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

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

[Out]

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

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Rubi [A]  time = 0.0463736, 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, 212, 206, 203} \[ \frac{\sec ^2(a+b x)}{2 b \csc ^{\frac{3}{2}}(a+b x)}+\frac{\tan ^{-1}\left (\sqrt{\csc (a+b x)}\right )}{4 b}+\frac{\tanh ^{-1}\left (\sqrt{\csc (a+b x)}\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 0.0312675, size = 33, normalized size = 0.53 \[ \frac{2 \, _2F_1\left (\frac{3}{4},2;\frac{7}{4};\sin ^2(a+b x)\right )}{3 b \csc ^{\frac{3}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Hypergeometric2F1[3/4, 2, 7/4, Sin[a + b*x]^2])/(3*b*Csc[a + b*x]^(3/2))

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Maple [A]  time = 1.461, size = 71, normalized size = 1.2 \begin{align*}{\frac{1}{8\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}b} \left ( - \left ( \ln \left ( \sqrt{\sin \left ( bx+a \right ) }-1 \right ) +2\,\arctan \left ( \sqrt{\sin \left ( bx+a \right ) } \right ) -\ln \left ( \sqrt{\sin \left ( bx+a \right ) }+1 \right ) \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{2}+4\, \left ( \sin \left ( bx+a \right ) \right ) ^{3/2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.45519, size = 85, normalized size = 1.37 \begin{align*} \frac{\frac{4}{{\left (\frac{1}{\sin \left (b x + a\right )^{2}} - 1\right )} \sqrt{\sin \left (b x + a\right )}} + 2 \, \arctan \left (\frac{1}{\sqrt{\sin \left (b x + a\right )}}\right ) + \log \left (\frac{1}{\sqrt{\sin \left (b x + a\right )}} + 1\right ) - \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(sec(b*x+a)^3/csc(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.30267, size = 400, normalized size = 6.45 \begin{align*} -\frac{2 \, \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} - \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 ) + \frac{8 \,{\left (\cos \left (b x + a\right )^{2} - 1\right )}}{\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(sec(b*x+a)^3/csc(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

-1/16*(2*arctan(1/2*(sin(b*x + a) - 1)/sqrt(sin(b*x + a)))*cos(b*x + a)^2 - 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*(cos(b*x + a)^2 - 1)/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 \frac{\sec ^{3}{\left (a + b x \right )}}{\sqrt{\csc{\left (a + b x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sec(b*x + a)^3/sqrt(csc(b*x + a)), x)