3.248 \(\int \frac{\csc ^3(a+b x)}{\sqrt{d \cos (a+b x)}} \, dx\)

Optimal. Leaf size=93 \[ -\frac{3 \tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b \sqrt{d}}-\frac{\csc ^2(a+b x) \sqrt{d \cos (a+b x)}}{2 b d}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b \sqrt{d}} \]

[Out]

(-3*ArcTan[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/(4*b*Sqrt[d]) - (3*ArcTanh[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/(4*b*Sqrt[
d]) - (Sqrt[d*Cos[a + b*x]]*Csc[a + b*x]^2)/(2*b*d)

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Rubi [A]  time = 0.0654275, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2565, 290, 329, 212, 206, 203} \[ -\frac{3 \tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b \sqrt{d}}-\frac{\csc ^2(a+b x) \sqrt{d \cos (a+b x)}}{2 b d}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b \sqrt{d}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^3/Sqrt[d*Cos[a + b*x]],x]

[Out]

(-3*ArcTan[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/(4*b*Sqrt[d]) - (3*ArcTanh[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/(4*b*Sqrt[
d]) - (Sqrt[d*Cos[a + b*x]]*Csc[a + b*x]^2)/(2*b*d)

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\csc ^3(a+b x)}{\sqrt{d \cos (a+b x)}} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (1-\frac{x^2}{d^2}\right )^2} \, dx,x,d \cos (a+b x)\right )}{b d}\\ &=-\frac{\sqrt{d \cos (a+b x)} \csc ^2(a+b x)}{2 b d}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (1-\frac{x^2}{d^2}\right )} \, dx,x,d \cos (a+b x)\right )}{4 b d}\\ &=-\frac{\sqrt{d \cos (a+b x)} \csc ^2(a+b x)}{2 b d}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{1-\frac{x^4}{d^2}} \, dx,x,\sqrt{d \cos (a+b x)}\right )}{2 b d}\\ &=-\frac{\sqrt{d \cos (a+b x)} \csc ^2(a+b x)}{2 b d}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{d-x^2} \, dx,x,\sqrt{d \cos (a+b x)}\right )}{4 b}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{d+x^2} \, dx,x,\sqrt{d \cos (a+b x)}\right )}{4 b}\\ &=-\frac{3 \tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b \sqrt{d}}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b \sqrt{d}}-\frac{\sqrt{d \cos (a+b x)} \csc ^2(a+b x)}{2 b d}\\ \end{align*}

Mathematica [C]  time = 0.22831, size = 69, normalized size = 0.74 \[ \frac{d \left (-\cot ^2(a+b x)\right )^{3/4} \left (\sqrt [4]{-\cot ^2(a+b x)}-\, _2F_1\left (\frac{3}{4},\frac{3}{4};\frac{7}{4};\csc ^2(a+b x)\right )\right )}{2 b (d \cos (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^3/Sqrt[d*Cos[a + b*x]],x]

[Out]

(d*(-Cot[a + b*x]^2)^(3/4)*((-Cot[a + b*x]^2)^(1/4) - Hypergeometric2F1[3/4, 3/4, 7/4, Csc[a + b*x]^2]))/(2*b*
(d*Cos[a + b*x])^(3/2))

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Maple [B]  time = 0.312, size = 297, normalized size = 3.2 \begin{align*} -{\frac{3}{8\,b}\ln \left ({ \left ( 4\,d\cos \left ( 1/2\,bx+a/2 \right ) +2\,\sqrt{d}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d}-2\,d \right ) \left ( \cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) -1 \right ) ^{-1}} \right ){\frac{1}{\sqrt{d}}}}+{\frac{1}{16\,bd}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d} \left ( \cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) -1 \right ) ^{-1}}-{\frac{3}{8\,b}\ln \left ({ \left ( -4\,d\cos \left ( 1/2\,bx+a/2 \right ) +2\,\sqrt{d}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d}-2\,d \right ) \left ( \cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) +1 \right ) ^{-1}} \right ){\frac{1}{\sqrt{d}}}}+{\frac{3}{4\,b}\ln \left ({ \left ( -2\,d+2\,\sqrt{-d}\sqrt{2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d-d} \right ) \left ( \cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{-1}} \right ){\frac{1}{\sqrt{-d}}}}-{\frac{1}{16\,bd}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d} \left ( \cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{8\,bd}\sqrt{2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d-d} \left ( \cos \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8/b/d^(1/2)*ln((4*d*cos(1/2*b*x+1/2*a)+2*d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-2*d)/(cos(1/2*b*x+1/2*
a)-1))+1/16/b/d/(cos(1/2*b*x+1/2*a)-1)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-3/8/b/d^(1/2)*ln((-4*d*cos(1/2*b*x+
1/2*a)+2*d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-2*d)/(cos(1/2*b*x+1/2*a)+1))+3/4/b/(-d)^(1/2)*ln((-2*d+2*
(-d)^(1/2)*(2*cos(1/2*b*x+1/2*a)^2*d-d)^(1/2))/cos(1/2*b*x+1/2*a))-1/16/b/d/(cos(1/2*b*x+1/2*a)+1)*(-2*sin(1/2
*b*x+1/2*a)^2*d+d)^(1/2)-1/8/b/d/cos(1/2*b*x+1/2*a)^2*(2*cos(1/2*b*x+1/2*a)^2*d-d)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3/(d*cos(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.66232, size = 922, normalized size = 9.91 \begin{align*} \left [\frac{6 \,{\left (\cos \left (b x + a\right )^{2} - 1\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{d \cos \left (b x + a\right )} \sqrt{-d}{\left (\cos \left (b x + a\right ) + 1\right )}}{2 \, d \cos \left (b x + a\right )}\right ) - 3 \,{\left (\cos \left (b x + a\right )^{2} - 1\right )} \sqrt{-d} \log \left (\frac{d \cos \left (b x + a\right )^{2} + 4 \, \sqrt{d \cos \left (b x + a\right )} \sqrt{-d}{\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, \sqrt{d \cos \left (b x + a\right )}}{16 \,{\left (b d \cos \left (b x + a\right )^{2} - b d\right )}}, -\frac{6 \,{\left (\cos \left (b x + a\right )^{2} - 1\right )} \sqrt{d} \arctan \left (\frac{\sqrt{d \cos \left (b x + a\right )}{\left (\cos \left (b x + a\right ) - 1\right )}}{2 \, \sqrt{d} \cos \left (b x + a\right )}\right ) - 3 \,{\left (\cos \left (b x + a\right )^{2} - 1\right )} \sqrt{d} \log \left (\frac{d \cos \left (b x + a\right )^{2} - 4 \, \sqrt{d \cos \left (b x + a\right )} \sqrt{d}{\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right ) - 8 \, \sqrt{d \cos \left (b x + a\right )}}{16 \,{\left (b d \cos \left (b x + a\right )^{2} - b d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3/(d*cos(b*x+a))^(1/2),x, algorithm="fricas")

[Out]

[1/16*(6*(cos(b*x + a)^2 - 1)*sqrt(-d)*arctan(1/2*sqrt(d*cos(b*x + a))*sqrt(-d)*(cos(b*x + a) + 1)/(d*cos(b*x
+ a))) - 3*(cos(b*x + a)^2 - 1)*sqrt(-d)*log((d*cos(b*x + a)^2 + 4*sqrt(d*cos(b*x + a))*sqrt(-d)*(cos(b*x + a)
- 1) - 6*d*cos(b*x + a) + d)/(cos(b*x + a)^2 + 2*cos(b*x + a) + 1)) + 8*sqrt(d*cos(b*x + a)))/(b*d*cos(b*x +
a)^2 - b*d), -1/16*(6*(cos(b*x + a)^2 - 1)*sqrt(d)*arctan(1/2*sqrt(d*cos(b*x + a))*(cos(b*x + a) - 1)/(sqrt(d)
*cos(b*x + a))) - 3*(cos(b*x + a)^2 - 1)*sqrt(d)*log((d*cos(b*x + a)^2 - 4*sqrt(d*cos(b*x + a))*sqrt(d)*(cos(b
*x + a) + 1) + 6*d*cos(b*x + a) + d)/(cos(b*x + a)^2 - 2*cos(b*x + a) + 1)) - 8*sqrt(d*cos(b*x + a)))/(b*d*cos
(b*x + a)^2 - b*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{3}{\left (a + b x \right )}}{\sqrt{d \cos{\left (a + b x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)**3/(d*cos(b*x+a))**(1/2),x)

[Out]

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

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Giac [A]  time = 1.13803, size = 123, normalized size = 1.32 \begin{align*} \frac{d^{3}{\left (\frac{2 \, \sqrt{d \cos \left (b x + a\right )}}{{\left (d^{2} \cos \left (b x + a\right )^{2} - d^{2}\right )} d^{2}} + \frac{3 \, \arctan \left (\frac{\sqrt{d \cos \left (b x + a\right )}}{\sqrt{-d}}\right )}{\sqrt{-d} d^{3}} - \frac{3 \, \arctan \left (\frac{\sqrt{d \cos \left (b x + a\right )}}{\sqrt{d}}\right )}{d^{\frac{7}{2}}}\right )}}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3/(d*cos(b*x+a))^(1/2),x, algorithm="giac")

[Out]

1/4*d^3*(2*sqrt(d*cos(b*x + a))/((d^2*cos(b*x + a)^2 - d^2)*d^2) + 3*arctan(sqrt(d*cos(b*x + a))/sqrt(-d))/(sq
rt(-d)*d^3) - 3*arctan(sqrt(d*cos(b*x + a))/sqrt(d))/d^(7/2))/b