3.271 \(\int \frac{(d \csc (a+b x))^{9/2}}{(c \sec (a+b x))^{5/2}} \, dx\)

Optimal. Leaf size=33 \[ -\frac{2 c d (d \csc (a+b x))^{7/2}}{7 b (c \sec (a+b x))^{7/2}} \]

[Out]

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

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Rubi [A]  time = 0.0540944, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.04, Rules used = {2619} \[ -\frac{2 c d (d \csc (a+b x))^{7/2}}{7 b (c \sec (a+b x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(d*Csc[a + b*x])^(9/2)/(c*Sec[a + b*x])^(5/2),x]

[Out]

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

Rule 2619

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*b*(a*Csc[e
 + f*x])^(m - 1)*(b*Sec[e + f*x])^(n - 1))/(f*(n - 1)), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n - 2, 0
] && NeQ[n, 1]

Rubi steps

\begin{align*} \int \frac{(d \csc (a+b x))^{9/2}}{(c \sec (a+b x))^{5/2}} \, dx &=-\frac{2 c d (d \csc (a+b x))^{7/2}}{7 b (c \sec (a+b x))^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.158198, size = 45, normalized size = 1.36 \[ -\frac{2 d^4 \cot ^3(a+b x) \sqrt{d \csc (a+b x)}}{7 b c^2 \sqrt{c \sec (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Csc[a + b*x])^(9/2)/(c*Sec[a + b*x])^(5/2),x]

[Out]

(-2*d^4*Cot[a + b*x]^3*Sqrt[d*Csc[a + b*x]])/(7*b*c^2*Sqrt[c*Sec[a + b*x]])

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Maple [A]  time = 0.138, size = 42, normalized size = 1.3 \begin{align*} -{\frac{2\,\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) }{7\,b} \left ({\frac{d}{\sin \left ( bx+a \right ) }} \right ) ^{{\frac{9}{2}}} \left ({\frac{c}{\cos \left ( bx+a \right ) }} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*csc(b*x+a))^(9/2)/(c*sec(b*x+a))^(5/2),x)

[Out]

-2/7/b*cos(b*x+a)*sin(b*x+a)*(d/sin(b*x+a))^(9/2)/(c/cos(b*x+a))^(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \csc \left (b x + a\right )\right )^{\frac{9}{2}}}{\left (c \sec \left (b x + a\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*csc(b*x+a))^(9/2)/(c*sec(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

integrate((d*csc(b*x + a))^(9/2)/(c*sec(b*x + a))^(5/2), x)

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Fricas [B]  time = 2.30939, size = 151, normalized size = 4.58 \begin{align*} \frac{2 \, d^{4} \sqrt{\frac{c}{\cos \left (b x + a\right )}} \sqrt{\frac{d}{\sin \left (b x + a\right )}} \cos \left (b x + a\right )^{4}}{7 \,{\left (b c^{3} \cos \left (b x + a\right )^{2} - b c^{3}\right )} \sin \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*csc(b*x+a))^(9/2)/(c*sec(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*csc(b*x+a))**(9/2)/(c*sec(b*x+a))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \csc \left (b x + a\right )\right )^{\frac{9}{2}}}{\left (c \sec \left (b x + a\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*csc(b*x+a))^(9/2)/(c*sec(b*x+a))^(5/2),x, algorithm="giac")

[Out]

integrate((d*csc(b*x + a))^(9/2)/(c*sec(b*x + a))^(5/2), x)