Integrand size = 25, antiderivative size = 143 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {(2 a+5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{2 a^{7/2} f}-\frac {2 a+5 b}{6 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\csc ^2(e+f x)}{2 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {2 a+5 b}{2 a^3 f \sqrt {a+b \sin ^2(e+f x)}} \]
1/2*(2*a+5*b)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/a^(1/2))/a^(7/2)/f+1/6*(-2* a-5*b)/a^2/f/(a+b*sin(f*x+e)^2)^(3/2)-1/2*csc(f*x+e)^2/a/f/(a+b*sin(f*x+e) ^2)^(3/2)+1/2*(-2*a-5*b)/a^3/f/(a+b*sin(f*x+e)^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.48 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {3 a \csc ^2(e+f x)+(2 a+5 b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},1+\frac {b \sin ^2(e+f x)}{a}\right )}{6 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]
-1/6*(3*a*Csc[e + f*x]^2 + (2*a + 5*b)*Hypergeometric2F1[-3/2, 1, -1/2, 1 + (b*Sin[e + f*x]^2)/a])/(a^2*f*(a + b*Sin[e + f*x]^2)^(3/2))
Time = 0.31 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.92, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3673, 87, 61, 61, 73, 221}
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.
\(\displaystyle \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (e+f x)^3 \left (a+b \sin (e+f x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3673 |
\(\displaystyle \frac {\int \frac {\csc ^4(e+f x) \left (1-\sin ^2(e+f x)\right )}{\left (b \sin ^2(e+f x)+a\right )^{5/2}}d\sin ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {-\frac {(2 a+5 b) \int \frac {\csc ^2(e+f x)}{\left (b \sin ^2(e+f x)+a\right )^{5/2}}d\sin ^2(e+f x)}{2 a}-\frac {\csc ^2(e+f x)}{a \left (a+b \sin ^2(e+f x)\right )^{3/2}}}{2 f}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {-\frac {(2 a+5 b) \left (\frac {\int \frac {\csc ^2(e+f x)}{\left (b \sin ^2(e+f x)+a\right )^{3/2}}d\sin ^2(e+f x)}{a}+\frac {2}{3 a \left (a+b \sin ^2(e+f x)\right )^{3/2}}\right )}{2 a}-\frac {\csc ^2(e+f x)}{a \left (a+b \sin ^2(e+f x)\right )^{3/2}}}{2 f}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {-\frac {(2 a+5 b) \left (\frac {\frac {\int \frac {\csc ^2(e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)}{a}+\frac {2}{a \sqrt {a+b \sin ^2(e+f x)}}}{a}+\frac {2}{3 a \left (a+b \sin ^2(e+f x)\right )^{3/2}}\right )}{2 a}-\frac {\csc ^2(e+f x)}{a \left (a+b \sin ^2(e+f x)\right )^{3/2}}}{2 f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {-\frac {(2 a+5 b) \left (\frac {\frac {2 \int \frac {1}{\frac {\sin ^4(e+f x)}{b}-\frac {a}{b}}d\sqrt {b \sin ^2(e+f x)+a}}{a b}+\frac {2}{a \sqrt {a+b \sin ^2(e+f x)}}}{a}+\frac {2}{3 a \left (a+b \sin ^2(e+f x)\right )^{3/2}}\right )}{2 a}-\frac {\csc ^2(e+f x)}{a \left (a+b \sin ^2(e+f x)\right )^{3/2}}}{2 f}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {-\frac {(2 a+5 b) \left (\frac {\frac {2}{a \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{a^{3/2}}}{a}+\frac {2}{3 a \left (a+b \sin ^2(e+f x)\right )^{3/2}}\right )}{2 a}-\frac {\csc ^2(e+f x)}{a \left (a+b \sin ^2(e+f x)\right )^{3/2}}}{2 f}\) |
(-(Csc[e + f*x]^2/(a*(a + b*Sin[e + f*x]^2)^(3/2))) - ((2*a + 5*b)*(2/(3*a *(a + b*Sin[e + f*x]^2)^(3/2)) + ((-2*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/S qrt[a]])/a^(3/2) + 2/(a*Sqrt[a + b*Sin[e + f*x]^2]))/a))/(2*a))/(2*f)
3.6.36.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^ (m_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x]^2, x]}, Simp[ff^((m + 1)/2)/(2*f) Subst[Int[x^((m - 1)/2)*((a + b*ff*x)^p/(1 - ff*x)^((m + 1 )/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && Integ erQ[(m - 1)/2]
Leaf count of result is larger than twice the leaf count of optimal. \(1037\) vs. \(2(123)=246\).
Time = 2.25 (sec) , antiderivative size = 1038, normalized size of antiderivative = 7.26
1/6/a^(13/2)/b^2/(cos(f*x+e)^6*b^2-2*cos(f*x+e)^4*a*b-3*cos(f*x+e)^4*b^2+c os(f*x+e)^2*a^2+4*cos(f*x+e)^2*a*b+3*cos(f*x+e)^2*b^2-a^2-2*a*b-b^2)*(3*a^ (11/2)*b^2*(a+b-b*cos(f*x+e)^2)^(1/2)-6*a^6*b^2*ln(2/sin(f*x+e)*(a^(1/2)*( a+b-b*cos(f*x+e)^2)^(1/2)+a))+3*(a+b-b*cos(f*x+e)^2)^(1/2)*a^(7/2)*b^4+8*a ^(11/2)*b^2*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)+20*a^(9/2)*(-b*cos(f*x +e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^3+6*a^(9/2)*b^3*(a+b-b*cos(f*x+e)^2)^(1/2)+ 12*a^(7/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^4-27*a^5*b^3*ln(2/sin (f*x+e)*(a^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+a))-36*a^4*b^4*ln(2/sin(f*x+e) *(a^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+a))-15*ln(2/sin(f*x+e)*(a^(1/2)*(a+b- b*cos(f*x+e)^2)^(1/2)+a))*a^3*b^5+3*ln(2/sin(f*x+e)*(a^(1/2)*(a+b-b*cos(f* x+e)^2)^(1/2)+a))*a^3*b^4*(2*a+5*b)*cos(f*x+e)^6+3*cos(f*x+e)^4*b^3*(2*a^( 9/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)+(a+b-b*cos(f*x+e)^2)^(1/2)*a^ (7/2)*b+4*a^(7/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b-4*ln(2/sin(f*x +e)*(a^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+a))*a^5-16*ln(2/sin(f*x+e)*(a^(1/2 )*(a+b-b*cos(f*x+e)^2)^(1/2)+a))*a^4*b-15*ln(2/sin(f*x+e)*(a^(1/2)*(a+b-b* cos(f*x+e)^2)^(1/2)+a))*a^3*b^2)-cos(f*x+e)^2*b^2*(8*a^(11/2)*(-b*cos(f*x+ e)^2+(a*b^2+b^3)/b^2)^(1/2)+6*(a+b-b*cos(f*x+e)^2)^(1/2)*a^(9/2)*b+26*a^(9 /2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b+6*(a+b-b*cos(f*x+e)^2)^(1/2) *a^(7/2)*b^2+24*a^(7/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^2-6*ln(2 /sin(f*x+e)*(a^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+a))*a^6-39*ln(2/sin(f*x...
Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (123) = 246\).
Time = 0.38 (sec) , antiderivative size = 666, normalized size of antiderivative = 4.66 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\left [\frac {3 \, {\left ({\left (2 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (f x + e\right )^{6} - {\left (4 \, a^{2} b + 16 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - 2 \, a^{3} - 9 \, a^{2} b - 12 \, a b^{2} - 5 \, b^{3} + {\left (2 \, a^{3} + 13 \, a^{2} b + 26 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \log \left (\frac {2 \, {\left (b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} - 2 \, a - b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) + 2 \, {\left (3 \, {\left (2 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{4} + 11 \, a^{3} + 26 \, a^{2} b + 15 \, a b^{2} - 2 \, {\left (4 \, a^{3} + 16 \, a^{2} b + 15 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{12 \, {\left (a^{4} b^{2} f \cos \left (f x + e\right )^{6} - {\left (2 \, a^{5} b + 3 \, a^{4} b^{2}\right )} f \cos \left (f x + e\right )^{4} + {\left (a^{6} + 4 \, a^{5} b + 3 \, a^{4} b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2}\right )} f\right )}}, -\frac {3 \, {\left ({\left (2 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (f x + e\right )^{6} - {\left (4 \, a^{2} b + 16 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - 2 \, a^{3} - 9 \, a^{2} b - 12 \, a b^{2} - 5 \, b^{3} + {\left (2 \, a^{3} + 13 \, a^{2} b + 26 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{a}\right ) - {\left (3 \, {\left (2 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{4} + 11 \, a^{3} + 26 \, a^{2} b + 15 \, a b^{2} - 2 \, {\left (4 \, a^{3} + 16 \, a^{2} b + 15 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{6 \, {\left (a^{4} b^{2} f \cos \left (f x + e\right )^{6} - {\left (2 \, a^{5} b + 3 \, a^{4} b^{2}\right )} f \cos \left (f x + e\right )^{4} + {\left (a^{6} + 4 \, a^{5} b + 3 \, a^{4} b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2}\right )} f\right )}}\right ] \]
[1/12*(3*((2*a*b^2 + 5*b^3)*cos(f*x + e)^6 - (4*a^2*b + 16*a*b^2 + 15*b^3) *cos(f*x + e)^4 - 2*a^3 - 9*a^2*b - 12*a*b^2 - 5*b^3 + (2*a^3 + 13*a^2*b + 26*a*b^2 + 15*b^3)*cos(f*x + e)^2)*sqrt(a)*log(2*(b*cos(f*x + e)^2 - 2*sq rt(-b*cos(f*x + e)^2 + a + b)*sqrt(a) - 2*a - b)/(cos(f*x + e)^2 - 1)) + 2 *(3*(2*a^2*b + 5*a*b^2)*cos(f*x + e)^4 + 11*a^3 + 26*a^2*b + 15*a*b^2 - 2* (4*a^3 + 16*a^2*b + 15*a*b^2)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/(a^4*b^2*f*cos(f*x + e)^6 - (2*a^5*b + 3*a^4*b^2)*f*cos(f*x + e)^4 + (a^6 + 4*a^5*b + 3*a^4*b^2)*f*cos(f*x + e)^2 - (a^6 + 2*a^5*b + a^4*b^2)*f ), -1/6*(3*((2*a*b^2 + 5*b^3)*cos(f*x + e)^6 - (4*a^2*b + 16*a*b^2 + 15*b^ 3)*cos(f*x + e)^4 - 2*a^3 - 9*a^2*b - 12*a*b^2 - 5*b^3 + (2*a^3 + 13*a^2*b + 26*a*b^2 + 15*b^3)*cos(f*x + e)^2)*sqrt(-a)*arctan(sqrt(-b*cos(f*x + e) ^2 + a + b)*sqrt(-a)/a) - (3*(2*a^2*b + 5*a*b^2)*cos(f*x + e)^4 + 11*a^3 + 26*a^2*b + 15*a*b^2 - 2*(4*a^3 + 16*a^2*b + 15*a*b^2)*cos(f*x + e)^2)*sqr t(-b*cos(f*x + e)^2 + a + b))/(a^4*b^2*f*cos(f*x + e)^6 - (2*a^5*b + 3*a^4 *b^2)*f*cos(f*x + e)^4 + (a^6 + 4*a^5*b + 3*a^4*b^2)*f*cos(f*x + e)^2 - (a ^6 + 2*a^5*b + a^4*b^2)*f)]
\[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\cot ^{3}{\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Time = 0.26 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.09 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\frac {6 \, \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {5}{2}}} + \frac {15 \, b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {7}{2}}} - \frac {6}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2}} - \frac {2}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a} - \frac {15 \, b}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{3}} - \frac {5 \, b}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2}} - \frac {3}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a \sin \left (f x + e\right )^{2}}}{6 \, f} \]
1/6*(6*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e))))/a^(5/2) + 15*b*arcsinh(a/( sqrt(a*b)*abs(sin(f*x + e))))/a^(7/2) - 6/(sqrt(b*sin(f*x + e)^2 + a)*a^2) - 2/((b*sin(f*x + e)^2 + a)^(3/2)*a) - 15*b/(sqrt(b*sin(f*x + e)^2 + a)*a ^3) - 5*b/((b*sin(f*x + e)^2 + a)^(3/2)*a^2) - 3/((b*sin(f*x + e)^2 + a)^( 3/2)*a*sin(f*x + e)^2))/f
\[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\cot \left (f x + e\right )^{3}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {{\mathrm {cot}\left (e+f\,x\right )}^3}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \]