: b \neq 0 \\\frac {x \sin ^{3}{\relax (a )}}{\sqrt {d \cos {\relax (a )}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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