3.3.29 \(\int (d \sec (a+b x))^{9/2} \sin ^3(a+b x) \, dx\) [229]

Optimal. Leaf size=43 \[ -\frac {2 d^3 (d \sec (a+b x))^{3/2}}{3 b}+\frac {2 d (d \sec (a+b x))^{7/2}}{7 b} \]

[Out]

-2/3*d^3*(d*sec(b*x+a))^(3/2)/b+2/7*d*(d*sec(b*x+a))^(7/2)/b

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Rubi [A]
time = 0.03, 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 = {2702, 14} \begin {gather*} \frac {2 d (d \sec (a+b x))^{7/2}}{7 b}-\frac {2 d^3 (d \sec (a+b x))^{3/2}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*Sec[a + b*x])^(9/2)*Sin[a + b*x]^3,x]

[Out]

(-2*d^3*(d*Sec[a + b*x])^(3/2))/(3*b) + (2*d*(d*Sec[a + b*x])^(7/2))/(7*b)

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 2702

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

Rubi steps

\begin {align*} \int (d \sec (a+b x))^{9/2} \sin ^3(a+b x) \, dx &=\frac {d^3 \text {Subst}\left (\int \sqrt {x} \left (-1+\frac {x^2}{d^2}\right ) \, dx,x,d \sec (a+b x)\right )}{b}\\ &=\frac {d^3 \text {Subst}\left (\int \left (-\sqrt {x}+\frac {x^{5/2}}{d^2}\right ) \, dx,x,d \sec (a+b x)\right )}{b}\\ &=-\frac {2 d^3 (d \sec (a+b x))^{3/2}}{3 b}+\frac {2 d (d \sec (a+b x))^{7/2}}{7 b}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 42, normalized size = 0.98 \begin {gather*} -\frac {d^4 (1+7 \cos (2 (a+b x))) \sec ^3(a+b x) \sqrt {d \sec (a+b x)}}{21 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[a + b*x])^(9/2)*Sin[a + b*x]^3,x]

[Out]

-1/21*(d^4*(1 + 7*Cos[2*(a + b*x)])*Sec[a + b*x]^3*Sqrt[d*Sec[a + b*x]])/b

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Maple [A]
time = 82.17, size = 36, normalized size = 0.84

method	result	size
default	\(-\frac {2 \left (7 \left (\cos ^{2}\left (b x +a \right )\right )-3\right ) \cos \left (b x +a \right ) \left (\frac {d}{\cos \left (b x +a \right )}\right )^{\frac {9}{2}}}{21 b}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(b*x+a))^(9/2)*sin(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

-2/21/b*(7*cos(b*x+a)^2-3)*cos(b*x+a)*(d/cos(b*x+a))^(9/2)

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Maxima [A]
time = 0.28, size = 38, normalized size = 0.88 \begin {gather*} -\frac {2 \, {\left (7 \, d^{2} \left (\frac {d}{\cos \left (b x + a\right )}\right )^{\frac {3}{2}} - 3 \, \left (\frac {d}{\cos \left (b x + a\right )}\right )^{\frac {7}{2}}\right )} d}{21 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(b*x+a))^(9/2)*sin(b*x+a)^3,x, algorithm="maxima")

[Out]

-2/21*(7*d^2*(d/cos(b*x + a))^(3/2) - 3*(d/cos(b*x + a))^(7/2))*d/b

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Fricas [A]
time = 2.10, size = 44, normalized size = 1.02 \begin {gather*} -\frac {2 \, {\left (7 \, d^{4} \cos \left (b x + a\right )^{2} - 3 \, d^{4}\right )} \sqrt {\frac {d}{\cos \left (b x + a\right )}}}{21 \, b \cos \left (b x + a\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(b*x+a))^(9/2)*sin(b*x+a)^3,x, algorithm="fricas")

[Out]

-2/21*(7*d^4*cos(b*x + a)^2 - 3*d^4)*sqrt(d/cos(b*x + a))/(b*cos(b*x + a)^3)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(b*x+a))**(9/2)*sin(b*x+a)**3,x)

[Out]

Timed out

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Giac [A]
time = 0.47, size = 49, normalized size = 1.14 \begin {gather*} -\frac {2 \, {\left (7 \, d^{5} \cos \left (b x + a\right )^{2} - 3 \, d^{5}\right )} \mathrm {sgn}\left (\cos \left (b x + a\right )\right )}{21 \, \sqrt {d \cos \left (b x + a\right )} b \cos \left (b x + a\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(b*x+a))^(9/2)*sin(b*x+a)^3,x, algorithm="giac")

[Out]

-2/21*(7*d^5*cos(b*x + a)^2 - 3*d^5)*sgn(cos(b*x + a))/(sqrt(d*cos(b*x + a))*b*cos(b*x + a)^3)

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Mupad [B]
time = 4.34, size = 95, normalized size = 2.21 \begin {gather*} -\frac {4\,d^4\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {d}{\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}}}\,\left (2\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+7\,{\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+7\right )}{21\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^3*(d/cos(a + b*x))^(9/2),x)

[Out]

-(4*d^4*exp(a*1i + b*x*1i)*(d/(exp(- a*1i - b*x*1i)/2 + exp(a*1i + b*x*1i)/2))^(1/2)*(2*exp(a*2i + b*x*2i) + 7
*exp(a*4i + b*x*4i) + 7))/(21*b*(exp(a*2i + b*x*2i) + 1)^3)

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