\(\int \frac {\sqrt {\tan (c+d x)}}{(a+i a \tan (c+d x))^{4/3}} \, dx\) [299]
Optimal result
Integrand size = 28, antiderivative size = 84 \[
\int \frac {\sqrt {\tan (c+d x)}}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {7}{3},1,\frac {5}{2},-i \tan (c+d x),i \tan (c+d x)\right ) \sqrt [3]{1+i \tan (c+d x)} \tan ^{\frac {3}{2}}(c+d x)}{3 a d \sqrt [3]{a+i a \tan (c+d x)}}
\]
[Out]
2/3*AppellF1(3/2,7/3,1,5/2,-I*tan(d*x+c),I*tan(d*x+c))*(1+I*tan(d*x+c))^(1/3)*tan(d*x+c)^(3/2)/a/d/(a+I*a*tan(
d*x+c))^(1/3)
Rubi [A] (verified)
Time = 0.18 (sec) , antiderivative size = 84, 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}
\[
\int \frac {\sqrt {\tan (c+d x)}}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {2 \sqrt [3]{1+i \tan (c+d x)} \tan ^{\frac {3}{2}}(c+d x) \operatorname {AppellF1}\left (\frac {3}{2},\frac {7}{3},1,\frac {5}{2},-i \tan (c+d x),i \tan (c+d x)\right )}{3 a d \sqrt [3]{a+i a \tan (c+d x)}}
\]
[In]
Int[Sqrt[Tan[c + d*x]]/(a + I*a*Tan[c + d*x])^(4/3),x]
[Out]
(2*AppellF1[3/2, 7/3, 1, 5/2, (-I)*Tan[c + d*x], I*Tan[c + d*x]]*(1 + I*Tan[c + d*x])^(1/3)*Tan[c + d*x]^(3/2)
)/(3*a*d*(a + I*a*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 {-\frac {i x}{a}}}{(a+x)^{7/3} \left (-a^2+a x\right )} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {\left (2 a^3\right ) \text {Subst}\left (\int \frac {x^2}{\left (a+i a x^2\right )^{7/3} \left (-a^2+i a^2 x^2\right )} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = -\frac {\left (2 a \sqrt [3]{1+i \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {x^2}{\left (1+i x^2\right )^{7/3} \left (-a^2+i a^2 x^2\right )} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d \sqrt [3]{a+i a \tan (c+d x)}} \\ & = \frac {2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {7}{3},1,\frac {5}{2},-i \tan (c+d x),i \tan (c+d x)\right ) \sqrt [3]{1+i \tan (c+d x)} \tan ^{\frac {3}{2}}(c+d x)}{3 a d \sqrt [3]{a+i a \tan (c+d x)}} \\
\end{align*}
Mathematica [F]
\[
\int \frac {\sqrt {\tan (c+d x)}}{(a+i a \tan (c+d x))^{4/3}} \, dx=\int \frac {\sqrt {\tan (c+d x)}}{(a+i a \tan (c+d x))^{4/3}} \, dx
\]
[In]
Integrate[Sqrt[Tan[c + d*x]]/(a + I*a*Tan[c + d*x])^(4/3),x]
[Out]
Integrate[Sqrt[Tan[c + d*x]]/(a + I*a*Tan[c + d*x])^(4/3), x]
Maple [F]
\[\int \frac {\sqrt {\tan }\left (d x +c \right )}{\left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}d x\]
[In]
int(tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(4/3),x)
[Out]
int(tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(4/3),x)
Fricas [F]
\[
\int \frac {\sqrt {\tan (c+d x)}}{(a+i a \tan (c+d x))^{4/3}} \, dx=\int { \frac {\sqrt {\tan \left (d x + c\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}}} \,d x }
\]
[In]
integrate(tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(4/3),x, algorithm="fricas")
[Out]
-1/32*(3*2^(2/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) +
1))*(-5*I*e^(6*I*d*x + 6*I*c) + 4*I*e^(5*I*d*x + 5*I*c) - 14*I*e^(4*I*d*x + 4*I*c) + 8*I*e^(3*I*d*x + 3*I*c) -
13*I*e^(2*I*d*x + 2*I*c) + 4*I*e^(I*d*x + I*c) - 4*I)*e^(4/3*I*d*x + 4/3*I*c) - 32*(a^2*d*e^(6*I*d*x + 6*I*c)
- 4*a^2*d*e^(5*I*d*x + 5*I*c) + 4*a^2*d*e^(4*I*d*x + 4*I*c))*integral(1/16*2^(2/3)*(a/(e^(2*I*d*x + 2*I*c) +
1))^(2/3)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(-4*I*e^(6*I*d*x + 6*I*c) + 48*I*e^(5*I
*d*x + 5*I*c) - 47*I*e^(4*I*d*x + 4*I*c) + 66*I*e^(3*I*d*x + 3*I*c) - 47*I*e^(2*I*d*x + 2*I*c) + 18*I*e^(I*d*x
+ I*c) - 4*I)*e^(4/3*I*d*x + 4/3*I*c)/(a^2*d*e^(7*I*d*x + 7*I*c) - 6*a^2*d*e^(6*I*d*x + 6*I*c) + 11*a^2*d*e^(
5*I*d*x + 5*I*c) - 2*a^2*d*e^(4*I*d*x + 4*I*c) - 12*a^2*d*e^(3*I*d*x + 3*I*c) + 8*a^2*d*e^(2*I*d*x + 2*I*c)),
x))/(a^2*d*e^(6*I*d*x + 6*I*c) - 4*a^2*d*e^(5*I*d*x + 5*I*c) + 4*a^2*d*e^(4*I*d*x + 4*I*c))
Sympy [F]
\[
\int \frac {\sqrt {\tan (c+d x)}}{(a+i a \tan (c+d x))^{4/3}} \, dx=\int \frac {\sqrt {\tan {\left (c + d x \right )}}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {4}{3}}}\, dx
\]
[In]
integrate(tan(d*x+c)**(1/2)/(a+I*a*tan(d*x+c))**(4/3),x)
[Out]
Integral(sqrt(tan(c + d*x))/(I*a*(tan(c + d*x) - I))**(4/3), x)
Maxima [F(-2)]
Exception generated. \[
\int \frac {\sqrt {\tan (c+d x)}}{(a+i a \tan (c+d x))^{4/3}} \, dx=\text {Exception raised: RuntimeError}
\]
[In]
integrate(tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(4/3),x, algorithm="maxima")
[Out]
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.
Giac [F]
\[
\int \frac {\sqrt {\tan (c+d x)}}{(a+i a \tan (c+d x))^{4/3}} \, dx=\int { \frac {\sqrt {\tan \left (d x + c\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}}} \,d x }
\]
[In]
integrate(tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(4/3),x, algorithm="giac")
[Out]
integrate(sqrt(tan(d*x + c))/(I*a*tan(d*x + c) + a)^(4/3), x)
Mupad [F(-1)]
Timed out. \[
\int \frac {\sqrt {\tan (c+d x)}}{(a+i a \tan (c+d x))^{4/3}} \, dx=\int \frac {\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{4/3}} \,d x
\]
[In]
int(tan(c + d*x)^(1/2)/(a + a*tan(c + d*x)*1i)^(4/3),x)
[Out]
int(tan(c + d*x)^(1/2)/(a + a*tan(c + d*x)*1i)^(4/3), x)