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

Optimal. Leaf size=135 \[ \frac {6 d^4 E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{5 b c^2 \sqrt {\sin (2 a+2 b x)} \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}+\frac {6 d^3 \sqrt {d \csc (a+b x)}}{5 b c (c \sec (a+b x))^{3/2}}-\frac {2 d (d \csc (a+b x))^{5/2}}{5 b c (c \sec (a+b x))^{3/2}} \]

[Out]

-2/5*d*(d*csc(b*x+a))^(5/2)/b/c/(c*sec(b*x+a))^(3/2)+6/5*d^3*(d*csc(b*x+a))^(1/2)/b/c/(c*sec(b*x+a))^(3/2)-6/5
*d^4*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticE(cos(a+1/4*Pi+b*x),2^(1/2))/b/c^2/(d*csc(b*x+a))^(
1/2)/(c*sec(b*x+a))^(1/2)/sin(2*b*x+2*a)^(1/2)

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Rubi [A]  time = 0.20, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2623, 2625, 2630, 2572, 2639} \[ \frac {6 d^4 E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{5 b c^2 \sqrt {\sin (2 a+2 b x)} \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}+\frac {6 d^3 \sqrt {d \csc (a+b x)}}{5 b c (c \sec (a+b x))^{3/2}}-\frac {2 d (d \csc (a+b x))^{5/2}}{5 b c (c \sec (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*d^3*Sqrt[d*Csc[a + b*x]])/(5*b*c*(c*Sec[a + b*x])^(3/2)) - (2*d*(d*Csc[a + b*x])^(5/2))/(5*b*c*(c*Sec[a + b
*x])^(3/2)) + (6*d^4*EllipticE[a - Pi/4 + b*x, 2])/(5*b*c^2*Sqrt[d*Csc[a + b*x]]*Sqrt[c*Sec[a + b*x]]*Sqrt[Sin
[2*a + 2*b*x]])

Rule 2572

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(Sqrt[a*Sin[e +
f*x]]*Sqrt[b*Cos[e + f*x]])/Sqrt[Sin[2*e + 2*f*x]], Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f},
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 2630

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

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin {align*} \int \frac {(d \csc (a+b x))^{7/2}}{(c \sec (a+b x))^{5/2}} \, dx &=-\frac {2 d (d \csc (a+b x))^{5/2}}{5 b c (c \sec (a+b x))^{3/2}}-\frac {\left (3 d^2\right ) \int \frac {(d \csc (a+b x))^{3/2}}{\sqrt {c \sec (a+b x)}} \, dx}{5 c^2}\\ &=\frac {6 d^3 \sqrt {d \csc (a+b x)}}{5 b c (c \sec (a+b x))^{3/2}}-\frac {2 d (d \csc (a+b x))^{5/2}}{5 b c (c \sec (a+b x))^{3/2}}+\frac {\left (6 d^4\right ) \int \frac {1}{\sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}} \, dx}{5 c^2}\\ &=\frac {6 d^3 \sqrt {d \csc (a+b x)}}{5 b c (c \sec (a+b x))^{3/2}}-\frac {2 d (d \csc (a+b x))^{5/2}}{5 b c (c \sec (a+b x))^{3/2}}+\frac {\left (6 d^4\right ) \int \sqrt {c \cos (a+b x)} \sqrt {d \sin (a+b x)} \, dx}{5 c^2 \sqrt {c \cos (a+b x)} \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {d \sin (a+b x)}}\\ &=\frac {6 d^3 \sqrt {d \csc (a+b x)}}{5 b c (c \sec (a+b x))^{3/2}}-\frac {2 d (d \csc (a+b x))^{5/2}}{5 b c (c \sec (a+b x))^{3/2}}+\frac {\left (6 d^4\right ) \int \sqrt {\sin (2 a+2 b x)} \, dx}{5 c^2 \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {\sin (2 a+2 b x)}}\\ &=\frac {6 d^3 \sqrt {d \csc (a+b x)}}{5 b c (c \sec (a+b x))^{3/2}}-\frac {2 d (d \csc (a+b x))^{5/2}}{5 b c (c \sec (a+b x))^{3/2}}+\frac {6 d^4 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{5 b c^2 \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {\sin (2 a+2 b x)}}\\ \end {align*}

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Mathematica [C]  time = 1.77, size = 101, normalized size = 0.75 \[ \frac {d^5 \sqrt {c \sec (a+b x)} \left (6 \sqrt [4]{-\cot ^2(a+b x)} \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {1}{2};\csc ^2(a+b x)\right )+(1-3 \cos (2 (a+b x))) \cot ^2(a+b x) \csc ^2(a+b x)\right )}{5 b c^3 (d \csc (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^5*((1 - 3*Cos[2*(a + b*x)])*Cot[a + b*x]^2*Csc[a + b*x]^2 + 6*(-Cot[a + b*x]^2)^(1/4)*Hypergeometric2F1[-1/
2, 1/4, 1/2, Csc[a + b*x]^2])*Sqrt[c*Sec[a + b*x]])/(5*b*c^3*(d*Csc[a + b*x])^(3/2))

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fricas [F]  time = 1.03, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d \csc \left (b x + a\right )} \sqrt {c \sec \left (b x + a\right )} d^{3} \csc \left (b x + a\right )^{3}}{c^{3} \sec \left (b x + a\right )^{3}}, x\right ) \]

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]

integral(sqrt(d*csc(b*x + a))*sqrt(c*sec(b*x + a))*d^3*csc(b*x + a)^3/(c^3*sec(b*x + a)^3), x)

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

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 [B]  time = 1.21, size = 977, normalized size = 7.24 \[ \text {result too large to display} \]

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]

1/5/b*(6*cos(b*x+a)^3*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/
2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticE(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))-3*co
s(b*x+a)^3*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos
(b*x+a))/sin(b*x+a))^(1/2)*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))+6*cos(b*x+a)^2*
((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/si
n(b*x+a))^(1/2)*EllipticE(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))-3*cos(b*x+a)^2*((1-cos(b*x
+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(
1/2)*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))-6*cos(b*x+a)*((1-cos(b*x+a)+sin(b*x+a
))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*Elliptic
E(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))+3*cos(b*x+a)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a)
)^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticF(((1-cos(b*x
+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))-6*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+
sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticE(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+
a))^(1/2),1/2*2^(1/2))+3*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^
(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))-3
*cos(b*x+a)^3*2^(1/2)-2^(1/2)*cos(b*x+a)^2+3*cos(b*x+a)*2^(1/2))*(d/sin(b*x+a))^(7/2)*sin(b*x+a)/(c/cos(b*x+a)
)^(5/2)/cos(b*x+a)^3*2^(1/2)

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

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 [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {d}{\sin \left (a+b\,x\right )}\right )}^{7/2}}{{\left (\frac {c}{\cos \left (a+b\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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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