Integrand size = 33, antiderivative size = 686 \[ \int \frac {\tan ^3(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx=-\frac {\sqrt {2 a-2 c-\sqrt {a^2+b^2-2 a c+c^2}} \sqrt {a^2-b^2-2 a c+c^2+(a-c) \sqrt {a^2+b^2-2 a c+c^2}} \text {arctanh}\left (\frac {b^2-(a-c) \left (a-c+\sqrt {a^2+b^2-2 a c+c^2}\right )-b \left (2 a-2 c-\sqrt {a^2+b^2-2 a c+c^2}\right ) \tan (d+e x)}{\sqrt {2} \sqrt {2 a-2 c-\sqrt {a^2+b^2-2 a c+c^2}} \sqrt {a^2-b^2-2 a c+c^2+(a-c) \sqrt {a^2+b^2-2 a c+c^2}} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} \left (a^2+b^2-2 a c+c^2\right )^{3/2} e}+\frac {\sqrt {2 a-2 c+\sqrt {a^2+b^2-2 a c+c^2}} \sqrt {a^2-b^2-2 a c+c^2-(a-c) \sqrt {a^2+b^2-2 a c+c^2}} \text {arctanh}\left (\frac {b^2-(a-c) \left (a-c-\sqrt {a^2+b^2-2 a c+c^2}\right )-b \left (2 a-2 c+\sqrt {a^2+b^2-2 a c+c^2}\right ) \tan (d+e x)}{\sqrt {2} \sqrt {2 a-2 c+\sqrt {a^2+b^2-2 a c+c^2}} \sqrt {a^2-b^2-2 a c+c^2-(a-c) \sqrt {a^2+b^2-2 a c+c^2}} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} \left (a^2+b^2-2 a c+c^2\right )^{3/2} e}+\frac {2 (2 a+b \tan (d+e x))}{\left (b^2-4 a c\right ) e \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}-\frac {2 \left (a \left (b^2-2 (a-c) c\right )+b c (a+c) \tan (d+e x)\right )}{\left (b^2+(a-c)^2\right ) \left (b^2-4 a c\right ) e \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}} \]
1/2*arctanh(1/2*(b^2-(a-c)*(a-c-(a^2-2*a*c+b^2+c^2)^(1/2))-b*(2*a-2*c+(a^2 -2*a*c+b^2+c^2)^(1/2))*tan(e*x+d))*2^(1/2)/(2*a-2*c+(a^2-2*a*c+b^2+c^2)^(1 /2))^(1/2)/(a^2-b^2-2*a*c+c^2-(a-c)*(a^2-2*a*c+b^2+c^2)^(1/2))^(1/2)/(a+b* tan(e*x+d)+c*tan(e*x+d)^2)^(1/2))*(2*a-2*c+(a^2-2*a*c+b^2+c^2)^(1/2))^(1/2 )*(a^2-b^2-2*a*c+c^2-(a-c)*(a^2-2*a*c+b^2+c^2)^(1/2))^(1/2)/(a^2-2*a*c+b^2 +c^2)^(3/2)/e*2^(1/2)-1/2*arctanh(1/2*(b^2-(a-c)*(a-c+(a^2-2*a*c+b^2+c^2)^ (1/2))-b*(2*a-2*c-(a^2-2*a*c+b^2+c^2)^(1/2))*tan(e*x+d))*2^(1/2)/(2*a-2*c- (a^2-2*a*c+b^2+c^2)^(1/2))^(1/2)/(a^2-b^2-2*a*c+c^2+(a-c)*(a^2-2*a*c+b^2+c ^2)^(1/2))^(1/2)/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(1/2))*(2*a-2*c-(a^2-2*a* c+b^2+c^2)^(1/2))^(1/2)*(a^2-b^2-2*a*c+c^2+(a-c)*(a^2-2*a*c+b^2+c^2)^(1/2) )^(1/2)/(a^2-2*a*c+b^2+c^2)^(3/2)/e*2^(1/2)+2*(2*a+b*tan(e*x+d))/(-4*a*c+b ^2)/e/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(1/2)-2*(a*(b^2-2*(a-c)*c)+b*c*(a+c) *tan(e*x+d))/(b^2+(a-c)^2)/(-4*a*c+b^2)/e/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^ (1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 7.33 (sec) , antiderivative size = 2339, normalized size of antiderivative = 3.41 \[ \int \frac {\tan ^3(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx=\text {Result too large to show} \]
(Sqrt[(a + c + a*Cos[2*(d + e*x)] - c*Cos[2*(d + e*x)] + b*Sin[2*(d + e*x) ])/(1 + Cos[2*(d + e*x)])]*((-2*a*(2*a^2 + b^2 - 2*a*c))/((a - c)*(a - I*b - c)*(-(a*b^2) - I*b^3 + 4*a^2*c + (4*I)*a*b*c + b^2*c - 4*a*c^2)) + ((Co s[2*(d + e*x)] - I*Sin[2*(d + e*x)])*(I*a^3*b + (2*I)*a^2*b*c + I*b^3*c - (3*I)*a*b*c^2 + 8*a^3*c*Cos[2*(d + e*x)] + 4*a*b^2*c*Cos[2*(d + e*x)] - 8* a^2*c^2*Cos[2*(d + e*x)] - I*a^3*b*Cos[4*(d + e*x)] - (2*I)*a^2*b*c*Cos[4* (d + e*x)] - I*b^3*c*Cos[4*(d + e*x)] + (3*I)*a*b*c^2*Cos[4*(d + e*x)] + ( 8*I)*a^3*c*Sin[2*(d + e*x)] + (4*I)*a*b^2*c*Sin[2*(d + e*x)] - (8*I)*a^2*c ^2*Sin[2*(d + e*x)] + a^3*b*Sin[4*(d + e*x)] + 2*a^2*b*c*Sin[4*(d + e*x)] + b^3*c*Sin[4*(d + e*x)] - 3*a*b*c^2*Sin[4*(d + e*x)]))/((a - c)*(a - I*b - c)*(a + I*b - c)*(-b^2 + 4*a*c)*(a + c + a*Cos[2*(d + e*x)] - c*Cos[2*(d + e*x)] + b*Sin[2*(d + e*x)]))))/e - ((b*ArcTanh[(2*Sqrt[c]*Tan[(d + e*x) /2])/(Sqrt[a]*(-1 + Tan[(d + e*x)/2]^2) - Sqrt[a*(-1 + Tan[(d + e*x)/2]^2) ^2 + 2*Tan[(d + e*x)/2]*(b + 2*c*Tan[(d + e*x)/2] - b*Tan[(d + e*x)/2]^2)] )]*(1 + Cos[d + e*x])*Sqrt[(1 + Cos[2*(d + e*x)])/(1 + Cos[d + e*x])^2]*Sq rt[(a + c + (a - c)*Cos[2*(d + e*x)] + b*Sin[2*(d + e*x)])/(1 + Cos[2*(d + e*x)])]*(-1 + Tan[(d + e*x)/2]^2)*(1 + Tan[(d + e*x)/2]^2)*Sqrt[(a*(-1 + Tan[(d + e*x)/2]^2)^2 + 2*Tan[(d + e*x)/2]*(b + 2*c*Tan[(d + e*x)/2] - b*T an[(d + e*x)/2]^2))/(1 + Tan[(d + e*x)/2]^2)^2])/(Sqrt[c]*Sqrt[a + c + (a - c)*Cos[2*(d + e*x)] + b*Sin[2*(d + e*x)]]*Sqrt[(-1 + Tan[(d + e*x)/2]...
Time = 4.27 (sec) , antiderivative size = 678, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {3042, 4183, 7276, 2009}
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.
\(\displaystyle \int \frac {\tan ^3(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (d+e x)^3}{\left (a+b \tan (d+e x)+c \tan (d+e x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4183 |
\(\displaystyle \frac {\int \frac {\tan ^3(d+e x)}{\left (\tan ^2(d+e x)+1\right ) \left (c \tan ^2(d+e x)+b \tan (d+e x)+a\right )^{3/2}}d\tan (d+e x)}{e}\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \frac {\int \left (\frac {\tan (d+e x)}{\left (c \tan ^2(d+e x)+b \tan (d+e x)+a\right )^{3/2}}-\frac {\tan (d+e x)}{\left (\tan ^2(d+e x)+1\right ) \left (c \tan ^2(d+e x)+b \tan (d+e x)+a\right )^{3/2}}\right )d\tan (d+e x)}{e}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {\sqrt {-\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c} \sqrt {(a-c) \sqrt {a^2-2 a c+b^2+c^2}+a^2-2 a c-b^2+c^2} \text {arctanh}\left (\frac {-b \left (-\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c\right ) \tan (d+e x)-(a-c) \left (\sqrt {a^2-2 a c+b^2+c^2}+a-c\right )+b^2}{\sqrt {2} \sqrt {-\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c} \sqrt {(a-c) \sqrt {a^2-2 a c+b^2+c^2}+a^2-2 a c-b^2+c^2} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} \left (a^2-2 a c+b^2+c^2\right )^{3/2}}+\frac {\sqrt {\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c} \sqrt {-(a-c) \sqrt {a^2-2 a c+b^2+c^2}+a^2-2 a c-b^2+c^2} \text {arctanh}\left (\frac {-b \left (\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c\right ) \tan (d+e x)-(a-c) \left (-\sqrt {a^2-2 a c+b^2+c^2}+a-c\right )+b^2}{\sqrt {2} \sqrt {\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c} \sqrt {-(a-c) \sqrt {a^2-2 a c+b^2+c^2}+a^2-2 a c-b^2+c^2} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} \left (a^2-2 a c+b^2+c^2\right )^{3/2}}+\frac {2 (2 a+b \tan (d+e x))}{\left (b^2-4 a c\right ) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}-\frac {2 \left (a \left (b^2-2 c (a-c)\right )+b c (a+c) \tan (d+e x)\right )}{\left ((a-c)^2+b^2\right ) \left (b^2-4 a c\right ) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}}{e}\) |
(-((Sqrt[2*a - 2*c - Sqrt[a^2 + b^2 - 2*a*c + c^2]]*Sqrt[a^2 - b^2 - 2*a*c + c^2 + (a - c)*Sqrt[a^2 + b^2 - 2*a*c + c^2]]*ArcTanh[(b^2 - (a - c)*(a - c + Sqrt[a^2 + b^2 - 2*a*c + c^2]) - b*(2*a - 2*c - Sqrt[a^2 + b^2 - 2*a *c + c^2])*Tan[d + e*x])/(Sqrt[2]*Sqrt[2*a - 2*c - Sqrt[a^2 + b^2 - 2*a*c + c^2]]*Sqrt[a^2 - b^2 - 2*a*c + c^2 + (a - c)*Sqrt[a^2 + b^2 - 2*a*c + c^ 2]]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/(Sqrt[2]*(a^2 + b^2 - 2 *a*c + c^2)^(3/2))) + (Sqrt[2*a - 2*c + Sqrt[a^2 + b^2 - 2*a*c + c^2]]*Sqr t[a^2 - b^2 - 2*a*c + c^2 - (a - c)*Sqrt[a^2 + b^2 - 2*a*c + c^2]]*ArcTanh [(b^2 - (a - c)*(a - c - Sqrt[a^2 + b^2 - 2*a*c + c^2]) - b*(2*a - 2*c + S qrt[a^2 + b^2 - 2*a*c + c^2])*Tan[d + e*x])/(Sqrt[2]*Sqrt[2*a - 2*c + Sqrt [a^2 + b^2 - 2*a*c + c^2]]*Sqrt[a^2 - b^2 - 2*a*c + c^2 - (a - c)*Sqrt[a^2 + b^2 - 2*a*c + c^2]]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/(Sqr t[2]*(a^2 + b^2 - 2*a*c + c^2)^(3/2)) + (2*(2*a + b*Tan[d + e*x]))/((b^2 - 4*a*c)*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2]) - (2*(a*(b^2 - 2*(a - c)*c) + b*c*(a + c)*Tan[d + e*x]))/((b^2 + (a - c)^2)*(b^2 - 4*a*c)*Sqrt[ a + b*Tan[d + e*x] + c*Tan[d + e*x]^2]))/e
3.1.21.3.1 Defintions of rubi rules used
Int[tan[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*((f_.)*tan[(d_.) + (e_.)*( x_)])^(n_.) + (c_.)*((f_.)*tan[(d_.) + (e_.)*(x_)])^(n2_.))^(p_), x_Symbol] :> Simp[f/e Subst[Int[(x/f)^m*((a + b*x^n + c*x^(2*n))^p/(f^2 + x^2)), x ], x, f*Tan[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[n 2, 2*n] && NeQ[b^2 - 4*a*c, 0]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 1.25 (sec) , antiderivative size = 13067312, normalized size of antiderivative = 19048.56
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 19371 vs. \(2 (629) = 1258\).
Time = 2.87 (sec) , antiderivative size = 19371, normalized size of antiderivative = 28.24 \[ \int \frac {\tan ^3(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {\tan ^3(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx=\int \frac {\tan ^{3}{\left (d + e x \right )}}{\left (a + b \tan {\left (d + e x \right )} + c \tan ^{2}{\left (d + e x \right )}\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {\tan ^3(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Timed out. \[ \int \frac {\tan ^3(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^3(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (d+e\,x\right )}^3}{{\left (c\,{\mathrm {tan}\left (d+e\,x\right )}^2+b\,\mathrm {tan}\left (d+e\,x\right )+a\right )}^{3/2}} \,d x \]