\(\int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx\) [675]

Optimal result

Integrand size = 25, antiderivative size = 165 \[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {3 \operatorname {AppellF1}\left (\frac {7}{3},\frac {1}{2},1,\frac {10}{3},-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right ) \tan ^{\frac {7}{3}}(c+d x) \sqrt {\frac {a+b \tan (c+d x)}{a}}}{14 d \sqrt {a+b \tan (c+d x)}}+\frac {3 \operatorname {AppellF1}\left (\frac {7}{3},\frac {1}{2},1,\frac {10}{3},-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right ) \tan ^{\frac {7}{3}}(c+d x) \sqrt {\frac {a+b \tan (c+d x)}{a}}}{14 d \sqrt {a+b \tan (c+d x)}} \] Output:

3/14*AppellF1(7/3,1,1/2,10/3,-I*tan(d*x+c),-b*tan(d*x+c)/a)*tan(d*x+c)^(7/ 
3)*((a+b*tan(d*x+c))/a)^(1/2)/d/(a+b*tan(d*x+c))^(1/2)+3/14*AppellF1(7/3,1 
,1/2,10/3,I*tan(d*x+c),-b*tan(d*x+c)/a)*tan(d*x+c)^(7/3)*((a+b*tan(d*x+c)) 
/a)^(1/2)/d/(a+b*tan(d*x+c))^(1/2)
 

Mathematica [F]

\[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx \] Input:

Integrate[Tan[c + d*x]^(4/3)/Sqrt[a + b*Tan[c + d*x]],x]
 

Output:

Integrate[Tan[c + d*x]^(4/3)/Sqrt[a + b*Tan[c + d*x]], x]
 

Rubi [A] (warning: unable to verify)

Time = 0.44 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 4058, 615, 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.

\(\displaystyle \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4058

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

\(\Big \downarrow \) 615

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {3 \tan ^{\frac {7}{3}}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} \operatorname {AppellF1}\left (\frac {7}{3},1,\frac {1}{2},\frac {10}{3},-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{14 \sqrt {a+b \tan (c+d x)}}+\frac {3 \tan ^{\frac {7}{3}}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} \operatorname {AppellF1}\left (\frac {7}{3},1,\frac {1}{2},\frac {10}{3},i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{14 \sqrt {a+b \tan (c+d x)}}}{d}\)

Input:

Int[Tan[c + d*x]^(4/3)/Sqrt[a + b*Tan[c + d*x]],x]
 

Output:

((3*AppellF1[7/3, 1, 1/2, 10/3, (-I)*Tan[c + d*x], -((b*Tan[c + d*x])/a)]* 
Tan[c + d*x]^(7/3)*Sqrt[1 + (b*Tan[c + d*x])/a])/(14*Sqrt[a + b*Tan[c + d* 
x]]) + (3*AppellF1[7/3, 1, 1/2, 10/3, I*Tan[c + d*x], -((b*Tan[c + d*x])/a 
)]*Tan[c + d*x]^(7/3)*Sqrt[1 + (b*Tan[c + d*x])/a])/(14*Sqrt[a + b*Tan[c + 
 d*x]]))/d
 

Defintions of rubi rules used

rule 615

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
 x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] 
 /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
 

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 4058

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

Maple [F]

\[\int \frac {\tan \left (d x +c \right )^{\frac {4}{3}}}{\sqrt {a +b \tan \left (d x +c \right )}}d x\]

Input:

int(tan(d*x+c)^(4/3)/(a+b*tan(d*x+c))^(1/2),x)
 

Output:

int(tan(d*x+c)^(4/3)/(a+b*tan(d*x+c))^(1/2),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate(tan(d*x+c)^(4/3)/(a+b*tan(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {\tan ^{\frac {4}{3}}{\left (c + d x \right )}}{\sqrt {a + b \tan {\left (c + d x \right )}}}\, dx \] Input:

integrate(tan(d*x+c)**(4/3)/(a+b*tan(d*x+c))**(1/2),x)
 

Output:

Integral(tan(c + d*x)**(4/3)/sqrt(a + b*tan(c + d*x)), x)
 

Maxima [F]

\[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\int { \frac {\tan \left (d x + c\right )^{\frac {4}{3}}}{\sqrt {b \tan \left (d x + c\right ) + a}} \,d x } \] Input:

integrate(tan(d*x+c)^(4/3)/(a+b*tan(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

integrate(tan(d*x + c)^(4/3)/sqrt(b*tan(d*x + c) + a), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate(tan(d*x+c)^(4/3)/(a+b*tan(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{4/3}}{\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}} \,d x \] Input:

int(tan(c + d*x)^(4/3)/(a + b*tan(c + d*x))^(1/2),x)
 

Output:

int(tan(c + d*x)^(4/3)/(a + b*tan(c + d*x))^(1/2), x)
 

Reduce [F]

\[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {\tan \left (d x +c \right )^{\frac {4}{3}} \sqrt {a +\tan \left (d x +c \right ) b}}{a +\tan \left (d x +c \right ) b}d x \] Input:

int(tan(d*x+c)^(4/3)/(a+b*tan(d*x+c))^(1/2),x)
 

Output:

int((tan(c + d*x)**(1/3)*sqrt(tan(c + d*x)*b + a)*tan(c + d*x))/(tan(c + d 
*x)*b + a),x)