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]

-2/5*c*d*(d*csc(b*x+a))^(5/2)*(c*sec(b*x+a))^(3/2)/b+16/15*c*d^3*(c*sec(b*x+a))^(3/2)*(d*csc(b*x+a))^(1/2)/b-6
4/15*c^3*d^3*(d*csc(b*x+a))^(1/2)/b/(c*sec(b*x+a))^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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 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]

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]

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*}

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Mathematica [A]  time = 0.19, 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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fricas [A]  time = 0.97, size = 89, normalized size = 0.84 \[ -\frac {2 \, {\left (32 \, c^{2} d^{3} \cos \left (b x + a\right )^{4} - 40 \, c^{2} d^{3} \cos \left (b x + a\right )^{2} + 5 \, c^{2} d^{3}\right )} \sqrt {\frac {c}{\cos \left (b x + a\right )}} \sqrt {\frac {d}{\sin \left (b x + a\right )}}}{15 \, {\left (b \cos \left (b x + a\right )^{3} - b \cos \left (b x + a\right )\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \csc \left (b x + a\right )\right )^{\frac {7}{2}} \left (c \sec \left (b x + a\right )\right )^{\frac {5}{2}}\,{d x} \]

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)

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maple [A]  time = 1.12, size = 64, normalized size = 0.60 \[ \frac {2 \left (32 \left (\cos ^{4}\left (b x +a \right )\right )-40 \left (\cos ^{2}\left (b x +a \right )\right )+5\right ) \cos \left (b x +a \right ) \left (\frac {d}{\sin \left (b x +a \right )}\right )^{\frac {7}{2}} \left (\frac {c}{\cos \left (b x +a \right )}\right )^{\frac {5}{2}} \sin \left (b x +a \right )}{15 b} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \csc \left (b x + a\right )\right )^{\frac {7}{2}} \left (c \sec \left (b x + a\right )\right )^{\frac {5}{2}}\,{d x} \]

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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mupad [B]  time = 2.32, size = 112, normalized size = 1.06 \[ \frac {16\,c^2\,d^3\,\sqrt {\frac {c}{\cos \left (a+b\,x\right )}}\,\sqrt {\frac {d}{\sin \left (a+b\,x\right )}}\,\left (5\,\cos \left (a+b\,x\right )-3\,\cos \left (3\,a+3\,b\,x\right )-4\,\cos \left (5\,a+5\,b\,x\right )+2\,\cos \left (7\,a+7\,b\,x\right )\right )}{15\,b\,\left (\cos \left (2\,a+2\,b\,x\right )+2\,\cos \left (4\,a+4\,b\,x\right )-\cos \left (6\,a+6\,b\,x\right )-2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(16*c^2*d^3*(c/cos(a + b*x))^(1/2)*(d/sin(a + b*x))^(1/2)*(5*cos(a + b*x) - 3*cos(3*a + 3*b*x) - 4*cos(5*a + 5
*b*x) + 2*cos(7*a + 7*b*x)))/(15*b*(cos(2*a + 2*b*x) + 2*cos(4*a + 4*b*x) - cos(6*a + 6*b*x) - 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))**(7/2)*(c*sec(b*x+a))**(5/2),x)

[Out]

Timed out

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