Integrand size = 33, antiderivative size = 748 \[ \int \tan ^3(d+e x) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)} \, dx=-\frac {\sqrt {a^2+b^2+c \left (c+\sqrt {a^2+b^2-2 a c+c^2}\right )-a \left (2 c+\sqrt {a^2+b^2-2 a c+c^2}\right )} \arctan \left (\frac {b^2+(a-c) \left (a-c-\sqrt {a^2+b^2-2 a c+c^2}\right )-b \sqrt {a^2+b^2-2 a c+c^2} \tan (d+e x)}{\sqrt {2} \sqrt [4]{a^2+b^2-2 a c+c^2} \sqrt {a^2+b^2+c \left (c+\sqrt {a^2+b^2-2 a c+c^2}\right )-a \left (2 c+\sqrt {a^2+b^2-2 a c+c^2}\right )} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} \sqrt [4]{a^2+b^2-2 a c+c^2} e}-\frac {b \text {arctanh}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{2 \sqrt {c} e}+\frac {b \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{16 c^{5/2} e}+\frac {\sqrt {a^2+b^2+c \left (c-\sqrt {a^2+b^2-2 a c+c^2}\right )-a \left (2 c-\sqrt {a^2+b^2-2 a c+c^2}\right )} \text {arctanh}\left (\frac {b^2+(a-c) \left (a-c+\sqrt {a^2+b^2-2 a c+c^2}\right )+b \sqrt {a^2+b^2-2 a c+c^2} \tan (d+e x)}{\sqrt {2} \sqrt [4]{a^2+b^2-2 a c+c^2} \sqrt {a^2+b^2+c \left (c-\sqrt {a^2+b^2-2 a c+c^2}\right )-a \left (2 c-\sqrt {a^2+b^2-2 a c+c^2}\right )} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} \sqrt [4]{a^2+b^2-2 a c+c^2} e}-\frac {\sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{e}-\frac {b (b+2 c \tan (d+e x)) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{8 c^2 e}+\frac {\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}}{3 c e} \]
1/16*b*(-4*a*c+b^2)*arctanh(1/2*(b+2*c*tan(e*x+d))/c^(1/2)/(a+b*tan(e*x+d) +c*tan(e*x+d)^2)^(1/2))/c^(5/2)/e-1/2*b*arctanh(1/2*(b+2*c*tan(e*x+d))/c^( 1/2)/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(1/2))/e/c^(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*(a^2-2*a*c+b^2+c^2)^(1/2)*tan(e*x +d))/(a^2-2*a*c+b^2+c^2)^(1/4)*2^(1/2)/(a^2+b^2+c*(c-(a^2-2*a*c+b^2+c^2)^( 1/2))-a*(2*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))*(a^2+b^2+c*(c-(a^2-2*a*c+b^2+c^2)^(1/2))-a*(2*c-(a^2-2*a*c+b^2 +c^2)^(1/2)))^(1/2)/(a^2-2*a*c+b^2+c^2)^(1/4)/e*2^(1/2)-1/2*arctan(1/2*(b^ 2+(a-c)*(a-c-(a^2-2*a*c+b^2+c^2)^(1/2))-b*(a^2-2*a*c+b^2+c^2)^(1/2)*tan(e* x+d))/(a^2-2*a*c+b^2+c^2)^(1/4)*2^(1/2)/(a^2+b^2+c*(c+(a^2-2*a*c+b^2+c^2)^ (1/2))-a*(2*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))*(a^2+b^2+c*(c+(a^2-2*a*c+b^2+c^2)^(1/2))-a*(2*c+(a^2-2*a*c+b^ 2+c^2)^(1/2)))^(1/2)/(a^2-2*a*c+b^2+c^2)^(1/4)/e*2^(1/2)-(a+b*tan(e*x+d)+c *tan(e*x+d)^2)^(1/2)/e-1/8*b*(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(1/2)*(b+2*c* tan(e*x+d))/c^2/e+1/3*(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(3/2)/c/e
Result contains complex when optimal does not.
Time = 2.57 (sec) , antiderivative size = 386, normalized size of antiderivative = 0.52 \[ \int \tan ^3(d+e x) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)} \, dx=\frac {3 \sqrt {a-i b-c} \text {arctanh}\left (\frac {2 a-i b+(b-2 i c) \tan (d+e x)}{2 \sqrt {a-i b-c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )+3 \sqrt {a+i b-c} \text {arctanh}\left (\frac {2 a+i b+(b+2 i c) \tan (d+e x)}{2 \sqrt {a+i b-c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )-\frac {3 b \text {arctanh}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {c}}+\frac {3 b \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{8 c^{5/2}}-6 \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}-\frac {3 b (b+2 c \tan (d+e x)) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}{4 c^2}+\frac {2 \left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}}{c}}{6 e} \]
(3*Sqrt[a - I*b - c]*ArcTanh[(2*a - I*b + (b - (2*I)*c)*Tan[d + e*x])/(2*S qrt[a - I*b - c]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])] + 3*Sqrt[a + I*b - c]*ArcTanh[(2*a + I*b + (b + (2*I)*c)*Tan[d + e*x])/(2*Sqrt[a + I* b - c]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])] - (3*b*ArcTanh[(b + 2 *c*Tan[d + e*x])/(2*Sqrt[c]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])]) /Sqrt[c] + (3*b*(b^2 - 4*a*c)*ArcTanh[(b + 2*c*Tan[d + e*x])/(2*Sqrt[c]*Sq rt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/(8*c^(5/2)) - 6*Sqrt[a + b*Ta n[d + e*x] + c*Tan[d + e*x]^2] - (3*b*(b + 2*c*Tan[d + e*x])*Sqrt[a + b*Ta n[d + e*x] + c*Tan[d + e*x]^2])/(4*c^2) + (2*(a + b*Tan[d + e*x] + c*Tan[d + e*x]^2)^(3/2))/c)/(6*e)
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.
\(\displaystyle \int \tan ^3(d+e x) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (d+e x)^3 \sqrt {a+b \tan (d+e x)+c \tan (d+e x)^2}dx\) |
\(\Big \downarrow \) 4183 |
\(\displaystyle \frac {\int \frac {\tan ^3(d+e x) \sqrt {c \tan ^2(d+e x)+b \tan (d+e x)+a}}{\tan ^2(d+e x)+1}d\tan (d+e x)}{e}\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \frac {\int \left (\tan (d+e x) \sqrt {c \tan ^2(d+e x)+b \tan (d+e x)+a}-\frac {\tan (d+e x) \sqrt {c \tan ^2(d+e x)+b \tan (d+e x)+a}}{\tan ^2(d+e x)+1}\right )d\tan (d+e x)}{e}\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \frac {\int \left (\tan (d+e x) \sqrt {c \tan ^2(d+e x)+b \tan (d+e x)+a}-\frac {\tan (d+e x) \sqrt {c \tan ^2(d+e x)+b \tan (d+e x)+a}}{\tan ^2(d+e x)+1}\right )d\tan (d+e x)}{e}\) |
3.1.3.3.1 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 = 2.79 (sec) , antiderivative size = 17766953, normalized size of antiderivative = 23752.61
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 2396 vs. \(2 (671) = 1342\).
Time = 0.90 (sec) , antiderivative size = 4793, normalized size of antiderivative = 6.41 \[ \int \tan ^3(d+e x) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)} \, dx=\text {Too large to display} \]
[-1/96*(24*c^3*e*sqrt(-(e^2*sqrt(-b^2/e^4) - a + c)/e^2)*log(-(2*(4*a^3*b^ 2 + 2*a*b^4 + 2*b^4*c - 4*a*b^2*c^2 - (4*a^4*b + 3*a^2*b^3 + b^5 + (4*a^2* b - b^3)*c^2 - 2*(4*a^3*b + 3*a*b^3)*c)*tan(e*x + d) + (2*(2*a^3*b + a*b^3 + b^3*c - 2*a*b*c^2)*e^2*tan(e*x + d) + (4*a^4 + 3*a^2*b^2 + b^4 + (4*a^2 - b^2)*c^2 - 2*(4*a^3 + 3*a*b^2)*c)*e^2)*sqrt(-b^2/e^4))*sqrt(c*tan(e*x + d)^2 + b*tan(e*x + d) + a) + ((4*a^3*b^2 + 3*a*b^4 - 8*a*b^2*c^2 - (4*a^2 *b^2 - 3*b^4)*c)*e*tan(e*x + d)^2 + 2*(4*a^4*b + 3*a^2*b^3 + b^5 + (4*a^2* b - b^3)*c^2 - 2*(4*a^3*b + 3*a*b^3)*c)*e*tan(e*x + d) - (4*a^3*b^2 + a*b^ 4 + (4*a^2*b^2 + b^4)*c)*e + ((b^4 + 2*(4*a^2 - b^2)*c^2 - 2*(4*a^3 + 5*a* b^2)*c)*e^3*tan(e*x + d)^2 - 4*(2*a^3*b + a*b^3 + b^3*c - 2*a*b*c^2)*e^3*t an(e*x + d) - (8*a^4 + 6*a^2*b^2 + b^4 - 2*(4*a^3 + a*b^2)*c)*e^3)*sqrt(-b ^2/e^4))*sqrt(-(e^2*sqrt(-b^2/e^4) - a + c)/e^2))/(tan(e*x + d)^2 + 1)) - 24*c^3*e*sqrt(-(e^2*sqrt(-b^2/e^4) - a + c)/e^2)*log(-(2*(4*a^3*b^2 + 2*a* b^4 + 2*b^4*c - 4*a*b^2*c^2 - (4*a^4*b + 3*a^2*b^3 + b^5 + (4*a^2*b - b^3) *c^2 - 2*(4*a^3*b + 3*a*b^3)*c)*tan(e*x + d) + (2*(2*a^3*b + a*b^3 + b^3*c - 2*a*b*c^2)*e^2*tan(e*x + d) + (4*a^4 + 3*a^2*b^2 + b^4 + (4*a^2 - b^2)* c^2 - 2*(4*a^3 + 3*a*b^2)*c)*e^2)*sqrt(-b^2/e^4))*sqrt(c*tan(e*x + d)^2 + b*tan(e*x + d) + a) - ((4*a^3*b^2 + 3*a*b^4 - 8*a*b^2*c^2 - (4*a^2*b^2 - 3 *b^4)*c)*e*tan(e*x + d)^2 + 2*(4*a^4*b + 3*a^2*b^3 + b^5 + (4*a^2*b - b^3) *c^2 - 2*(4*a^3*b + 3*a*b^3)*c)*e*tan(e*x + d) - (4*a^3*b^2 + a*b^4 + (...
\[ \int \tan ^3(d+e x) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)} \, dx=\int \sqrt {a + b \tan {\left (d + e x \right )} + c \tan ^{2}{\left (d + e x \right )}} \tan ^{3}{\left (d + e x \right )}\, dx \]
\[ \int \tan ^3(d+e x) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)} \, dx=\int { \sqrt {c \tan \left (e x + d\right )^{2} + b \tan \left (e x + d\right ) + a} \tan \left (e x + d\right )^{3} \,d x } \]
Timed out. \[ \int \tan ^3(d+e x) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \tan ^3(d+e x) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)} \, dx=\int {\mathrm {tan}\left (d+e\,x\right )}^3\,\sqrt {c\,{\mathrm {tan}\left (d+e\,x\right )}^2+b\,\mathrm {tan}\left (d+e\,x\right )+a} \,d x \]