Integrand size = 21, antiderivative size = 123 \[ \int \csc ^5(e+f x) \sqrt {b \sec (e+f x)} \, dx=\frac {21 \sqrt {b} \arctan \left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{32 f}-\frac {21 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{32 f}-\frac {7 \cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{16 b f}-\frac {\cot ^4(e+f x) (b \sec (e+f x))^{7/2}}{4 b^3 f} \] Output:
21/32*b^(1/2)*arctan((b*sec(f*x+e))^(1/2)/b^(1/2))/f-21/32*b^(1/2)*arctanh ((b*sec(f*x+e))^(1/2)/b^(1/2))/f-7/16*cot(f*x+e)^2*(b*sec(f*x+e))^(3/2)/b/ f-1/4*cot(f*x+e)^4*(b*sec(f*x+e))^(7/2)/b^3/f
Time = 1.25 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.87 \[ \int \csc ^5(e+f x) \sqrt {b \sec (e+f x)} \, dx=\frac {b \left (-28 \csc ^2(e+f x)-16 \csc ^4(e+f x)+42 \arctan \left (\sqrt {\sec (e+f x)}\right ) \sqrt {\sec (e+f x)}+21 \left (\log \left (1-\sqrt {\sec (e+f x)}\right )-\log \left (1+\sqrt {\sec (e+f x)}\right )\right ) \sqrt {\sec (e+f x)}\right )}{64 f \sqrt {b \sec (e+f x)}} \] Input:
Integrate[Csc[e + f*x]^5*Sqrt[b*Sec[e + f*x]],x]
Output:
(b*(-28*Csc[e + f*x]^2 - 16*Csc[e + f*x]^4 + 42*ArcTan[Sqrt[Sec[e + f*x]]] *Sqrt[Sec[e + f*x]] + 21*(Log[1 - Sqrt[Sec[e + f*x]]] - Log[1 + Sqrt[Sec[e + f*x]]])*Sqrt[Sec[e + f*x]]))/(64*f*Sqrt[b*Sec[e + f*x]])
Time = 0.48 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3102, 25, 27, 252, 252, 266, 827, 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.
\(\displaystyle \int \csc ^5(e+f x) \sqrt {b \sec (e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc (e+f x)^5 \sqrt {b \sec (e+f x)}dx\) |
\(\Big \downarrow \) 3102 |
\(\displaystyle \frac {\int -\frac {b^6 (b \sec (e+f x))^{9/2}}{\left (b^2-b^2 \sec ^2(e+f x)\right )^3}d(b \sec (e+f x))}{b^5 f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {b^6 (b \sec (e+f x))^{9/2}}{\left (b^2-b^2 \sec ^2(e+f x)\right )^3}d(b \sec (e+f x))}{b^5 f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {b \int \frac {(b \sec (e+f x))^{9/2}}{\left (b^2-b^2 \sec ^2(e+f x)\right )^3}d(b \sec (e+f x))}{f}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {b \left (\frac {(b \sec (e+f x))^{7/2}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}-\frac {7}{8} \int \frac {(b \sec (e+f x))^{5/2}}{\left (b^2-b^2 \sec ^2(e+f x)\right )^2}d(b \sec (e+f x))\right )}{f}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {b \left (\frac {(b \sec (e+f x))^{7/2}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}-\frac {7}{8} \left (\frac {(b \sec (e+f x))^{3/2}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {3}{4} \int \frac {\sqrt {b \sec (e+f x)}}{b^2-b^2 \sec ^2(e+f x)}d(b \sec (e+f x))\right )\right )}{f}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle -\frac {b \left (\frac {(b \sec (e+f x))^{7/2}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}-\frac {7}{8} \left (\frac {(b \sec (e+f x))^{3/2}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {3}{2} \int \frac {b^2 \sec ^2(e+f x)}{b^2-b^4 \sec ^4(e+f x)}d\sqrt {b \sec (e+f x)}\right )\right )}{f}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle -\frac {b \left (\frac {(b \sec (e+f x))^{7/2}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}-\frac {7}{8} \left (\frac {(b \sec (e+f x))^{3/2}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {3}{2} \left (\frac {1}{2} \int \frac {1}{b-b^2 \sec ^2(e+f x)}d\sqrt {b \sec (e+f x)}-\frac {1}{2} \int \frac {1}{b^2 \sec ^2(e+f x)+b}d\sqrt {b \sec (e+f x)}\right )\right )\right )}{f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {b \left (\frac {(b \sec (e+f x))^{7/2}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}-\frac {7}{8} \left (\frac {(b \sec (e+f x))^{3/2}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {3}{2} \left (\frac {1}{2} \int \frac {1}{b-b^2 \sec ^2(e+f x)}d\sqrt {b \sec (e+f x)}-\frac {\arctan \left (\sqrt {b} \sec (e+f x)\right )}{2 \sqrt {b}}\right )\right )\right )}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {b \left (\frac {(b \sec (e+f x))^{7/2}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}-\frac {7}{8} \left (\frac {(b \sec (e+f x))^{3/2}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {3}{2} \left (\frac {\text {arctanh}\left (\sqrt {b} \sec (e+f x)\right )}{2 \sqrt {b}}-\frac {\arctan \left (\sqrt {b} \sec (e+f x)\right )}{2 \sqrt {b}}\right )\right )\right )}{f}\) |
Input:
Int[Csc[e + f*x]^5*Sqrt[b*Sec[e + f*x]],x]
Output:
-((b*((b*Sec[e + f*x])^(7/2)/(4*(b^2 - b^2*Sec[e + f*x]^2)^2) - (7*((-3*(- 1/2*ArcTan[Sqrt[b]*Sec[e + f*x]]/Sqrt[b] + ArcTanh[Sqrt[b]*Sec[e + f*x]]/( 2*Sqrt[b])))/2 + (b*Sec[e + f*x])^(3/2)/(2*(b^2 - b^2*Sec[e + f*x]^2))))/8 ))/f)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_S ymbol] :> Simp[1/(f*a^n) Subst[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/ 2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1 )/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Leaf count of result is larger than twice the leaf count of optimal. \(308\) vs. \(2(99)=198\).
Time = 2.80 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.51
method | result | size |
default | \(\frac {\left (\left (28 \cos \left (f x +e \right )^{2}-44\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+\left (-11 \cos \left (f x +e \right )+11\right ) \sin \left (f x +e \right )^{2} \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right )+\left (32 \cos \left (f x +e \right )-32\right ) \sin \left (f x +e \right )^{2} \ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \cos \left (f x +e \right )+2}{\cos \left (f x +e \right )+1}\right )+\left (-21 \cos \left (f x +e \right )+21\right ) \sin \left (f x +e \right )^{2} \arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )\right ) \sqrt {b \sec \left (f x +e \right )}\, \cot \left (f x +e \right ) \csc \left (f x +e \right )^{3}}{64 f \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\) | \(309\) |
Input:
int(csc(f*x+e)^5*(b*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
Output:
1/64/f*((28*cos(f*x+e)^2-44)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+(-11*cos (f*x+e)+11)*sin(f*x+e)^2*ln((2*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^( 1/2)+2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)+1)/(cos(f*x+e)+1))+ (32*cos(f*x+e)-32)*sin(f*x+e)^2*ln(2*(2*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e )+1)^2)^(1/2)+2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)+1)/(cos(f* x+e)+1))+(-21*cos(f*x+e)+21)*sin(f*x+e)^2*arctan(1/2/(-cos(f*x+e)/(cos(f*x +e)+1)^2)^(1/2)))*(b*sec(f*x+e))^(1/2)/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2 )*cot(f*x+e)*csc(f*x+e)^3
Leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (99) = 198\).
Time = 0.14 (sec) , antiderivative size = 457, normalized size of antiderivative = 3.72 \[ \int \csc ^5(e+f x) \sqrt {b \sec (e+f x)} \, dx=\left [\frac {42 \, {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt {-b} \arctan \left (\frac {2 \, \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b \cos \left (f x + e\right ) + b}\right ) + 21 \, {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt {-b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \, {\left (7 \, \cos \left (f x + e\right )^{3} - 11 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{128 \, {\left (f \cos \left (f x + e\right )^{4} - 2 \, f \cos \left (f x + e\right )^{2} + f\right )}}, \frac {42 \, {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt {b} \arctan \left (\frac {2 \, \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b \cos \left (f x + e\right ) - b}\right ) + 21 \, {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt {b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \, {\left (7 \, \cos \left (f x + e\right )^{3} - 11 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{128 \, {\left (f \cos \left (f x + e\right )^{4} - 2 \, f \cos \left (f x + e\right )^{2} + f\right )}}\right ] \] Input:
integrate(csc(f*x+e)^5*(b*sec(f*x+e))^(1/2),x, algorithm="fricas")
Output:
[1/128*(42*(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(-b)*arctan(2*sqrt( -b)*sqrt(b/cos(f*x + e))*cos(f*x + e)/(b*cos(f*x + e) + b)) + 21*(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(-b)*log((b*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 - cos(f*x + e))*sqrt(-b)*sqrt(b/cos(f*x + e)) - 6*b*cos(f*x + e) + b)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) + 8*(7*cos(f*x + e)^3 - 11*cos( f*x + e))*sqrt(b/cos(f*x + e)))/(f*cos(f*x + e)^4 - 2*f*cos(f*x + e)^2 + f ), 1/128*(42*(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(b)*arctan(2*sqrt (b)*sqrt(b/cos(f*x + e))*cos(f*x + e)/(b*cos(f*x + e) - b)) + 21*(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(b)*log((b*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 + cos(f*x + e))*sqrt(b)*sqrt(b/cos(f*x + e)) + 6*b*cos(f*x + e) + b )/(cos(f*x + e)^2 - 2*cos(f*x + e) + 1)) + 8*(7*cos(f*x + e)^3 - 11*cos(f* x + e))*sqrt(b/cos(f*x + e)))/(f*cos(f*x + e)^4 - 2*f*cos(f*x + e)^2 + f)]
\[ \int \csc ^5(e+f x) \sqrt {b \sec (e+f x)} \, dx=\int \sqrt {b \sec {\left (e + f x \right )}} \csc ^{5}{\left (e + f x \right )}\, dx \] Input:
integrate(csc(f*x+e)**5*(b*sec(f*x+e))**(1/2),x)
Output:
Integral(sqrt(b*sec(e + f*x))*csc(e + f*x)**5, x)
Time = 0.11 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.12 \[ \int \csc ^5(e+f x) \sqrt {b \sec (e+f x)} \, dx=\frac {b {\left (\frac {42 \, \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b}}\right )}{\sqrt {b}} + \frac {21 \, \log \left (-\frac {\sqrt {b} - \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b} + \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right )}{\sqrt {b}} + \frac {4 \, {\left (7 \, b^{2} \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {3}{2}} - 11 \, \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {7}{2}}\right )}}{b^{4} - \frac {2 \, b^{4}}{\cos \left (f x + e\right )^{2}} + \frac {b^{4}}{\cos \left (f x + e\right )^{4}}}\right )}}{64 \, f} \] Input:
integrate(csc(f*x+e)^5*(b*sec(f*x+e))^(1/2),x, algorithm="maxima")
Output:
1/64*b*(42*arctan(sqrt(b/cos(f*x + e))/sqrt(b))/sqrt(b) + 21*log(-(sqrt(b) - sqrt(b/cos(f*x + e)))/(sqrt(b) + sqrt(b/cos(f*x + e))))/sqrt(b) + 4*(7* b^2*(b/cos(f*x + e))^(3/2) - 11*(b/cos(f*x + e))^(7/2))/(b^4 - 2*b^4/cos(f *x + e)^2 + b^4/cos(f*x + e)^4))/f
Time = 0.13 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.03 \[ \int \csc ^5(e+f x) \sqrt {b \sec (e+f x)} \, dx=\frac {b^{6} {\left (\frac {21 \, \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{5}} - \frac {21 \, \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {b}}\right )}{b^{\frac {11}{2}}} + \frac {2 \, {\left (7 \, \sqrt {b \cos \left (f x + e\right )} b^{2} \cos \left (f x + e\right )^{2} - 11 \, \sqrt {b \cos \left (f x + e\right )} b^{2}\right )}}{{\left (b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )}^{2} b^{4}}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{32 \, f} \] Input:
integrate(csc(f*x+e)^5*(b*sec(f*x+e))^(1/2),x, algorithm="giac")
Output:
1/32*b^6*(21*arctan(sqrt(b*cos(f*x + e))/sqrt(-b))/(sqrt(-b)*b^5) - 21*arc tan(sqrt(b*cos(f*x + e))/sqrt(b))/b^(11/2) + 2*(7*sqrt(b*cos(f*x + e))*b^2 *cos(f*x + e)^2 - 11*sqrt(b*cos(f*x + e))*b^2)/((b^2*cos(f*x + e)^2 - b^2) ^2*b^4))*sgn(cos(f*x + e))/f
Timed out. \[ \int \csc ^5(e+f x) \sqrt {b \sec (e+f x)} \, dx=\int \frac {\sqrt {\frac {b}{\cos \left (e+f\,x\right )}}}{{\sin \left (e+f\,x\right )}^5} \,d x \] Input:
int((b/cos(e + f*x))^(1/2)/sin(e + f*x)^5,x)
Output:
int((b/cos(e + f*x))^(1/2)/sin(e + f*x)^5, x)
\[ \int \csc ^5(e+f x) \sqrt {b \sec (e+f x)} \, dx=\sqrt {b}\, \left (\int \sqrt {\sec \left (f x +e \right )}\, \csc \left (f x +e \right )^{5}d x \right ) \] Input:
int(csc(f*x+e)^5*(b*sec(f*x+e))^(1/2),x)
Output:
sqrt(b)*int(sqrt(sec(e + f*x))*csc(e + f*x)**5,x)