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

Optimal. Leaf size=115 \[ \frac{5 \tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b d^{3/2}}-\frac{5 \tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b d^{3/2}}+\frac{5}{2 b d \sqrt{d \cos (a+b x)}}-\frac{\csc ^2(a+b x)}{2 b d \sqrt{d \cos (a+b x)}} \]

[Out]

(5*ArcTan[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/(4*b*d^(3/2)) - (5*ArcTanh[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/(4*b*d^(3/2
)) + 5/(2*b*d*Sqrt[d*Cos[a + b*x]]) - Csc[a + b*x]^2/(2*b*d*Sqrt[d*Cos[a + b*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.0825338, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2565, 290, 325, 329, 298, 203, 206} \[ \frac{5 \tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b d^{3/2}}-\frac{5 \tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b d^{3/2}}+\frac{5}{2 b d \sqrt{d \cos (a+b x)}}-\frac{\csc ^2(a+b x)}{2 b d \sqrt{d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*ArcTan[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/(4*b*d^(3/2)) - (5*ArcTanh[Sqrt[d*Cos[a + b*x]]/Sqrt[d]])/(4*b*d^(3/2
)) + 5/(2*b*d*Sqrt[d*Cos[a + b*x]]) - Csc[a + b*x]^2/(2*b*d*Sqrt[d*Cos[a + b*x]])

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -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 298

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

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])

Rubi steps

\begin{align*} \int \frac{\csc ^3(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{1}{x^{3/2} \left (1-\frac{x^2}{d^2}\right )^2} \, dx,x,d \cos (a+b x)\right )}{b d}\\ &=-\frac{\csc ^2(a+b x)}{2 b d \sqrt{d \cos (a+b x)}}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{x^{3/2} \left (1-\frac{x^2}{d^2}\right )} \, dx,x,d \cos (a+b x)\right )}{4 b d}\\ &=\frac{5}{2 b d \sqrt{d \cos (a+b x)}}-\frac{\csc ^2(a+b x)}{2 b d \sqrt{d \cos (a+b x)}}-\frac{5 \operatorname{Subst}\left (\int \frac{\sqrt{x}}{1-\frac{x^2}{d^2}} \, dx,x,d \cos (a+b x)\right )}{4 b d^3}\\ &=\frac{5}{2 b d \sqrt{d \cos (a+b x)}}-\frac{\csc ^2(a+b x)}{2 b d \sqrt{d \cos (a+b x)}}-\frac{5 \operatorname{Subst}\left (\int \frac{x^2}{1-\frac{x^4}{d^2}} \, dx,x,\sqrt{d \cos (a+b x)}\right )}{2 b d^3}\\ &=\frac{5}{2 b d \sqrt{d \cos (a+b x)}}-\frac{\csc ^2(a+b x)}{2 b d \sqrt{d \cos (a+b x)}}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{d-x^2} \, dx,x,\sqrt{d \cos (a+b x)}\right )}{4 b d}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{d+x^2} \, dx,x,\sqrt{d \cos (a+b x)}\right )}{4 b d}\\ &=\frac{5 \tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b d^{3/2}}-\frac{5 \tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{4 b d^{3/2}}+\frac{5}{2 b d \sqrt{d \cos (a+b x)}}-\frac{\csc ^2(a+b x)}{2 b d \sqrt{d \cos (a+b x)}}\\ \end{align*}

Mathematica [C]  time = 0.245834, size = 91, normalized size = 0.79 \[ \frac{5 \cot ^2(a+b x) \, _2F_1\left (\frac{1}{4},\frac{1}{4};\frac{5}{4};\csc ^2(a+b x)\right )-\left (-\cot ^2(a+b x)\right )^{3/4} \left (\cot ^2(a+b x)-4\right )}{2 b d \left (-\cot ^2(a+b x)\right )^{3/4} \sqrt{d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.338, size = 705, normalized size = 6.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/d^(7/2)/(-d)^(1/2)/sin(1/2*b*x+1/2*a)^2/(2*sin(1/2*b*x+1/2*a)^4-3*sin(1/2*b*x+1/2*a)^2+1)*(-(-2*sin(1/2*b*
x+1/2*a)^2*d+d)^(1/2)*d^(3/2)*(-d)^(1/2)-sin(1/2*b*x+1/2*a)^6*(20*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*sin(
1/2*b*x+1/2*a)^2*d+d)^(1/2)-d))*d^(5/2)+10*ln(2/(cos(1/2*b*x+1/2*a)-1)*(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)-d))*(-d)^(1/2)*d^2+10*ln(2/(cos(1/2*b*x+1/2*a)+1)*(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)-d))*(-d)^(1/2)*d^2)+5*(6*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*sin(1/2*
b*x+1/2*a)^2*d+d)^(1/2)-d))*d^(5/2)-4*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*d^(3/2)*(-d)^(1/2)+3*ln(2/(cos(1/2*b
*x+1/2*a)-1)*(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)-d))*(-d)^(1/2)*d^2+3*ln(2/(co
s(1/2*b*x+1/2*a)+1)*(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)-d))*(-d)^(1/2)*d^2)*si
n(1/2*b*x+1/2*a)^4-5*(2*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-d))*d^(5/2)-4*
(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*d^(3/2)*(-d)^(1/2)+ln(2/(cos(1/2*b*x+1/2*a)-1)*(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)-d))*(-d)^(1/2)*d^2+ln(2/(cos(1/2*b*x+1/2*a)+1)*(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)-d))*(-d)^(1/2)*d^2)*sin(1/2*b*x+1/2*a)^2)/b

________________________________________________________________________________________

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))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.71421, size = 1091, normalized size = 9.49 \begin{align*} \left [\frac{10 \,{\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\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 ) - 5 \,{\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\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 )}{\left (5 \, \cos \left (b x + a\right )^{2} - 4\right )}}{16 \,{\left (b d^{2} \cos \left (b x + a\right )^{3} - b d^{2} \cos \left (b x + a\right )\right )}}, \frac{10 \,{\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\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 ) + 5 \,{\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\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 )}{\left (5 \, \cos \left (b x + a\right )^{2} - 4\right )}}{16 \,{\left (b d^{2} \cos \left (b x + a\right )^{3} - b d^{2} \cos \left (b x + a\right )\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))^(3/2),x, algorithm="fricas")

[Out]

[1/16*(10*(cos(b*x + a)^3 - cos(b*x + a))*sqrt(-d)*arctan(1/2*sqrt(d*cos(b*x + a))*sqrt(-d)*(cos(b*x + a) + 1)
/(d*cos(b*x + a))) - 5*(cos(b*x + a)^3 - cos(b*x + a))*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))*(5*cos(b*x + a)^2 - 4))/(b*d^2*cos(b*x + a)^3 - b*d^2*cos(b*x + a)), 1/16*(10*(cos(b*x + a)^3 - cos(b*x
+ a))*sqrt(d)*arctan(1/2*sqrt(d*cos(b*x + a))*(cos(b*x + a) - 1)/(sqrt(d)*cos(b*x + a))) + 5*(cos(b*x + a)^3
- cos(b*x + a))*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))*(5*cos(b*x + a)^2 - 4))/(b*d^2*cos
(b*x + a)^3 - b*d^2*cos(b*x + a))]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{3}{\left (a + b x \right )}}{\left (d \cos{\left (a + b x \right )}\right )^{\frac{3}{2}}}\, 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))**(3/2),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.14128, size = 162, normalized size = 1.41 \begin{align*} \frac{d^{3}{\left (\frac{5 \, \arctan \left (\frac{\sqrt{d \cos \left (b x + a\right )}}{\sqrt{-d}}\right )}{\sqrt{-d} d^{4}} + \frac{5 \, \arctan \left (\frac{\sqrt{d \cos \left (b x + a\right )}}{\sqrt{d}}\right )}{d^{\frac{9}{2}}} + \frac{2 \,{\left (5 \, d^{2} \cos \left (b x + a\right )^{2} - 4 \, d^{2}\right )}}{{\left (\sqrt{d \cos \left (b x + a\right )} d^{2} \cos \left (b x + a\right )^{2} - \sqrt{d \cos \left (b x + a\right )} d^{2}\right )} d^{4}}\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))^(3/2),x, algorithm="giac")

[Out]

1/4*d^3*(5*arctan(sqrt(d*cos(b*x + a))/sqrt(-d))/(sqrt(-d)*d^4) + 5*arctan(sqrt(d*cos(b*x + a))/sqrt(d))/d^(9/
2) + 2*(5*d^2*cos(b*x + a)^2 - 4*d^2)/((sqrt(d*cos(b*x + a))*d^2*cos(b*x + a)^2 - sqrt(d*cos(b*x + a))*d^2)*d^
4))/b