\(\int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {4}{3}}(c+d x)} \, dx\) [259]

Optimal result

Integrand size = 28, antiderivative size = 80 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {4}{3}}(c+d x)} \, dx=-\frac {3 a \operatorname {AppellF1}\left (-\frac {1}{3},-\frac {1}{2},1,\frac {2}{3},-i \tan (c+d x),i \tan (c+d x)\right ) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {1+i \tan (c+d x)} \sqrt [3]{\tan (c+d x)}} \]

[Out]

-3*a*AppellF1(-1/3,-1/2,1,2/3,-I*tan(d*x+c),I*tan(d*x+c))*(a+I*a*tan(d*x+c))^(1/2)/d/(1+I*tan(d*x+c))^(1/2)/ta
n(d*x+c)^(1/3)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3645, 129, 525, 524} \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {4}{3}}(c+d x)} \, dx=-\frac {3 a \sqrt {a+i a \tan (c+d x)} \operatorname {AppellF1}\left (-\frac {1}{3},-\frac {1}{2},1,\frac {2}{3},-i \tan (c+d x),i \tan (c+d x)\right )}{d \sqrt {1+i \tan (c+d x)} \sqrt [3]{\tan (c+d x)}} \]

[In]

Int[(a + I*a*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(4/3),x]

[Out]

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

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 3645

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (i a^2\right ) \text {Subst}\left (\int \frac {\sqrt {a+x}}{\left (-\frac {i x}{a}\right )^{4/3} \left (-a^2+a x\right )} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {\left (3 a^3\right ) \text {Subst}\left (\int \frac {\sqrt {a+i a x^3}}{x^2 \left (-a^2+i a^2 x^3\right )} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{d} \\ & = -\frac {\left (3 a^3 \sqrt {a+i a \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {\sqrt {1+i x^3}}{x^2 \left (-a^2+i a^2 x^3\right )} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{d \sqrt {1+i \tan (c+d x)}} \\ & = -\frac {3 a \operatorname {AppellF1}\left (-\frac {1}{3},-\frac {1}{2},1,\frac {2}{3},-i \tan (c+d x),i \tan (c+d x)\right ) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {1+i \tan (c+d x)} \sqrt [3]{\tan (c+d x)}} \\ \end{align*}

Mathematica [F]

\[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {4}{3}}(c+d x)} \, dx=\int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {4}{3}}(c+d x)} \, dx \]

[In]

Integrate[(a + I*a*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(4/3),x]

[Out]

Integrate[(a + I*a*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(4/3), x]

Maple [F]

\[\int \frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{\tan \left (d x +c \right )^{\frac {4}{3}}}d x\]

[In]

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

[Out]

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

Fricas [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {4}{3}}(c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F]

\[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {4}{3}}(c+d x)} \, dx=\int \frac {\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}{\tan ^{\frac {4}{3}}{\left (c + d x \right )}}\, dx \]

[In]

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

[Out]

Integral((I*a*(tan(c + d*x) - I))**(3/2)/tan(c + d*x)**(4/3), x)

Maxima [F]

\[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {4}{3}}(c+d x)} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\tan \left (d x + c\right )^{\frac {4}{3}}} \,d x } \]

[In]

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

[Out]

integrate((I*a*tan(d*x + c) + a)^(3/2)/tan(d*x + c)^(4/3), x)

Giac [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {4}{3}}(c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {4}{3}}(c+d x)} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{4/3}} \,d x \]

[In]

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

[Out]

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