3.275 \(\int \sqrt{\csc (a+b x)} \sec ^3(a+b x) \, dx\)

Optimal. Leaf size=62 \[ -\frac{3 \tan ^{-1}\left (\sqrt{\csc (a+b x)}\right )}{4 b}+\frac{\sec ^2(a+b x)}{2 b \sqrt{\csc (a+b x)}}+\frac{3 \tanh ^{-1}\left (\sqrt{\csc (a+b x)}\right )}{4 b} \]

[Out]

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

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Rubi [A]  time = 0.0532984, antiderivative size = 62, 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 = {2621, 288, 329, 298, 203, 206} \[ -\frac{3 \tan ^{-1}\left (\sqrt{\csc (a+b x)}\right )}{4 b}+\frac{\sec ^2(a+b x)}{2 b \sqrt{\csc (a+b x)}}+\frac{3 \tanh ^{-1}\left (\sqrt{\csc (a+b x)}\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Csc[a + b*x]]*Sec[a + b*x]^3,x]

[Out]

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

Rule 2621

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 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 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 \sqrt{\csc (a+b x)} \sec ^3(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^{5/2}}{\left (-1+x^2\right )^2} \, dx,x,\csc (a+b x)\right )}{b}\\ &=\frac{\sec ^2(a+b x)}{2 b \sqrt{\csc (a+b x)}}-\frac{3 \operatorname{Subst}\left (\int \frac{\sqrt{x}}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{4 b}\\ &=\frac{\sec ^2(a+b x)}{2 b \sqrt{\csc (a+b x)}}-\frac{3 \operatorname{Subst}\left (\int \frac{x^2}{-1+x^4} \, dx,x,\sqrt{\csc (a+b x)}\right )}{2 b}\\ &=\frac{\sec ^2(a+b x)}{2 b \sqrt{\csc (a+b x)}}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{\csc (a+b x)}\right )}{4 b}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{\csc (a+b x)}\right )}{4 b}\\ &=-\frac{3 \tan ^{-1}\left (\sqrt{\csc (a+b x)}\right )}{4 b}+\frac{3 \tanh ^{-1}\left (\sqrt{\csc (a+b x)}\right )}{4 b}+\frac{\sec ^2(a+b x)}{2 b \sqrt{\csc (a+b x)}}\\ \end{align*}

Mathematica [A]  time = 0.113689, size = 73, normalized size = 1.18 \[ \frac{\sqrt{\sin (a+b x)} \sqrt{\csc (a+b x)} \left (2 \sqrt{\sin (a+b x)} \sec ^2(a+b x)+3 \left (\tan ^{-1}\left (\sqrt{\sin (a+b x)}\right )+\tanh ^{-1}\left (\sqrt{\sin (a+b x)}\right )\right )\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Csc[a + b*x]]*Sec[a + b*x]^3,x]

[Out]

(Sqrt[Csc[a + b*x]]*(3*(ArcTan[Sqrt[Sin[a + b*x]]] + ArcTanh[Sqrt[Sin[a + b*x]]]) + 2*Sec[a + b*x]^2*Sqrt[Sin[
a + b*x]])*Sqrt[Sin[a + b*x]])/(4*b)

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Maple [A]  time = 1.51, size = 73, normalized size = 1.2 \begin{align*}{\frac{1}{8\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}b} \left ( - \left ( -3\,\ln \left ( \sqrt{\sin \left ( bx+a \right ) }+1 \right ) +3\,\ln \left ( \sqrt{\sin \left ( bx+a \right ) }-1 \right ) -6\,\arctan \left ( \sqrt{\sin \left ( bx+a \right ) } \right ) \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{2}+4\,\sqrt{\sin \left ( bx+a \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(-(-3*ln(sin(b*x+a)^(1/2)+1)+3*ln(sin(b*x+a)^(1/2)-1)-6*arctan(sin(b*x+a)^(1/2)))*cos(b*x+a)^2+4*sin(b*x+a
)^(1/2))/cos(b*x+a)^2/b

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Maxima [A]  time = 1.4409, size = 88, normalized size = 1.42 \begin{align*} \frac{\frac{4}{{\left (\frac{1}{\sin \left (b x + a\right )^{2}} - 1\right )} \sin \left (b x + a\right )^{\frac{3}{2}}} - 6 \, \arctan \left (\frac{1}{\sqrt{\sin \left (b x + a\right )}}\right ) + 3 \, \log \left (\frac{1}{\sqrt{\sin \left (b x + a\right )}} + 1\right ) - 3 \, \log \left (\frac{1}{\sqrt{\sin \left (b x + a\right )}} - 1\right )}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4/((1/sin(b*x + a)^2 - 1)*sin(b*x + a)^(3/2)) - 6*arctan(1/sqrt(sin(b*x + a))) + 3*log(1/sqrt(sin(b*x + a
)) + 1) - 3*log(1/sqrt(sin(b*x + a)) - 1))/b

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Fricas [B]  time = 1.29247, size = 373, normalized size = 6.02 \begin{align*} \frac{6 \, \arctan \left (\frac{\sin \left (b x + a\right ) - 1}{2 \, \sqrt{\sin \left (b x + a\right )}}\right ) \cos \left (b x + a\right )^{2} + 3 \, \cos \left (b x + a\right )^{2} \log \left (\frac{\cos \left (b x + a\right )^{2} + \frac{4 \,{\left (\cos \left (b x + a\right )^{2} - \sin \left (b x + a\right ) - 1\right )}}{\sqrt{\sin \left (b x + a\right )}} - 6 \, \sin \left (b x + a\right ) - 2}{\cos \left (b x + a\right )^{2} + 2 \, \sin \left (b x + a\right ) - 2}\right ) + 8 \, \sqrt{\sin \left (b x + a\right )}}{16 \, b \cos \left (b x + a\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(csc(a + b*x))*sec(a + b*x)**3, x)

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Giac [A]  time = 1.26178, size = 99, normalized size = 1.6 \begin{align*} -\frac{{\left (\frac{4 \, \sqrt{\sin \left (b x + a\right )}}{\sin \left (b x + a\right )^{2} - 1} - 6 \, \arctan \left (\sqrt{\sin \left (b x + a\right )}\right ) - 3 \, \log \left (\sqrt{\sin \left (b x + a\right )} + 1\right ) + 3 \, \log \left ({\left | \sqrt{\sin \left (b x + a\right )} - 1 \right |}\right )\right )} \mathrm{sgn}\left (\sin \left (b x + a\right )\right )}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(4*sqrt(sin(b*x + a))/(sin(b*x + a)^2 - 1) - 6*arctan(sqrt(sin(b*x + a))) - 3*log(sqrt(sin(b*x + a)) + 1)
+ 3*log(abs(sqrt(sin(b*x + a)) - 1)))*sgn(sin(b*x + a))/b