3.7.77 \(\int \frac {\sqrt [3]{\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}} \, dx\) [677]

3.7.77.1 Optimal result

Integrand size = 25, antiderivative size = 163 \[ \int \frac {\sqrt [3]{\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {3 \operatorname {AppellF1}\left (\frac {4}{3},1,\frac {1}{2},\frac {7}{3},-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right ) \tan ^{\frac {4}{3}}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{8 d \sqrt {a+b \tan (c+d x)}}+\frac {3 \operatorname {AppellF1}\left (\frac {4}{3},1,\frac {1}{2},\frac {7}{3},i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right ) \tan ^{\frac {4}{3}}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{8 d \sqrt {a+b \tan (c+d x)}} \]

output

3/8*AppellF1(4/3,1,1/2,7/3,-I*tan(d*x+c),-b*tan(d*x+c)/a)*(1+b*tan(d*x+c)/ 
a)^(1/2)*tan(d*x+c)^(4/3)/d/(a+b*tan(d*x+c))^(1/2)+3/8*AppellF1(4/3,1,1/2, 
7/3,I*tan(d*x+c),-b*tan(d*x+c)/a)*(1+b*tan(d*x+c)/a)^(1/2)*tan(d*x+c)^(4/3 
)/d/(a+b*tan(d*x+c))^(1/2)
 

3.7.77.2 Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(6076\) vs. \(2(163)=326\).

Time = 58.56 (sec) , antiderivative size = 6076, normalized size of antiderivative = 37.28 \[ \int \frac {\sqrt [3]{\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Result too large to show} \]

input

Integrate[Tan[c + d*x]^(1/3)/Sqrt[a + b*Tan[c + d*x]],x]
 

output

Result too large to show
 

3.7.77.3 Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt [3]{\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}dx\)

\(\Big \downarrow \) 4058

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

\(\Big \downarrow \) 615

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {3 \tan ^{\frac {4}{3}}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} \operatorname {AppellF1}\left (\frac {4}{3},1,\frac {1}{2},\frac {7}{3},-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{8 \sqrt {a+b \tan (c+d x)}}+\frac {3 \tan ^{\frac {4}{3}}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} \operatorname {AppellF1}\left (\frac {4}{3},1,\frac {1}{2},\frac {7}{3},i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{8 \sqrt {a+b \tan (c+d x)}}}{d}\)

input

Int[Tan[c + d*x]^(1/3)/Sqrt[a + b*Tan[c + d*x]],x]
 

output

((3*AppellF1[4/3, 1, 1/2, 7/3, (-I)*Tan[c + d*x], -((b*Tan[c + d*x])/a)]*T 
an[c + d*x]^(4/3)*Sqrt[1 + (b*Tan[c + d*x])/a])/(8*Sqrt[a + b*Tan[c + d*x] 
]) + (3*AppellF1[4/3, 1, 1/2, 7/3, I*Tan[c + d*x], -((b*Tan[c + d*x])/a)]* 
Tan[c + d*x]^(4/3)*Sqrt[1 + (b*Tan[c + d*x])/a])/(8*Sqrt[a + b*Tan[c + d*x 
]]))/d
 

3.7.77.3.1 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]
 

3.7.77.4 Maple [F]

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

input

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

output

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

3.7.77.5 Fricas [F(-1)]

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

input

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

output

Timed out
 

3.7.77.6 Sympy [F]

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

input

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

output

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

3.7.77.7 Maxima [F]

\[ \int \frac {\sqrt [3]{\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}} \, dx=\int { \frac {\tan \left (d x + c\right )^{\frac {1}{3}}}{\sqrt {b \tan \left (d x + c\right ) + a}} \,d x } \]

input

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

output

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

3.7.77.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.7.77.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [3]{\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{1/3}}{\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}} \,d x \]

input

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

output

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