\(\int \frac {\tan ^3(d+e x)}{(a+b \tan (d+e x)+c \tan ^2(d+e x))^{3/2}} \, dx\) [19]

Optimal result

Integrand size = 33, antiderivative size = 686 \[ \int \frac {\tan ^3(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx=-\frac {\sqrt {2 a-2 c-\sqrt {a^2+b^2-2 a c+c^2}} \sqrt {a^2-b^2-2 a c+c^2+(a-c) \sqrt {a^2+b^2-2 a c+c^2}} \text {arctanh}\left (\frac {b^2-(a-c) \left (a-c+\sqrt {a^2+b^2-2 a c+c^2}\right )-b \left (2 a-2 c-\sqrt {a^2+b^2-2 a c+c^2}\right ) \tan (d+e x)}{\sqrt {2} \sqrt {2 a-2 c-\sqrt {a^2+b^2-2 a c+c^2}} \sqrt {a^2-b^2-2 a c+c^2+(a-c) \sqrt {a^2+b^2-2 a c+c^2}} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} \left (a^2+b^2-2 a c+c^2\right )^{3/2} e}+\frac {\sqrt {2 a-2 c+\sqrt {a^2+b^2-2 a c+c^2}} \sqrt {a^2-b^2-2 a c+c^2-(a-c) \sqrt {a^2+b^2-2 a c+c^2}} \text {arctanh}\left (\frac {b^2-(a-c) \left (a-c-\sqrt {a^2+b^2-2 a c+c^2}\right )-b \left (2 a-2 c+\sqrt {a^2+b^2-2 a c+c^2}\right ) \tan (d+e x)}{\sqrt {2} \sqrt {2 a-2 c+\sqrt {a^2+b^2-2 a c+c^2}} \sqrt {a^2-b^2-2 a c+c^2-(a-c) \sqrt {a^2+b^2-2 a c+c^2}} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} \left (a^2+b^2-2 a c+c^2\right )^{3/2} e}+\frac {2 (2 a+b \tan (d+e x))}{\left (b^2-4 a c\right ) e \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}-\frac {2 \left (a \left (b^2-2 (a-c) c\right )+b c (a+c) \tan (d+e x)\right )}{\left (b^2+(a-c)^2\right ) \left (b^2-4 a c\right ) e \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}} \] Output:

-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)*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^(1/2)/(a^2- 
2*a*c+b^2+c^2)^(3/2)/e+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)*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))*ta 
n(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^(1/2)/(a^2-2*a*c+b^2+c^2)^(3/2)/e+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)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 6.81 (sec) , antiderivative size = 2339, normalized size of antiderivative = 3.41 \[ \int \frac {\tan ^3(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx=\text {Result too large to show} \] Input:

Integrate[Tan[d + e*x]^3/(a + b*Tan[d + e*x] + c*Tan[d + e*x]^2)^(3/2),x]
 

Output:

(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]...
 

Rubi [A] (verified)

Time = 2.87 (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

\(\displaystyle \frac {-\frac {\sqrt {-\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c} \sqrt {(a-c) \sqrt {a^2-2 a c+b^2+c^2}+a^2-2 a c-b^2+c^2} \text {arctanh}\left (\frac {-b \left (-\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c\right ) \tan (d+e x)-(a-c) \left (\sqrt {a^2-2 a c+b^2+c^2}+a-c\right )+b^2}{\sqrt {2} \sqrt {-\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c} \sqrt {(a-c) \sqrt {a^2-2 a c+b^2+c^2}+a^2-2 a c-b^2+c^2} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} \left (a^2-2 a c+b^2+c^2\right )^{3/2}}+\frac {\sqrt {\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c} \sqrt {-(a-c) \sqrt {a^2-2 a c+b^2+c^2}+a^2-2 a c-b^2+c^2} \text {arctanh}\left (\frac {-b \left (\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c\right ) \tan (d+e x)-(a-c) \left (-\sqrt {a^2-2 a c+b^2+c^2}+a-c\right )+b^2}{\sqrt {2} \sqrt {\sqrt {a^2-2 a c+b^2+c^2}+2 a-2 c} \sqrt {-(a-c) \sqrt {a^2-2 a c+b^2+c^2}+a^2-2 a c-b^2+c^2} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} \left (a^2-2 a c+b^2+c^2\right )^{3/2}}+\frac {2 (2 a+b \tan (d+e x))}{\left (b^2-4 a c\right ) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}-\frac {2 \left (a \left (b^2-2 c (a-c)\right )+b c (a+c) \tan (d+e x)\right )}{\left ((a-c)^2+b^2\right ) \left (b^2-4 a c\right ) \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}}{e}\)

Input:

Int[Tan[d + e*x]^3/(a + b*Tan[d + e*x] + c*Tan[d + e*x]^2)^(3/2),x]
 

Output:

(-((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
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4183

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]
 

rule 7276

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]
 

Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 1.66 (sec) , antiderivative size = 13067312, normalized size of antiderivative = 19048.56

\[\text {output too large to display}\]

Input:

int(tan(e*x+d)^3/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(3/2),x)
 

Output:

result too large to display
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 19371 vs. \(2 (629) = 1258\).

Time = 2.73 (sec) , antiderivative size = 19371, normalized size of antiderivative = 28.24 \[ \int \frac {\tan ^3(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(tan(e*x+d)^3/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(3/2),x, algorithm= 
"fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {\tan ^3(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx=\int \frac {\tan ^{3}{\left (d + e x \right )}}{\left (a + b \tan {\left (d + e x \right )} + c \tan ^{2}{\left (d + e x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(tan(e*x+d)**3/(a+b*tan(e*x+d)+c*tan(e*x+d)**2)**(3/2),x)
 

Output:

Integral(tan(d + e*x)**3/(a + b*tan(d + e*x) + c*tan(d + e*x)**2)**(3/2), 
x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^3(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(tan(e*x+d)^3/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(3/2),x, algorithm= 
"maxima")
 

Output:

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
 

Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^3(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(tan(e*x+d)^3/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(3/2),x, algorithm= 
"giac")
                                                                                    
                                                                                    
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^3(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (d+e\,x\right )}^3}{{\left (c\,{\mathrm {tan}\left (d+e\,x\right )}^2+b\,\mathrm {tan}\left (d+e\,x\right )+a\right )}^{3/2}} \,d x \] Input:

int(tan(d + e*x)^3/(a + b*tan(d + e*x) + c*tan(d + e*x)^2)^(3/2),x)
 

Output:

int(tan(d + e*x)^3/(a + b*tan(d + e*x) + c*tan(d + e*x)^2)^(3/2), x)
 

Reduce [F]

\[ \int \frac {\tan ^3(d+e x)}{\left (a+b \tan (d+e x)+c \tan ^2(d+e x)\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

int(tan(e*x+d)^3/(a+b*tan(e*x+d)+c*tan(e*x+d)^2)^(3/2),x)
 

Output:

( - 2*sqrt(tan(d + e*x)**2*c + tan(d + e*x)*b + a)*tan(d + e*x)*b - 4*sqrt 
(tan(d + e*x)**2*c + tan(d + e*x)*b + a)*a - 4*int((sqrt(tan(d + e*x)**2*c 
 + tan(d + e*x)*b + a)*tan(d + e*x))/(tan(d + e*x)**4*c**2 + 2*tan(d + e*x 
)**3*b*c + 2*tan(d + e*x)**2*a*c + tan(d + e*x)**2*b**2 + 2*tan(d + e*x)*a 
*b + a**2),x)*tan(d + e*x)**2*a*c**2*e + int((sqrt(tan(d + e*x)**2*c + tan 
(d + e*x)*b + a)*tan(d + e*x))/(tan(d + e*x)**4*c**2 + 2*tan(d + e*x)**3*b 
*c + 2*tan(d + e*x)**2*a*c + tan(d + e*x)**2*b**2 + 2*tan(d + e*x)*a*b + a 
**2),x)*tan(d + e*x)**2*b**2*c*e - 4*int((sqrt(tan(d + e*x)**2*c + tan(d + 
 e*x)*b + a)*tan(d + e*x))/(tan(d + e*x)**4*c**2 + 2*tan(d + e*x)**3*b*c + 
 2*tan(d + e*x)**2*a*c + tan(d + e*x)**2*b**2 + 2*tan(d + e*x)*a*b + a**2) 
,x)*tan(d + e*x)*a*b*c*e + int((sqrt(tan(d + e*x)**2*c + tan(d + e*x)*b + 
a)*tan(d + e*x))/(tan(d + e*x)**4*c**2 + 2*tan(d + e*x)**3*b*c + 2*tan(d + 
 e*x)**2*a*c + tan(d + e*x)**2*b**2 + 2*tan(d + e*x)*a*b + a**2),x)*tan(d 
+ e*x)*b**3*e - 4*int((sqrt(tan(d + e*x)**2*c + tan(d + e*x)*b + a)*tan(d 
+ e*x))/(tan(d + e*x)**4*c**2 + 2*tan(d + e*x)**3*b*c + 2*tan(d + e*x)**2* 
a*c + tan(d + e*x)**2*b**2 + 2*tan(d + e*x)*a*b + a**2),x)*a**2*c*e + int( 
(sqrt(tan(d + e*x)**2*c + tan(d + e*x)*b + a)*tan(d + e*x))/(tan(d + e*x)* 
*4*c**2 + 2*tan(d + e*x)**3*b*c + 2*tan(d + e*x)**2*a*c + tan(d + e*x)**2* 
b**2 + 2*tan(d + e*x)*a*b + a**2),x)*a*b**2*e)/(e*(4*tan(d + e*x)**2*a*c** 
2 - tan(d + e*x)**2*b**2*c + 4*tan(d + e*x)*a*b*c - tan(d + e*x)*b**3 +...