3.262 \(\int \frac{(d \csc (a+b x))^{11/2}}{(c \sec (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=110 \[ \frac{8 d^5 \sqrt{d \csc (a+b x)}}{45 b c \sqrt{c \sec (a+b x)}}+\frac{2 d^3 (d \csc (a+b x))^{5/2}}{45 b c \sqrt{c \sec (a+b x)}}-\frac{2 d (d \csc (a+b x))^{9/2}}{9 b c \sqrt{c \sec (a+b x)}} \]

[Out]

(8*d^5*Sqrt[d*Csc[a + b*x]])/(45*b*c*Sqrt[c*Sec[a + b*x]]) + (2*d^3*(d*Csc[a + b*x])^(5/2))/(45*b*c*Sqrt[c*Sec
[a + b*x]]) - (2*d*(d*Csc[a + b*x])^(9/2))/(9*b*c*Sqrt[c*Sec[a + b*x]])

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Rubi [A]  time = 0.153996, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2623, 2625, 2619} \[ \frac{8 d^5 \sqrt{d \csc (a+b x)}}{45 b c \sqrt{c \sec (a+b x)}}+\frac{2 d^3 (d \csc (a+b x))^{5/2}}{45 b c \sqrt{c \sec (a+b x)}}-\frac{2 d (d \csc (a+b x))^{9/2}}{9 b c \sqrt{c \sec (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[(d*Csc[a + b*x])^(11/2)/(c*Sec[a + b*x])^(3/2),x]

[Out]

(8*d^5*Sqrt[d*Csc[a + b*x]])/(45*b*c*Sqrt[c*Sec[a + b*x]]) + (2*d^3*(d*Csc[a + b*x])^(5/2))/(45*b*c*Sqrt[c*Sec
[a + b*x]]) - (2*d*(d*Csc[a + b*x])^(9/2))/(9*b*c*Sqrt[c*Sec[a + b*x]])

Rule 2623

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

Rule 2625

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*(m - 1)), x] + Dist[(a^2*(m + n - 2))/(m - 1), Int[(a*Csc[e +
f*x])^(m - 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && IntegersQ[2*m, 2*n] &&
!GtQ[n, m]

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))^{11/2}}{(c \sec (a+b x))^{3/2}} \, dx &=-\frac{2 d (d \csc (a+b x))^{9/2}}{9 b c \sqrt{c \sec (a+b x)}}-\frac{d^2 \int (d \csc (a+b x))^{7/2} \sqrt{c \sec (a+b x)} \, dx}{9 c^2}\\ &=\frac{2 d^3 (d \csc (a+b x))^{5/2}}{45 b c \sqrt{c \sec (a+b x)}}-\frac{2 d (d \csc (a+b x))^{9/2}}{9 b c \sqrt{c \sec (a+b x)}}-\frac{\left (4 d^4\right ) \int (d \csc (a+b x))^{3/2} \sqrt{c \sec (a+b x)} \, dx}{45 c^2}\\ &=\frac{8 d^5 \sqrt{d \csc (a+b x)}}{45 b c \sqrt{c \sec (a+b x)}}+\frac{2 d^3 (d \csc (a+b x))^{5/2}}{45 b c \sqrt{c \sec (a+b x)}}-\frac{2 d (d \csc (a+b x))^{9/2}}{9 b c \sqrt{c \sec (a+b x)}}\\ \end{align*}

Mathematica [A]  time = 0.280943, size = 57, normalized size = 0.52 \[ \frac{2 d^3 (2 \cos (2 (a+b x))-7) \cot ^2(a+b x) (d \csc (a+b x))^{5/2}}{45 b c \sqrt{c \sec (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Csc[a + b*x])^(11/2)/(c*Sec[a + b*x])^(3/2),x]

[Out]

(2*d^3*(-7 + 2*Cos[2*(a + b*x)])*Cot[a + b*x]^2*(d*Csc[a + b*x])^(5/2))/(45*b*c*Sqrt[c*Sec[a + b*x]])

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Maple [A]  time = 0.158, size = 54, normalized size = 0.5 \begin{align*}{\frac{ \left ( 8\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-18 \right ) \cos \left ( bx+a \right ) \sin \left ( bx+a \right ) }{45\,b} \left ({\frac{d}{\sin \left ( bx+a \right ) }} \right ) ^{{\frac{11}{2}}} \left ({\frac{c}{\cos \left ( bx+a \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*csc(b*x+a))^(11/2)/(c*sec(b*x+a))^(3/2),x)

[Out]

2/45/b*(4*cos(b*x+a)^2-9)*(d/sin(b*x+a))^(11/2)*cos(b*x+a)*sin(b*x+a)/(c/cos(b*x+a))^(3/2)

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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{11}{2}}}{\left (c \sec \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*csc(b*x+a))^(11/2)/(c*sec(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

integrate((d*csc(b*x + a))^(11/2)/(c*sec(b*x + a))^(3/2), x)

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Fricas [A]  time = 2.21246, size = 203, normalized size = 1.85 \begin{align*} \frac{2 \,{\left (4 \, d^{5} \cos \left (b x + a\right )^{5} - 9 \, d^{5} \cos \left (b x + a\right )^{3}\right )} \sqrt{\frac{c}{\cos \left (b x + a\right )}} \sqrt{\frac{d}{\sin \left (b x + a\right )}}}{45 \,{\left (b c^{2} \cos \left (b x + a\right )^{4} - 2 \, b c^{2} \cos \left (b x + a\right )^{2} + b c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*csc(b*x+a))^(11/2)/(c*sec(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

2/45*(4*d^5*cos(b*x + a)^5 - 9*d^5*cos(b*x + a)^3)*sqrt(c/cos(b*x + a))*sqrt(d/sin(b*x + a))/(b*c^2*cos(b*x +
a)^4 - 2*b*c^2*cos(b*x + a)^2 + b*c^2)

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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))**(11/2)/(c*sec(b*x+a))**(3/2),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{11}{2}}}{\left (c \sec \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*csc(b*x+a))^(11/2)/(c*sec(b*x+a))^(3/2),x, algorithm="giac")

[Out]

integrate((d*csc(b*x + a))^(11/2)/(c*sec(b*x + a))^(3/2), x)