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

Optimal result

Integrand size = 25, antiderivative size = 163 \[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {3 \operatorname {AppellF1}\left (\frac {7}{3},1,\frac {1}{2},\frac {10}{3},-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right ) \tan ^{\frac {7}{3}}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{14 d \sqrt {a+b \tan (c+d x)}}+\frac {3 \operatorname {AppellF1}\left (\frac {7}{3},1,\frac {1}{2},\frac {10}{3},i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right ) \tan ^{\frac {7}{3}}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{14 d \sqrt {a+b \tan (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3656, 926, 129, 525, 524} \[ \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {3 \tan ^{\frac {7}{3}}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} \operatorname {AppellF1}\left (\frac {7}{3},1,\frac {1}{2},\frac {10}{3},-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{14 d \sqrt {a+b \tan (c+d x)}}+\frac {3 \tan ^{\frac {7}{3}}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} \operatorname {AppellF1}\left (\frac {7}{3},1,\frac {1}{2},\frac {10}{3},i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{14 d \sqrt {a+b \tan (c+d x)}} \]

[In]

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

[Out]

(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*d*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*d*Sqrt[a + b*Tan[c + d*x]])

Rule 129

Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]
}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + b*(x^k/e))^m*(c + d*(x^k/e))^n, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 926

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)^n, 1/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[c*d^2 + a*e^2,
0] &&  !IntegerQ[m] &&  !IntegerQ[n]

Rule 3656

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Wit
h[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^{4/3}}{\sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (\frac {i x^{4/3}}{2 (i-x) \sqrt {a+b x}}+\frac {i x^{4/3}}{2 (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {i \text {Subst}\left (\int \frac {x^{4/3}}{(i-x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {i \text {Subst}\left (\int \frac {x^{4/3}}{(i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d} \\ & = \frac {(3 i) \text {Subst}\left (\int \frac {x^6}{\left (i-x^3\right ) \sqrt {a+b x^3}} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{2 d}+\frac {(3 i) \text {Subst}\left (\int \frac {x^6}{\left (i+x^3\right ) \sqrt {a+b x^3}} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{2 d} \\ & = \frac {\left (3 i \sqrt {1+\frac {b \tan (c+d x)}{a}}\right ) \text {Subst}\left (\int \frac {x^6}{\left (i-x^3\right ) \sqrt {1+\frac {b x^3}{a}}} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{2 d \sqrt {a+b \tan (c+d x)}}+\frac {\left (3 i \sqrt {1+\frac {b \tan (c+d x)}{a}}\right ) \text {Subst}\left (\int \frac {x^6}{\left (i+x^3\right ) \sqrt {1+\frac {b x^3}{a}}} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{2 d \sqrt {a+b \tan (c+d x)}} \\ & = \frac {3 \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 ) \tan ^{\frac {7}{3}}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{14 d \sqrt {a+b \tan (c+d x)}}+\frac {3 \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 ) \tan ^{\frac {7}{3}}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{14 d \sqrt {a+b \tan (c+d x)}} \\ \end{align*}

Mathematica [F]

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

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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