3.210 \(\int \frac {\sin ^3(a+b x)}{(d \cos (a+b x))^{11/2}} \, dx\)

Optimal. Leaf size=45 \[ \frac {2}{9 b d (d \cos (a+b x))^{9/2}}-\frac {2}{5 b d^3 (d \cos (a+b x))^{5/2}} \]

[Out]

2/9/b/d/(d*cos(b*x+a))^(9/2)-2/5/b/d^3/(d*cos(b*x+a))^(5/2)

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Rubi [A]  time = 0.05, antiderivative size = 45, 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} \[ \frac {2}{9 b d (d \cos (a+b x))^{9/2}}-\frac {2}{5 b d^3 (d \cos (a+b x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^3/(d*Cos[a + b*x])^(11/2),x]

[Out]

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

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)}{(d \cos (a+b x))^{11/2}} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {1-\frac {x^2}{d^2}}{x^{11/2}} \, dx,x,d \cos (a+b x)\right )}{b d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {1}{x^{11/2}}-\frac {1}{d^2 x^{7/2}}\right ) \, dx,x,d \cos (a+b x)\right )}{b d}\\ &=\frac {2}{9 b d (d \cos (a+b x))^{9/2}}-\frac {2}{5 b d^3 (d \cos (a+b x))^{5/2}}\\ \end {align*}

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Mathematica [B]  time = 0.53, size = 94, normalized size = 2.09 \[ \frac {2 \tan ^4(a+b x) \left (4 \sqrt [4]{\cos ^2(a+b x)}+4 \left (\sqrt [4]{\cos ^2(a+b x)}-1\right ) \csc ^4(a+b x)+\left (9-8 \sqrt [4]{\cos ^2(a+b x)}\right ) \csc ^2(a+b x)\right )}{45 b d^5 \sqrt {d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^3/(d*Cos[a + b*x])^(11/2),x]

[Out]

(2*(4*(Cos[a + b*x]^2)^(1/4) + (9 - 8*(Cos[a + b*x]^2)^(1/4))*Csc[a + b*x]^2 + 4*(-1 + (Cos[a + b*x]^2)^(1/4))
*Csc[a + b*x]^4)*Tan[a + b*x]^4)/(45*b*d^5*Sqrt[d*Cos[a + b*x]])

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fricas [A]  time = 0.43, size = 38, normalized size = 0.84 \[ -\frac {2 \, \sqrt {d \cos \left (b x + a\right )} {\left (9 \, \cos \left (b x + a\right )^{2} - 5\right )}}{45 \, b d^{6} \cos \left (b x + a\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3/(d*cos(b*x+a))^(11/2),x, algorithm="fricas")

[Out]

-2/45*sqrt(d*cos(b*x + a))*(9*cos(b*x + a)^2 - 5)/(b*d^6*cos(b*x + a)^5)

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giac [A]  time = 1.40, size = 51, normalized size = 1.13 \[ -\frac {2 \, {\left (9 \, b^{5} d^{5} \cos \left (b x + a\right )^{2} - 5 \, b^{5} d^{5}\right )}}{45 \, \sqrt {d \cos \left (b x + a\right )} b^{6} d^{10} \cos \left (b x + a\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3/(d*cos(b*x+a))^(11/2),x, algorithm="giac")

[Out]

-2/45*(9*b^5*d^5*cos(b*x + a)^2 - 5*b^5*d^5)/(sqrt(d*cos(b*x + a))*b^6*d^10*cos(b*x + a)^4)

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maple [B]  time = 0.29, size = 124, normalized size = 2.76 \[ \frac {8 \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}\, \left (9 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-9 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+1\right )}{45 d^{6} \left (32 \left (\sin ^{10}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-80 \left (\sin ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+80 \left (\sin ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-40 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+10 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^3/(d*cos(b*x+a))^(11/2),x)

[Out]

8/45/d^6/(32*sin(1/2*b*x+1/2*a)^10-80*sin(1/2*b*x+1/2*a)^8+80*sin(1/2*b*x+1/2*a)^6-40*sin(1/2*b*x+1/2*a)^4+10*
sin(1/2*b*x+1/2*a)^2-1)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*(9*sin(1/2*b*x+1/2*a)^4-9*sin(1/2*b*x+1/2*a)^2+1)/
b

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maxima [A]  time = 0.54, size = 37, normalized size = 0.82 \[ -\frac {2 \, {\left (9 \, d^{2} \cos \left (b x + a\right )^{2} - 5 \, d^{2}\right )}}{45 \, \left (d \cos \left (b x + a\right )\right )^{\frac {9}{2}} b d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3/(d*cos(b*x+a))^(11/2),x, algorithm="maxima")

[Out]

-2/45*(9*d^2*cos(b*x + a)^2 - 5*d^2)/((d*cos(b*x + a))^(9/2)*b*d^3)

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mupad [B]  time = 5.33, size = 279, normalized size = 6.20 \[ \frac {16\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {d\,\left (\frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}-b\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}}{2}\right )}}{5\,b\,d^6\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^2}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {d\,\left (\frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}-b\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}}{2}\right )}\,464{}\mathrm {i}}{45\,b\,d^6\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^3}-\frac {128\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {d\,\left (\frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}-b\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}}{2}\right )}}{9\,b\,d^6\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^4}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {d\,\left (\frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}-b\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}}{2}\right )}\,64{}\mathrm {i}}{9\,b\,d^6\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^3/(d*cos(a + b*x))^(11/2),x)

[Out]

(16*exp(a*1i + b*x*1i)*(d*(exp(- a*1i - b*x*1i)/2 + exp(a*1i + b*x*1i)/2))^(1/2))/(5*b*d^6*(exp(a*2i + b*x*2i)
*1i + 1i)^2) - (exp(a*1i + b*x*1i)*(d*(exp(- a*1i - b*x*1i)/2 + exp(a*1i + b*x*1i)/2))^(1/2)*464i)/(45*b*d^6*(
exp(a*2i + b*x*2i)*1i + 1i)^3) - (128*exp(a*1i + b*x*1i)*(d*(exp(- a*1i - b*x*1i)/2 + exp(a*1i + b*x*1i)/2))^(
1/2))/(9*b*d^6*(exp(a*2i + b*x*2i)*1i + 1i)^4) + (exp(a*1i + b*x*1i)*(d*(exp(- a*1i - b*x*1i)/2 + exp(a*1i + b
*x*1i)/2))^(1/2)*64i)/(9*b*d^6*(exp(a*2i + b*x*2i)*1i + 1i)^5)

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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(sin(b*x+a)**3/(d*cos(b*x+a))**(11/2),x)

[Out]

Timed out

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