Integrand size = 14, antiderivative size = 110 \[ \int \left (b \tan ^4(c+d x)\right )^{3/2} \, dx=\frac {b \cot (c+d x) \sqrt {b \tan ^4(c+d x)}}{d}-b x \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)}-\frac {b \tan (c+d x) \sqrt {b \tan ^4(c+d x)}}{3 d}+\frac {b \tan ^3(c+d x) \sqrt {b \tan ^4(c+d x)}}{5 d} \] Output:
b*cot(d*x+c)*(tan(d*x+c)^4*b)^(1/2)/d-b*x*cot(d*x+c)^2*(tan(d*x+c)^4*b)^(1 /2)-1/3*b*tan(d*x+c)*(tan(d*x+c)^4*b)^(1/2)/d+1/5*b*tan(d*x+c)^3*(tan(d*x+ c)^4*b)^(1/2)/d
Time = 0.50 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.60 \[ \int \left (b \tan ^4(c+d x)\right )^{3/2} \, dx=\frac {\cot (c+d x) \left (3-5 \cot ^2(c+d x)+15 \cot ^4(c+d x)-15 \arctan (\tan (c+d x)) \cot ^5(c+d x)\right ) \left (b \tan ^4(c+d x)\right )^{3/2}}{15 d} \] Input:
Integrate[(b*Tan[c + d*x]^4)^(3/2),x]
Output:
(Cot[c + d*x]*(3 - 5*Cot[c + d*x]^2 + 15*Cot[c + d*x]^4 - 15*ArcTan[Tan[c + d*x]]*Cot[c + d*x]^5)*(b*Tan[c + d*x]^4)^(3/2))/(15*d)
Time = 0.38 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.62, 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.
\(\displaystyle \int \left (b \tan ^4(c+d x)\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (b \tan (c+d x)^4\right )^{3/2}dx\) |
\(\Big \downarrow \) 4141 |
\(\displaystyle b \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)} \int \tan ^6(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)} \int \tan (c+d x)^6dx\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle b \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)} \left (\frac {\tan ^5(c+d x)}{5 d}-\int \tan ^4(c+d x)dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)} \left (\frac {\tan ^5(c+d x)}{5 d}-\int \tan (c+d x)^4dx\right )\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle b \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)} \left (\int \tan ^2(c+d x)dx+\frac {\tan ^5(c+d x)}{5 d}-\frac {\tan ^3(c+d x)}{3 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)} \left (\int \tan (c+d x)^2dx+\frac {\tan ^5(c+d x)}{5 d}-\frac {\tan ^3(c+d x)}{3 d}\right )\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle b \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)} \left (-\int 1dx+\frac {\tan ^5(c+d x)}{5 d}-\frac {\tan ^3(c+d x)}{3 d}+\frac {\tan (c+d x)}{d}\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle b \left (\frac {\tan ^5(c+d x)}{5 d}-\frac {\tan ^3(c+d x)}{3 d}+\frac {\tan (c+d x)}{d}-x\right ) \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)}\) |
Input:
Int[(b*Tan[c + d*x]^4)^(3/2),x]
Output:
b*Cot[c + d*x]^2*Sqrt[b*Tan[c + d*x]^4]*(-x + Tan[c + d*x]/d - Tan[c + d*x ]^3/(3*d) + Tan[c + d*x]^5/(5*d))
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 = 0.05 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.58
method | result | size |
derivativedivides | \(-\frac {\left (b \tan \left (d x +c \right )^{4}\right )^{\frac {3}{2}} \left (-3 \tan \left (d x +c \right )^{5}+5 \tan \left (d x +c \right )^{3}+15 \arctan \left (\tan \left (d x +c \right )\right )-15 \tan \left (d x +c \right )\right )}{15 d \tan \left (d x +c \right )^{6}}\) | \(64\) |
default | \(-\frac {\left (b \tan \left (d x +c \right )^{4}\right )^{\frac {3}{2}} \left (-3 \tan \left (d x +c \right )^{5}+5 \tan \left (d x +c \right )^{3}+15 \arctan \left (\tan \left (d x +c \right )\right )-15 \tan \left (d x +c \right )\right )}{15 d \tan \left (d x +c \right )^{6}}\) | \(64\) |
risch | \(\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}}\, x}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {2 i b \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}}\, \left (45 \,{\mathrm e}^{8 i \left (d x +c \right )}+90 \,{\mathrm e}^{6 i \left (d x +c \right )}+140 \,{\mathrm e}^{4 i \left (d x +c \right )}+70 \,{\mathrm e}^{2 i \left (d x +c \right )}+23\right )}{15 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} d}\) | \(170\) |
Input:
int((b*tan(d*x+c)^4)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/15/d*(b*tan(d*x+c)^4)^(3/2)*(-3*tan(d*x+c)^5+5*tan(d*x+c)^3+15*arctan(t an(d*x+c))-15*tan(d*x+c))/tan(d*x+c)^6
Time = 0.12 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.56 \[ \int \left (b \tan ^4(c+d x)\right )^{3/2} \, dx=\frac {{\left (3 \, b \tan \left (d x + c\right )^{5} - 5 \, b \tan \left (d x + c\right )^{3} - 15 \, b d x + 15 \, b \tan \left (d x + c\right )\right )} \sqrt {b \tan \left (d x + c\right )^{4}}}{15 \, d \tan \left (d x + c\right )^{2}} \] Input:
integrate((b*tan(d*x+c)^4)^(3/2),x, algorithm="fricas")
Output:
1/15*(3*b*tan(d*x + c)^5 - 5*b*tan(d*x + c)^3 - 15*b*d*x + 15*b*tan(d*x + c))*sqrt(b*tan(d*x + c)^4)/(d*tan(d*x + c)^2)
\[ \int \left (b \tan ^4(c+d x)\right )^{3/2} \, dx=\int \left (b \tan ^{4}{\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx \] Input:
integrate((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 = 53, normalized size of antiderivative = 0.48 \[ \int \left (b \tan ^4(c+d x)\right )^{3/2} \, dx=\frac {3 \, b^{\frac {3}{2}} \tan \left (d x + c\right )^{5} - 5 \, b^{\frac {3}{2}} \tan \left (d x + c\right )^{3} - 15 \, {\left (d x + c\right )} b^{\frac {3}{2}} + 15 \, b^{\frac {3}{2}} \tan \left (d x + c\right )}{15 \, d} \] Input:
integrate((b*tan(d*x+c)^4)^(3/2),x, algorithm="maxima")
Output:
1/15*(3*b^(3/2)*tan(d*x + c)^5 - 5*b^(3/2)*tan(d*x + c)^3 - 15*(d*x + c)*b ^(3/2) + 15*b^(3/2)*tan(d*x + c))/d
Time = 0.16 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.54 \[ \int \left (b \tan ^4(c+d x)\right )^{3/2} \, dx=-\frac {1}{15} \, b^{\frac {3}{2}} {\left (\frac {15 \, {\left (d x + c\right )}}{d} - \frac {3 \, d^{4} \tan \left (d x + c\right )^{5} - 5 \, d^{4} \tan \left (d x + c\right )^{3} + 15 \, d^{4} \tan \left (d x + c\right )}{d^{5}}\right )} \] Input:
integrate((b*tan(d*x+c)^4)^(3/2),x, algorithm="giac")
Output:
-1/15*b^(3/2)*(15*(d*x + c)/d - (3*d^4*tan(d*x + c)^5 - 5*d^4*tan(d*x + c) ^3 + 15*d^4*tan(d*x + c))/d^5)
Timed out. \[ \int \left (b \tan ^4(c+d x)\right )^{3/2} \, dx=\int {\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^4\right )}^{3/2} \,d x \] Input:
int((b*tan(c + d*x)^4)^(3/2),x)
Output:
int((b*tan(c + d*x)^4)^(3/2), x)
Time = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.37 \[ \int \left (b \tan ^4(c+d x)\right )^{3/2} \, dx=\frac {\sqrt {b}\, b \left (3 \tan \left (d x +c \right )^{5}-5 \tan \left (d x +c \right )^{3}+15 \tan \left (d x +c \right )-15 d x \right )}{15 d} \] Input:
int((b*tan(d*x+c)^4)^(3/2),x)
Output:
(sqrt(b)*b*(3*tan(c + d*x)**5 - 5*tan(c + d*x)**3 + 15*tan(c + d*x) - 15*d *x))/(15*d)