\(\int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} \, dx\) [616]

Optimal result

Integrand size = 25, antiderivative size = 226 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} \, dx=\frac {(i a-b)^{3/2} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left (3 a^2-8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 \sqrt {b} d}+\frac {(i a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d} \] Output:

(I*a-b)^(3/2)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2) 
)/d+1/4*(3*a^2-8*b^2)*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1 
/2))/b^(1/2)/d+(I*a+b)^(3/2)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*t 
an(d*x+c))^(1/2))/d+3/4*a*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)/d+1/2*ta 
n(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(3/2)/d
 

Mathematica [A] (verified)

Time = 4.16 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.19 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} \, dx=\frac {-4 \sqrt [4]{-1} \sqrt {-a+i b} (i a+b) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+4 (-1)^{3/4} (a+i b)^{3/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+\frac {\sqrt {a} \left (3 a^2-8 b^2\right ) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {b} \sqrt {a+b \tan (c+d x)}}+\frac {\sqrt {\tan (c+d x)} \left (5 a^2+7 a b \tan (c+d x)+2 b^2 \tan ^2(c+d x)\right )}{\sqrt {a+b \tan (c+d x)}}}{4 d} \] Input:

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

Output:

(-4*(-1)^(1/4)*Sqrt[-a + I*b]*(I*a + b)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]* 
Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + 4*(-1)^(3/4)*(a + I*b)^(3/ 
2)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + 
 d*x]]] + (Sqrt[a]*(3*a^2 - 8*b^2)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sq 
rt[a]]*Sqrt[1 + (b*Tan[c + d*x])/a])/(Sqrt[b]*Sqrt[a + b*Tan[c + d*x]]) + 
(Sqrt[Tan[c + d*x]]*(5*a^2 + 7*a*b*Tan[c + d*x] + 2*b^2*Tan[c + d*x]^2))/S 
qrt[a + b*Tan[c + d*x]])/(4*d)
 

Rubi [A] (verified)

Time = 1.15 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4053, 27, 3042, 4130, 27, 3042, 4138, 2035, 2257, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (c+d x)^{3/2} (a+b \tan (c+d x))^{3/2}dx\)

\(\Big \downarrow \) 4053

\(\displaystyle \frac {1}{2} \int -\frac {\sqrt {a+b \tan (c+d x)} \left (-3 a \tan ^2(c+d x)+4 b \tan (c+d x)+a\right )}{2 \sqrt {\tan (c+d x)}}dx+\frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}-\frac {1}{4} \int \frac {\sqrt {a+b \tan (c+d x)} \left (-3 a \tan ^2(c+d x)+4 b \tan (c+d x)+a\right )}{\sqrt {\tan (c+d x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}-\frac {1}{4} \int \frac {\sqrt {a+b \tan (c+d x)} \left (-3 a \tan (c+d x)^2+4 b \tan (c+d x)+a\right )}{\sqrt {\tan (c+d x)}}dx\)

\(\Big \downarrow \) 4130

\(\displaystyle \frac {1}{4} \left (\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\int \frac {5 a^2+16 b \tan (c+d x) a-\left (3 a^2-8 b^2\right ) \tan ^2(c+d x)}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )+\frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {1}{2} \int \frac {5 a^2+16 b \tan (c+d x) a-\left (3 a^2-8 b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )+\frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {1}{2} \int \frac {5 a^2+16 b \tan (c+d x) a-\left (3 a^2-8 b^2\right ) \tan (c+d x)^2}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )+\frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 4138

\(\displaystyle \frac {1}{4} \left (\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {\int \frac {5 a^2+16 b \tan (c+d x) a-\left (3 a^2-8 b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{2 d}\right )+\frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 2035

\(\displaystyle \frac {1}{4} \left (\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {\int \frac {5 a^2+16 b \tan (c+d x) a-\left (3 a^2-8 b^2\right ) \tan ^2(c+d x)}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{d}\right )+\frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 2257

\(\displaystyle \frac {1}{4} \left (\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {\int \left (\frac {8 b^2-3 a^2}{\sqrt {a+b \tan (c+d x)}}+\frac {8 \left (a^2+2 b \tan (c+d x) a-b^2\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}\right )d\sqrt {\tan (c+d x)}}{d}\right )+\frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}+\frac {1}{4} \left (\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {-\frac {\left (3 a^2-8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b}}-4 (-b+i a)^{3/2} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-4 (b+i a)^{3/2} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\right )\)

Input:

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

Output:

(Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2))/(2*d) + (-((-4*(I*a - b)^( 
3/2)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] - 
 ((3*a^2 - 8*b^2)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + 
d*x]]])/Sqrt[b] - 4*(I*a + b)^(3/2)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d* 
x]])/Sqrt[a + b*Tan[c + d*x]]])/d) + (3*a*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Ta 
n[c + d*x]])/d)/4
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

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

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

rule 2257

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol 
] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a 
, c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
 

rule 3042

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

rule 4053

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a + b*Tan[e + f*x])^(m - 1)*((c + 
d*Tan[e + f*x])^n/(f*(m + n - 1))), x] + Simp[1/(m + n - 1)   Int[(a + b*Ta 
n[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n - 1)*Simp[a^2*c*(m + n - 1) - b 
*(b*c*(m - 1) + a*d*n) + (2*a*b*c + a^2*d - b^2*d)*(m + n - 1)*Tan[e + f*x] 
 + b*(b*c*n + a*d*(2*m + n - 2))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, 
 c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2 
, 0] && GtQ[m, 1] && GtQ[n, 0] && IntegerQ[2*n]
 

rule 4130

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_. 
) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d*Tan[ 
e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a 
+ b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C 
*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f*x] - (C* 
m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, 
b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && 
 NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[ 
c, 0] && NeQ[a, 0])))
 

rule 4138

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S 
imp[ff/f   Subst[Int[(a + b*ff*x)^m*(c + d*ff*x)^n*((A + B*ff*x + C*ff^2*x^ 
2)/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f 
, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + 
 d^2, 0]
 

Maple [B] (warning: unable to verify)

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

Time = 0.70 (sec) , antiderivative size = 1345246, normalized size of antiderivative = 5952.42

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

Input:

int(tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))^(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. 3660 vs. \(2 (180) = 360\).

Time = 0.92 (sec) , antiderivative size = 7326, normalized size of antiderivative = 32.42 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**2,x)*b + int 
(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x),x)*a