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

Optimal. Leaf size=93 \[ \frac {2 d^2 \sqrt {\sin (2 a+2 b x)} F\left (\left .a+b x-\frac {\pi }{4}\right |2\right ) \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}{3 b}-\frac {2 c d (d \csc (a+b x))^{3/2}}{3 b \sqrt {c \sec (a+b x)}} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2625, 2630, 2573, 2641} \[ \frac {2 d^2 \sqrt {\sin (2 a+2 b x)} F\left (\left .a+b x-\frac {\pi }{4}\right |2\right ) \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}{3 b}-\frac {2 c d (d \csc (a+b x))^{3/2}}{3 b \sqrt {c \sec (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

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 2641

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

Rubi steps

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

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Mathematica [C]  time = 0.89, size = 109, normalized size = 1.17 \[ -\frac {d (\cos (a+b x)+\cos (3 (a+b x))) \sec ^2(a+b x) \sqrt {c \sec (a+b x)} (d \csc (a+b x))^{3/2} \left (\left (-\cot ^2(a+b x)\right )^{3/4} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};\csc ^2(a+b x)\right )+\cot ^2(a+b x)\right )}{3 b \left (\csc ^2(a+b x)-2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(d*(Cos[a + b*x] + Cos[3*(a + b*x)])*(d*Csc[a + b*x])^(3/2)*(Cot[a + b*x]^2 + (-Cot[a + b*x]^2)^(3/4)*Hyp
ergeometric2F1[1/2, 3/4, 3/2, Csc[a + b*x]^2])*Sec[a + b*x]^2*Sqrt[c*Sec[a + b*x]])/(b*(-2 + Csc[a + b*x]^2))

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fricas [F]  time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {d \csc \left (b x + a\right )} \sqrt {c \sec \left (b x + a\right )} d^{2} \csc \left (b x + a\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d*csc(b*x + a))*sqrt(c*sec(b*x + a))*d^2*csc(b*x + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.23, size = 281, normalized size = 3.02 \[ \frac {\left (2 \sin \left (b x +a \right ) \cos \left (b x +a \right ) \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticF \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )+2 \sin \left (b x +a \right ) \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticF \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )-\cos \left (b x +a \right ) \sqrt {2}\right ) \left (\frac {d}{\sin \left (b x +a \right )}\right )^{\frac {5}{2}} \sqrt {\frac {c}{\cos \left (b x +a \right )}}\, \sin \left (b x +a \right ) \sqrt {2}}{3 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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