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

Optimal. Leaf size=78 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b d^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b d^{3/2}}+\frac{2}{b d \sqrt{d \cos (a+b x)}} \]

[Out]

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

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Rubi [A]  time = 0.0666221, antiderivative size = 78, 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, 325, 329, 298, 203, 206} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b d^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b d^{3/2}}+\frac{2}{b d \sqrt{d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[Sqrt[d*Cos[a + b*x]]/Sqrt[d]]/(b*d^(3/2)) - ArcTanh[Sqrt[d*Cos[a + b*x]]/Sqrt[d]]/(b*d^(3/2)) + 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 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 (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 )} \, dx,x,d \cos (a+b x)\right )}{b d}\\ &=\frac{2}{b d \sqrt{d \cos (a+b x)}}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{x}}{1-\frac{x^2}{d^2}} \, dx,x,d \cos (a+b x)\right )}{b d^3}\\ &=\frac{2}{b d \sqrt{d \cos (a+b x)}}-\frac{2 \operatorname{Subst}\left (\int \frac{x^2}{1-\frac{x^4}{d^2}} \, dx,x,\sqrt{d \cos (a+b x)}\right )}{b d^3}\\ &=\frac{2}{b d \sqrt{d \cos (a+b x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{d-x^2} \, dx,x,\sqrt{d \cos (a+b x)}\right )}{b d}+\frac{\operatorname{Subst}\left (\int \frac{1}{d+x^2} \, dx,x,\sqrt{d \cos (a+b x)}\right )}{b d}\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b d^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b d^{3/2}}+\frac{2}{b d \sqrt{d \cos (a+b x)}}\\ \end{align*}

Mathematica [C]  time = 0.0551461, size = 36, normalized size = 0.46 \[ \frac{2 \, _2F_1\left (-\frac{1}{4},1;\frac{3}{4};\cos ^2(a+b x)\right )}{b d \sqrt{d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.297, size = 422, normalized size = 5.4 \begin{align*}{\frac{1}{2\,b} \left ( - \left ( 4\,\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 ){d}^{5/2}+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 ) \sqrt{-d}{d}^{2}+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 ) \sqrt{-d}{d}^{2} \right ) \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}+2\,\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 ){d}^{5/2}-4\,\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d}{d}^{3/2}\sqrt{-d}+\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 ) \sqrt{-d}{d}^{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 ) \sqrt{-d}{d}^{2} \right ){d}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{-d}}} \left ( 2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-(4*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)+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+2*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)^2+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)/d^(7/2)/(-d)^(1/2)/(2*sin(1/2*b*x+1/2*a)^2-1)/b

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.72727, size = 853, normalized size = 10.94 \begin{align*} \left [\frac{2 \, \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 ) \cos \left (b x + a\right ) - \sqrt{-d} \cos \left (b x + a\right ) \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 )}}{4 \, b d^{2} \cos \left (b x + a\right )}, \frac{2 \, \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 ) \cos \left (b x + a\right ) + \sqrt{d} \cos \left (b x + a\right ) \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 )}}{4 \, b d^{2} \cos \left (b x + a\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.12894, size = 89, normalized size = 1.14 \begin{align*} \frac{d{\left (\frac{\arctan \left (\frac{\sqrt{d \cos \left (b x + a\right )}}{\sqrt{-d}}\right )}{\sqrt{-d} d^{2}} + \frac{\arctan \left (\frac{\sqrt{d \cos \left (b x + a\right )}}{\sqrt{d}}\right )}{d^{\frac{5}{2}}} + \frac{2}{\sqrt{d \cos \left (b x + a\right )} d^{2}}\right )}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*(arctan(sqrt(d*cos(b*x + a))/sqrt(-d))/(sqrt(-d)*d^2) + arctan(sqrt(d*cos(b*x + a))/sqrt(d))/d^(5/2) + 2/(sq
rt(d*cos(b*x + a))*d^2))/b