\(\int \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} \, dx\) [608]

Optimal result

Integrand size = 25, antiderivative size = 231 \[ \int \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=-\frac {\sqrt {i a-b} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {\left (a^2+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 b^{3/2} d}+\frac {\sqrt {i a+b} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d} \] Output:

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

Mathematica [A] (verified)

Time = 1.24 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.22 \[ \int \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d}+\frac {-\frac {a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {-2 \sqrt [4]{-1} \sqrt {-a+i b} b \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-2 \sqrt [4]{-1} \sqrt {a+i b} b \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\frac {\sqrt {a} \left (a^2+8 b^2\right ) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{2 \sqrt {b} \sqrt {a+b \tan (c+d x)}}}{d}}{2 b} \] Input:

Integrate[Tan[c + d*x]^(5/2)*Sqrt[a + b*Tan[c + d*x]],x]
 

Output:

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

Rubi [A] (verified)

Time = 1.12 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4049, 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.

\(\displaystyle \int \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (c+d x)^{5/2} \sqrt {a+b \tan (c+d x)}dx\)

\(\Big \downarrow \) 4049

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4130

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4138

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 2257

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

\(\Big \downarrow \) 2009

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

Input:

Int[Tan[c + d*x]^(5/2)*Sqrt[a + b*Tan[c + d*x]],x]
 

Output:

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

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 4049

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((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 - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n 
- 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ 
e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2, x], x], 
x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2 
, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || I 
ntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])) 
)
 

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 = 1.90 (sec) , antiderivative size = 1092020, normalized size of antiderivative = 4727.36

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

Input:

int(tan(d*x+c)^(5/2)*(a+b*tan(d*x+c))^(1/2),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. 2797 vs. \(2 (185) = 370\).

Time = 0.79 (sec) , antiderivative size = 5600, normalized size of antiderivative = 24.24 \[ \int \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(5/2)*(a+b*tan(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\int \sqrt {a + b \tan {\left (c + d x \right )}} \tan ^{\frac {5}{2}}{\left (c + d x \right )}\, dx \] Input:

integrate(tan(d*x+c)**(5/2)*(a+b*tan(d*x+c))**(1/2),x)
 

Output:

Integral(sqrt(a + b*tan(c + d*x))*tan(c + d*x)**(5/2), x)
 

Maxima [F]

\[ \int \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\int { \sqrt {b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{\frac {5}{2}} \,d x } \] Input:

integrate(tan(d*x+c)^(5/2)*(a+b*tan(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(b*tan(d*x + c) + a)*tan(d*x + c)^(5/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(tan(d*x+c)^(5/2)*(a+b*tan(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const ve 
cteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )} \,d x \] Input:

int(tan(c + d*x)^(5/2)*(a + b*tan(c + d*x))^(1/2),x)
 

Output:

int(tan(c + d*x)^(5/2)*(a + b*tan(c + d*x))^(1/2), x)
 

Reduce [F]

\[ \int \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\int \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{2}d x \] Input:

int(tan(d*x+c)^(5/2)*(a+b*tan(d*x+c))^(1/2),x)
 

Output:

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