3.247 \(\int (d \csc (a+b x))^{7/2} (c \sec (a+b x))^{5/2} \, dx\)

Optimal. Leaf size=106 \[ -\frac{64 c^3 d^3 \sqrt{d \csc (a+b x)}}{15 b \sqrt{c \sec (a+b x)}}+\frac{16 c d^3 (c \sec (a+b x))^{3/2} \sqrt{d \csc (a+b x)}}{15 b}-\frac{2 c d (c \sec (a+b x))^{3/2} (d \csc (a+b x))^{5/2}}{5 b} \]

[Out]

(-64*c^3*d^3*Sqrt[d*Csc[a + b*x]])/(15*b*Sqrt[c*Sec[a + b*x]]) + (16*c*d^3*Sqrt[d*Csc[a + b*x]]*(c*Sec[a + b*x
])^(3/2))/(15*b) - (2*c*d*(d*Csc[a + b*x])^(5/2)*(c*Sec[a + b*x])^(3/2))/(5*b)

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Rubi [A]  time = 0.162101, antiderivative size = 106, 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 = {2625, 2626, 2619} \[ -\frac{64 c^3 d^3 \sqrt{d \csc (a+b x)}}{15 b \sqrt{c \sec (a+b x)}}+\frac{16 c d^3 (c \sec (a+b x))^{3/2} \sqrt{d \csc (a+b x)}}{15 b}-\frac{2 c d (c \sec (a+b x))^{3/2} (d \csc (a+b x))^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[(d*Csc[a + b*x])^(7/2)*(c*Sec[a + b*x])^(5/2),x]

[Out]

(-64*c^3*d^3*Sqrt[d*Csc[a + b*x]])/(15*b*Sqrt[c*Sec[a + b*x]]) + (16*c*d^3*Sqrt[d*Csc[a + b*x]]*(c*Sec[a + b*x
])^(3/2))/(15*b) - (2*c*d*(d*Csc[a + b*x])^(5/2)*(c*Sec[a + b*x])^(3/2))/(5*b)

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 2626

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

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 (d \csc (a+b x))^{7/2} (c \sec (a+b x))^{5/2} \, dx &=-\frac{2 c d (d \csc (a+b x))^{5/2} (c \sec (a+b x))^{3/2}}{5 b}+\frac{1}{5} \left (8 d^2\right ) \int (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{5/2} \, dx\\ &=\frac{16 c d^3 \sqrt{d \csc (a+b x)} (c \sec (a+b x))^{3/2}}{15 b}-\frac{2 c d (d \csc (a+b x))^{5/2} (c \sec (a+b x))^{3/2}}{5 b}+\frac{1}{15} \left (32 c^2 d^2\right ) \int (d \csc (a+b x))^{3/2} \sqrt{c \sec (a+b x)} \, dx\\ &=-\frac{64 c^3 d^3 \sqrt{d \csc (a+b x)}}{15 b \sqrt{c \sec (a+b x)}}+\frac{16 c d^3 \sqrt{d \csc (a+b x)} (c \sec (a+b x))^{3/2}}{15 b}-\frac{2 c d (d \csc (a+b x))^{5/2} (c \sec (a+b x))^{3/2}}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.193343, size = 57, normalized size = 0.54 \[ -\frac{2 c d^3 (c \sec (a+b x))^{3/2} \sqrt{d \csc (a+b x)} \left (32 \cos ^2(a+b x)+3 \cot ^2(a+b x)-5\right )}{15 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Csc[a + b*x])^(7/2)*(c*Sec[a + b*x])^(5/2),x]

[Out]

(-2*c*d^3*(-5 + 32*Cos[a + b*x]^2 + 3*Cot[a + b*x]^2)*Sqrt[d*Csc[a + b*x]]*(c*Sec[a + b*x])^(3/2))/(15*b)

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Maple [A]  time = 0.165, size = 64, normalized size = 0.6 \begin{align*}{\frac{ \left ( 64\, \left ( \cos \left ( bx+a \right ) \right ) ^{4}-80\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}+10 \right ) \cos \left ( bx+a \right ) \sin \left ( bx+a \right ) }{15\,b} \left ({\frac{d}{\sin \left ( bx+a \right ) }} \right ) ^{{\frac{7}{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))^(7/2)*(c*sec(b*x+a))^(5/2),x)

[Out]

2/15/b*(32*cos(b*x+a)^4-40*cos(b*x+a)^2+5)*cos(b*x+a)*(d/sin(b*x+a))^(7/2)*(c/cos(b*x+a))^(5/2)*sin(b*x+a)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \csc \left (b x + a\right )\right )^{\frac{7}{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))^(7/2)*(c*sec(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

integrate((d*csc(b*x + a))^(7/2)*(c*sec(b*x + a))^(5/2), x)

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Fricas [A]  time = 1.87092, size = 207, normalized size = 1.95 \begin{align*} -\frac{2 \,{\left (32 \, c^{2} d^{3} \cos \left (b x + a\right )^{4} - 40 \, c^{2} d^{3} \cos \left (b x + a\right )^{2} + 5 \, c^{2} d^{3}\right )} \sqrt{\frac{c}{\cos \left (b x + a\right )}} \sqrt{\frac{d}{\sin \left (b x + a\right )}}}{15 \,{\left (b \cos \left (b x + a\right )^{3} - b \cos \left (b x + a\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*csc(b*x+a))^(7/2)*(c*sec(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

-2/15*(32*c^2*d^3*cos(b*x + a)^4 - 40*c^2*d^3*cos(b*x + a)^2 + 5*c^2*d^3)*sqrt(c/cos(b*x + a))*sqrt(d/sin(b*x
+ a))/(b*cos(b*x + a)^3 - b*cos(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))**(7/2)*(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 \left (d \csc \left (b x + a\right )\right )^{\frac{7}{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))^(7/2)*(c*sec(b*x+a))^(5/2),x, algorithm="giac")

[Out]

integrate((d*csc(b*x + a))^(7/2)*(c*sec(b*x + a))^(5/2), x)