Integrand size = 14, antiderivative size = 119 \[ \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{3/2}} \, dx=\frac {\cot (c+d x)}{3 b d \sqrt {b \tan ^4(c+d x)}}-\frac {\cot ^3(c+d x)}{5 b d \sqrt {b \tan ^4(c+d x)}}-\frac {\tan (c+d x)}{b d \sqrt {b \tan ^4(c+d x)}}-\frac {x \tan ^2(c+d x)}{b \sqrt {b \tan ^4(c+d x)}} \] Output:
1/3*cot(d*x+c)/b/d/(tan(d*x+c)^4*b)^(1/2)-1/5*cot(d*x+c)^3/b/d/(tan(d*x+c) ^4*b)^(1/2)-tan(d*x+c)/b/d/(tan(d*x+c)^4*b)^(1/2)-x*tan(d*x+c)^2/b/(tan(d* x+c)^4*b)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.38 \[ \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{3/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},-\tan ^2(c+d x)\right ) \tan (c+d x)}{5 d \left (b \tan ^4(c+d x)\right )^{3/2}} \] Input:
Integrate[(b*Tan[c + d*x]^4)^(-3/2),x]
Output:
-1/5*(Hypergeometric2F1[-5/2, 1, -3/2, -Tan[c + d*x]^2]*Tan[c + d*x])/(d*( b*Tan[c + d*x]^4)^(3/2))
Time = 0.38 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.60, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {3042, 4141, 3042, 3954, 3042, 3954, 3042, 3954, 24}
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.
\(\displaystyle \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (b \tan (c+d x)^4\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4141 |
\(\displaystyle \frac {\tan ^2(c+d x) \int \cot ^6(c+d x)dx}{b \sqrt {b \tan ^4(c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan ^2(c+d x) \int \tan \left (c+d x+\frac {\pi }{2}\right )^6dx}{b \sqrt {b \tan ^4(c+d x)}}\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle \frac {\tan ^2(c+d x) \left (-\int \cot ^4(c+d x)dx-\frac {\cot ^5(c+d x)}{5 d}\right )}{b \sqrt {b \tan ^4(c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan ^2(c+d x) \left (-\int \tan \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {\cot ^5(c+d x)}{5 d}\right )}{b \sqrt {b \tan ^4(c+d x)}}\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle \frac {\tan ^2(c+d x) \left (\int \cot ^2(c+d x)dx-\frac {\cot ^5(c+d x)}{5 d}+\frac {\cot ^3(c+d x)}{3 d}\right )}{b \sqrt {b \tan ^4(c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan ^2(c+d x) \left (\int \tan \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {\cot ^5(c+d x)}{5 d}+\frac {\cot ^3(c+d x)}{3 d}\right )}{b \sqrt {b \tan ^4(c+d x)}}\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle \frac {\tan ^2(c+d x) \left (-\int 1dx-\frac {\cot ^5(c+d x)}{5 d}+\frac {\cot ^3(c+d x)}{3 d}-\frac {\cot (c+d x)}{d}\right )}{b \sqrt {b \tan ^4(c+d x)}}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\tan ^2(c+d x) \left (-\frac {\cot ^5(c+d x)}{5 d}+\frac {\cot ^3(c+d x)}{3 d}-\frac {\cot (c+d x)}{d}-x\right )}{b \sqrt {b \tan ^4(c+d x)}}\) |
Input:
Int[(b*Tan[c + d*x]^4)^(-3/2),x]
Output:
((-x - Cot[c + d*x]/d + Cot[c + d*x]^3/(3*d) - Cot[c + d*x]^5/(5*d))*Tan[c + d*x]^2)/(b*Sqrt[b*Tan[c + d*x]^4])
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d *x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2 Int[(b*Tan[c + d*x])^(n - 2), x] , x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^ n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p])) Int[ActivateTrig[u]*(Ta n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] && !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / ; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
Time = 0.06 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.53
method | result | size |
derivativedivides | \(-\frac {\tan \left (d x +c \right ) \left (15 \arctan \left (\tan \left (d x +c \right )\right ) \tan \left (d x +c \right )^{5}+15 \tan \left (d x +c \right )^{4}-5 \tan \left (d x +c \right )^{2}+3\right )}{15 d \left (b \tan \left (d x +c \right )^{4}\right )^{\frac {3}{2}}}\) | \(63\) |
default | \(-\frac {\tan \left (d x +c \right ) \left (15 \arctan \left (\tan \left (d x +c \right )\right ) \tan \left (d x +c \right )^{5}+15 \tan \left (d x +c \right )^{4}-5 \tan \left (d x +c \right )^{2}+3\right )}{15 d \left (b \tan \left (d x +c \right )^{4}\right )^{\frac {3}{2}}}\) | \(63\) |
risch | \(\frac {\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} x}{b \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}}}+\frac {2 i \left (45 \,{\mathrm e}^{8 i \left (d x +c \right )}-90 \,{\mathrm e}^{6 i \left (d x +c \right )}+140 \,{\mathrm e}^{4 i \left (d x +c \right )}-70 \,{\mathrm e}^{2 i \left (d x +c \right )}+23\right )}{15 b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}}\, d}\) | \(174\) |
Input:
int(1/(b*tan(d*x+c)^4)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/15/d*tan(d*x+c)*(15*arctan(tan(d*x+c))*tan(d*x+c)^5+15*tan(d*x+c)^4-5*t an(d*x+c)^2+3)/(b*tan(d*x+c)^4)^(3/2)
Time = 0.09 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.52 \[ \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{3/2}} \, dx=-\frac {{\left (15 \, d x \tan \left (d x + c\right )^{5} + 15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3\right )} \sqrt {b \tan \left (d x + c\right )^{4}}}{15 \, b^{2} d \tan \left (d x + c\right )^{7}} \] Input:
integrate(1/(b*tan(d*x+c)^4)^(3/2),x, algorithm="fricas")
Output:
-1/15*(15*d*x*tan(d*x + c)^5 + 15*tan(d*x + c)^4 - 5*tan(d*x + c)^2 + 3)*s qrt(b*tan(d*x + c)^4)/(b^2*d*tan(d*x + c)^7)
\[ \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{3/2}} \, dx=\int \frac {1}{\left (b \tan ^{4}{\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/(b*tan(d*x+c)**4)**(3/2),x)
Output:
Integral((b*tan(c + d*x)**4)**(-3/2), x)
Time = 0.11 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.42 \[ \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{3/2}} \, dx=-\frac {\frac {15 \, {\left (d x + c\right )}}{b^{\frac {3}{2}}} + \frac {15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3}{b^{\frac {3}{2}} \tan \left (d x + c\right )^{5}}}{15 \, d} \] Input:
integrate(1/(b*tan(d*x+c)^4)^(3/2),x, algorithm="maxima")
Output:
-1/15*(15*(d*x + c)/b^(3/2) + (15*tan(d*x + c)^4 - 5*tan(d*x + c)^2 + 3)/( b^(3/2)*tan(d*x + c)^5))/d
Time = 0.19 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.47 \[ \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{3/2}} \, dx=-\frac {\frac {15 \, {\left (d x + c\right )}}{b d} + \frac {15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3}{b d \tan \left (d x + c\right )^{5}}}{15 \, \sqrt {b}} \] Input:
integrate(1/(b*tan(d*x+c)^4)^(3/2),x, algorithm="giac")
Output:
-1/15*(15*(d*x + c)/(b*d) + (15*tan(d*x + c)^4 - 5*tan(d*x + c)^2 + 3)/(b* d*tan(d*x + c)^5))/sqrt(b)
Timed out. \[ \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{3/2}} \, dx=\int \frac {1}{{\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^4\right )}^{3/2}} \,d x \] Input:
int(1/(b*tan(c + d*x)^4)^(3/2),x)
Output:
int(1/(b*tan(c + d*x)^4)^(3/2), x)
Time = 0.15 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.44 \[ \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{3/2}} \, dx=\frac {\sqrt {b}\, \left (-15 \tan \left (d x +c \right )^{5} d x -15 \tan \left (d x +c \right )^{4}+5 \tan \left (d x +c \right )^{2}-3\right )}{15 \tan \left (d x +c \right )^{5} b^{2} d} \] Input:
int(1/(b*tan(d*x+c)^4)^(3/2),x)
Output:
(sqrt(b)*( - 15*tan(c + d*x)**5*d*x - 15*tan(c + d*x)**4 + 5*tan(c + d*x)* *2 - 3))/(15*tan(c + d*x)**5*b**2*d)