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

Optimal. Leaf size=77 \[ -\frac{d^{3/2} \tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b}-\frac{d^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b}+\frac{2 d \sqrt{d \cos (a+b x)}}{b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0625691, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {2565, 321, 329, 212, 206, 203} \[ -\frac{d^{3/2} \tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b}-\frac{d^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b}+\frac{2 d \sqrt{d \cos (a+b x)}}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 (d \cos (a+b x))^{3/2} \csc (a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^{3/2}}{1-\frac{x^2}{d^2}} \, dx,x,d \cos (a+b x)\right )}{b d}\\ &=\frac{2 d \sqrt{d \cos (a+b x)}}{b}-\frac{d \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (1-\frac{x^2}{d^2}\right )} \, dx,x,d \cos (a+b x)\right )}{b}\\ &=\frac{2 d \sqrt{d \cos (a+b x)}}{b}-\frac{(2 d) \operatorname{Subst}\left (\int \frac{1}{1-\frac{x^4}{d^2}} \, dx,x,\sqrt{d \cos (a+b x)}\right )}{b}\\ &=\frac{2 d \sqrt{d \cos (a+b x)}}{b}-\frac{d^2 \operatorname{Subst}\left (\int \frac{1}{d-x^2} \, dx,x,\sqrt{d \cos (a+b x)}\right )}{b}-\frac{d^2 \operatorname{Subst}\left (\int \frac{1}{d+x^2} \, dx,x,\sqrt{d \cos (a+b x)}\right )}{b}\\ &=-\frac{d^{3/2} \tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b}-\frac{d^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b}+\frac{2 d \sqrt{d \cos (a+b x)}}{b}\\ \end{align*}

Mathematica [A]  time = 0.0649017, size = 65, normalized size = 0.84 \[ \frac{(d \cos (a+b x))^{3/2} \left (2 \sqrt{\cos (a+b x)}-\tan ^{-1}\left (\sqrt{\cos (a+b x)}\right )-\tanh ^{-1}\left (\sqrt{\cos (a+b x)}\right )\right )}{b \cos ^{\frac{3}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.141, size = 204, normalized size = 2.7 \begin{align*} -{\frac{1}{2\,b}{d}^{{\frac{3}{2}}}\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d}+2\,d\cos \left ( 1/2\,bx+a/2 \right ) -d}{\cos \left ( 1/2\,bx+a/2 \right ) -1}} \right ) }-{\frac{1}{2\,b}{d}^{{\frac{3}{2}}}\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d}-2\,d\cos \left ( 1/2\,bx+a/2 \right ) -d}{\cos \left ( 1/2\,bx+a/2 \right ) +1}} \right ) }+{\frac{{d}^{2}}{b}\ln \left ( 2\,{\frac{\sqrt{-d}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d}-d}{\cos \left ( 1/2\,bx+a/2 \right ) }} \right ){\frac{1}{\sqrt{-d}}}}+2\,{\frac{d\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d}}{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 3.79855, size = 702, normalized size = 9.12 \begin{align*} \left [\frac{2 \, \sqrt{-d} d \arctan \left (\frac{2 \, \sqrt{d \cos \left (b x + a\right )} \sqrt{-d}}{d \cos \left (b x + a\right ) + d}\right ) + \sqrt{-d} 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 )} d}{4 \, b}, \frac{2 \, d^{\frac{3}{2}} \arctan \left (\frac{2 \, \sqrt{d \cos \left (b x + a\right )} \sqrt{d}}{d \cos \left (b x + a\right ) - d}\right ) + d^{\frac{3}{2}} \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 )} d}{4 \, b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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