3.239 \(\int \frac {\csc ^2(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=94 \[ -\frac {3 E\left (\left .\frac {1}{2} (a+b x)\right |2\right ) \sqrt {d \cos (a+b x)}}{b d^2 \sqrt {\cos (a+b x)}}+\frac {3 \sin (a+b x)}{b d \sqrt {d \cos (a+b x)}}-\frac {\csc (a+b x)}{b d \sqrt {d \cos (a+b x)}} \]

[Out]

-csc(b*x+a)/b/d/(d*cos(b*x+a))^(1/2)+3*sin(b*x+a)/b/d/(d*cos(b*x+a))^(1/2)-3*(cos(1/2*a+1/2*b*x)^2)^(1/2)/cos(
1/2*a+1/2*b*x)*EllipticE(sin(1/2*a+1/2*b*x),2^(1/2))*(d*cos(b*x+a))^(1/2)/b/d^2/cos(b*x+a)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2570, 2636, 2640, 2639} \[ -\frac {3 E\left (\left .\frac {1}{2} (a+b x)\right |2\right ) \sqrt {d \cos (a+b x)}}{b d^2 \sqrt {\cos (a+b x)}}+\frac {3 \sin (a+b x)}{b d \sqrt {d \cos (a+b x)}}-\frac {\csc (a+b x)}{b d \sqrt {d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[a + b*x]/(b*d*Sqrt[d*Cos[a + b*x]])) - (3*Sqrt[d*Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2])/(b*d^2*Sqrt[Co
s[a + b*x]]) + (3*Sin[a + b*x])/(b*d*Sqrt[d*Cos[a + b*x]])

Rule 2570

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((b*Cos[e + f
*x])^(n + 1)*(a*Sin[e + f*x])^(m + 1))/(a*b*f*(m + 1)), x] + Dist[(m + n + 2)/(a^2*(m + 1)), Int[(b*Cos[e + f*
x])^n*(a*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n]

Rule 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

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]

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 65, normalized size = 0.69 \[ \frac {2 \sin (a+b x)-\cos (a+b x) \cot (a+b x)-3 \sqrt {\cos (a+b x)} E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{b d \sqrt {d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Cos[a + b*x]*Cot[a + b*x]) - 3*Sqrt[Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2] + 2*Sin[a + b*x])/(b*d*Sqrt[d*C
os[a + b*x]])

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d \cos \left (b x + a\right )} \csc \left (b x + a\right )^{2}}{d^{2} \cos \left (b x + a\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d*cos(b*x + a))*csc(b*x + a)^2/(d^2*cos(b*x + a)^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (b x + a\right )^{2}}{\left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(csc(b*x + a)^2/(d*cos(b*x + a))^(3/2), x)

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maple [A]  time = 0.33, size = 209, normalized size = 2.22 \[ -\frac {\sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, \left (-2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +\left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d \right )^{\frac {3}{2}} \left (6 \sqrt {2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \EllipticE \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+12 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-12 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+1\right )}{2 d^{3} \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{5} \left (2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )^{2} \cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )}\, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^2/(d*cos(b*x+a))^(3/2),x)

[Out]

-1/2*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)/d^3/sin(1/2*b*x+1/2*a)^5/(2*sin(1/2*b*x+1/2*a)^
2-1)^2/cos(1/2*b*x+1/2*a)*(-2*sin(1/2*b*x+1/2*a)^4*d+sin(1/2*b*x+1/2*a)^2*d)^(3/2)*(6*(2*sin(1/2*b*x+1/2*a)^2-
1)^(1/2)*(sin(1/2*b*x+1/2*a)^2)^(1/2)*EllipticE(cos(1/2*b*x+1/2*a),2^(1/2))*cos(1/2*b*x+1/2*a)+12*sin(1/2*b*x+
1/2*a)^4-12*sin(1/2*b*x+1/2*a)^2+1)/(d*(2*cos(1/2*b*x+1/2*a)^2-1))^(1/2)/b

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (b x + a\right )^{2}}{\left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(csc(b*x + a)^2/(d*cos(b*x + a))^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\sin \left (a+b\,x\right )}^2\,{\left (d\,\cos \left (a+b\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(a + b*x)^2*(d*cos(a + b*x))^(3/2)),x)

[Out]

int(1/(sin(a + b*x)^2*(d*cos(a + b*x))^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{2}{\left (a + b x \right )}}{\left (d \cos {\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)**2/(d*cos(b*x+a))**(3/2),x)

[Out]

Integral(csc(a + b*x)**2/(d*cos(a + b*x))**(3/2), x)

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