Integrand size = 19, antiderivative size = 62 \[ \int \frac {\sec ^3(a+b x)}{\sqrt {\csc (a+b x)}} \, dx=\frac {\arctan \left (\sqrt {\csc (a+b x)}\right )}{4 b}+\frac {\text {arctanh}\left (\sqrt {\csc (a+b x)}\right )}{4 b}+\frac {\sec ^2(a+b x)}{2 b \csc ^{\frac {3}{2}}(a+b x)} \] Output:
1/4*arctan(csc(b*x+a)^(1/2))/b+1/4*arctanh(csc(b*x+a)^(1/2))/b+1/2*sec(b*x +a)^2/b/csc(b*x+a)^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.05 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.53 \[ \int \frac {\sec ^3(a+b x)}{\sqrt {\csc (a+b x)}} \, dx=\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},2,\frac {7}{4},\sin ^2(a+b x)\right )}{3 b \csc ^{\frac {3}{2}}(a+b x)} \] Input:
Integrate[Sec[a + b*x]^3/Sqrt[Csc[a + b*x]],x]
Output:
(2*Hypergeometric2F1[3/4, 2, 7/4, Sin[a + b*x]^2])/(3*b*Csc[a + b*x]^(3/2) )
Time = 0.24 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3042, 3101, 252, 266, 756, 216, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sec ^3(a+b x)}{\sqrt {\csc (a+b x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (a+b x)^3}{\sqrt {\csc (a+b x)}}dx\) |
\(\Big \downarrow \) 3101 |
\(\displaystyle -\frac {\int \frac {\csc ^{\frac {3}{2}}(a+b x)}{\left (1-\csc ^2(a+b x)\right )^2}d\csc (a+b x)}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {\frac {\sqrt {\csc (a+b x)}}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {1}{4} \int \frac {1}{\sqrt {\csc (a+b x)} \left (1-\csc ^2(a+b x)\right )}d\csc (a+b x)}{b}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle -\frac {\frac {\sqrt {\csc (a+b x)}}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {1}{2} \int \frac {1}{1-\csc ^2(a+b x)}d\sqrt {\csc (a+b x)}}{b}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle -\frac {\frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{1-\csc (a+b x)}d\sqrt {\csc (a+b x)}-\frac {1}{2} \int \frac {1}{\csc (a+b x)+1}d\sqrt {\csc (a+b x)}\right )+\frac {\sqrt {\csc (a+b x)}}{2 \left (1-\csc ^2(a+b x)\right )}}{b}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {\frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{1-\csc (a+b x)}d\sqrt {\csc (a+b x)}-\frac {1}{2} \arctan \left (\sqrt {\csc (a+b x)}\right )\right )+\frac {\sqrt {\csc (a+b x)}}{2 \left (1-\csc ^2(a+b x)\right )}}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\frac {1}{2} \left (-\frac {1}{2} \arctan \left (\sqrt {\csc (a+b x)}\right )-\frac {1}{2} \text {arctanh}\left (\sqrt {\csc (a+b x)}\right )\right )+\frac {\sqrt {\csc (a+b x)}}{2 \left (1-\csc ^2(a+b x)\right )}}{b}\) |
Input:
Int[Sec[a + b*x]^3/Sqrt[Csc[a + b*x]],x]
Output:
-(((-1/2*ArcTan[Sqrt[Csc[a + b*x]]] - ArcTanh[Sqrt[Csc[a + b*x]]]/2)/2 + S qrt[Csc[a + b*x]]/(2*(1 - Csc[a + b*x]^2)))/b)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_S ymbol] :> Simp[-(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] && IntegerQ[(n + 1)/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Time = 0.59 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.15
method | result | size |
default | \(\frac {-\left (-\ln \left (\sqrt {\sin \left (b x +a \right )}+1\right )+\ln \left (\sqrt {\sin \left (b x +a \right )}-1\right )+2 \arctan \left (\sqrt {\sin \left (b x +a \right )}\right )\right ) \cos \left (b x +a \right )^{2}+4 \sin \left (b x +a \right )^{\frac {3}{2}}}{8 \cos \left (b x +a \right )^{2} b}\) | \(71\) |
Input:
int(sec(b*x+a)^3/csc(b*x+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/8*(-(-ln(sin(b*x+a)^(1/2)+1)+ln(sin(b*x+a)^(1/2)-1)+2*arctan(sin(b*x+a)^ (1/2)))*cos(b*x+a)^2+4*sin(b*x+a)^(3/2))/cos(b*x+a)^2/b
Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (50) = 100\).
Time = 0.12 (sec) , antiderivative size = 141, normalized size of antiderivative = 2.27 \[ \int \frac {\sec ^3(a+b x)}{\sqrt {\csc (a+b x)}} \, dx=-\frac {2 \, \arctan \left (\frac {\sin \left (b x + a\right ) - 1}{2 \, \sqrt {\sin \left (b x + a\right )}}\right ) \cos \left (b x + a\right )^{2} - \cos \left (b x + a\right )^{2} \log \left (\frac {\cos \left (b x + a\right )^{2} + \frac {4 \, {\left (\cos \left (b x + a\right )^{2} - \sin \left (b x + a\right ) - 1\right )}}{\sqrt {\sin \left (b x + a\right )}} - 6 \, \sin \left (b x + a\right ) - 2}{\cos \left (b x + a\right )^{2} + 2 \, \sin \left (b x + a\right ) - 2}\right ) + \frac {8 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )}}{\sqrt {\sin \left (b x + a\right )}}}{16 \, b \cos \left (b x + a\right )^{2}} \] Input:
integrate(sec(b*x+a)^3/csc(b*x+a)^(1/2),x, algorithm="fricas")
Output:
-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)
\[ \int \frac {\sec ^3(a+b x)}{\sqrt {\csc (a+b x)}} \, dx=\int \frac {\sec ^{3}{\left (a + b x \right )}}{\sqrt {\csc {\left (a + b x \right )}}}\, dx \] Input:
integrate(sec(b*x+a)**3/csc(b*x+a)**(1/2),x)
Output:
Integral(sec(a + b*x)**3/sqrt(csc(a + b*x)), x)
Time = 0.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.02 \[ \int \frac {\sec ^3(a+b x)}{\sqrt {\csc (a+b x)}} \, dx=\frac {\frac {4}{{\left (\frac {1}{\sin \left (b x + a\right )^{2}} - 1\right )} \sqrt {\sin \left (b x + a\right )}} + 2 \, \arctan \left (\frac {1}{\sqrt {\sin \left (b x + a\right )}}\right ) + \log \left (\frac {1}{\sqrt {\sin \left (b x + a\right )}} + 1\right ) - \log \left (\frac {1}{\sqrt {\sin \left (b x + a\right )}} - 1\right )}{8 \, b} \] Input:
integrate(sec(b*x+a)^3/csc(b*x+a)^(1/2),x, algorithm="maxima")
Output:
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
Time = 0.12 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.03 \[ \int \frac {\sec ^3(a+b x)}{\sqrt {\csc (a+b x)}} \, dx=-\frac {\frac {4 \, \sin \left (b x + a\right )^{\frac {3}{2}}}{\sin \left (b x + a\right )^{2} - 1} + 2 \, \arctan \left (\sqrt {\sin \left (b x + a\right )}\right ) - \log \left (\sqrt {\sin \left (b x + a\right )} + 1\right ) + \log \left ({\left | \sqrt {\sin \left (b x + a\right )} - 1 \right |}\right )}{8 \, b} \] Input:
integrate(sec(b*x+a)^3/csc(b*x+a)^(1/2),x, algorithm="giac")
Output:
-1/8*(4*sin(b*x + a)^(3/2)/(sin(b*x + a)^2 - 1) + 2*arctan(sqrt(sin(b*x + a))) - log(sqrt(sin(b*x + a)) + 1) + log(abs(sqrt(sin(b*x + a)) - 1)))/b
Timed out. \[ \int \frac {\sec ^3(a+b x)}{\sqrt {\csc (a+b x)}} \, dx=\int \frac {1}{{\cos \left (a+b\,x\right )}^3\,\sqrt {\frac {1}{\sin \left (a+b\,x\right )}}} \,d x \] Input:
int(1/(cos(a + b*x)^3*(1/sin(a + b*x))^(1/2)),x)
Output:
int(1/(cos(a + b*x)^3*(1/sin(a + b*x))^(1/2)), x)
\[ \int \frac {\sec ^3(a+b x)}{\sqrt {\csc (a+b x)}} \, dx=\int \frac {\sqrt {\csc \left (b x +a \right )}\, \sec \left (b x +a \right )^{3}}{\csc \left (b x +a \right )}d x \] Input:
int(sec(b*x+a)^3/csc(b*x+a)^(1/2),x)
Output:
int((sqrt(csc(a + b*x))*sec(a + b*x)**3)/csc(a + b*x),x)