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

Optimal. Leaf size=33 \[ -\frac {2 c d (d \csc (a+b x))^{3/2}}{3 b (c \sec (a+b x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2619} \[ -\frac {2 c d (d \csc (a+b x))^{3/2}}{3 b (c \sec (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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))^{5/2}}{\sqrt {c \sec (a+b x)}} \, dx &=-\frac {2 c d (d \csc (a+b x))^{3/2}}{3 b (c \sec (a+b x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 33, normalized size = 1.00 \[ -\frac {2 c d (d \csc (a+b x))^{3/2}}{3 b (c \sec (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.77, size = 51, normalized size = 1.55 \[ -\frac {2 \, d^{2} \sqrt {\frac {c}{\cos \left (b x + a\right )}} \sqrt {\frac {d}{\sin \left (b x + a\right )}} \cos \left (b x + a\right )^{2}}{3 \, b c \sin \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*csc(b*x+a))^(5/2)/(c*sec(b*x+a))^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \csc \left (b x + a\right )\right )^{\frac {5}{2}}}{\sqrt {c \sec \left (b x + a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*csc(b*x+a))^(5/2)/(c*sec(b*x+a))^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 1.16, size = 42, normalized size = 1.27 \[ -\frac {2 \left (\frac {d}{\sin \left (b x +a \right )}\right )^{\frac {5}{2}} \cos \left (b x +a \right ) \sin \left (b x +a \right )}{3 b \sqrt {\frac {c}{\cos \left (b x +a \right )}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*csc(b*x+a))^(5/2)/(c*sec(b*x+a))^(1/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \csc \left (b x + a\right )\right )^{\frac {5}{2}}}{\sqrt {c \sec \left (b x + a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*csc(b*x+a))^(5/2)/(c*sec(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.80, size = 49, normalized size = 1.48 \[ -\frac {d^2\,\sin \left (2\,a+2\,b\,x\right )\,\sqrt {\frac {d}{\sin \left (a+b\,x\right )}}}{3\,b\,{\sin \left (a+b\,x\right )}^2\,\sqrt {\frac {c}{\cos \left (a+b\,x\right )}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d/sin(a + b*x))^(5/2)/(c/cos(a + b*x))^(1/2),x)

[Out]

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

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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((d*csc(b*x+a))**(5/2)/(c*sec(b*x+a))**(1/2),x)

[Out]

Timed out

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