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

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

[Out]

2/3*c*d*(c*sec(b*x+a))^(3/2)/b/(d*csc(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} \[ \frac {2 c d (c \sec (a+b x))^{3/2}}{3 b (d \csc (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.11, size = 33, normalized size = 1.00 \[ \frac {2 c d (c \sec (a+b x))^{3/2}}{3 b (d \csc (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.78, size = 58, normalized size = 1.76 \[ -\frac {2 \, {\left (c^{2} \cos \left (b x + a\right )^{2} - c^{2}\right )} \sqrt {\frac {c}{\cos \left (b x + a\right )}} \sqrt {\frac {d}{\sin \left (b x + a\right )}}}{3 \, b d \cos \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.99, size = 42, normalized size = 1.27 \[ \frac {2 \left (\frac {c}{\cos \left (b x +a \right )}\right )^{\frac {5}{2}} \cos \left (b x +a \right ) \sin \left (b x +a \right )}{3 b \sqrt {\frac {d}{\sin \left (b x +a \right )}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.86, size = 66, normalized size = 2.00 \[ \frac {c^2\,\sqrt {\frac {c}{\cos \left (a+b\,x\right )}}\,\sqrt {\frac {d}{\sin \left (a+b\,x\right )}}\,\left (\cos \left (a+b\,x\right )-\cos \left (3\,a+3\,b\,x\right )\right )}{3\,b\,d\,\left (\cos \left (2\,a+2\,b\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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