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

Optimal. Leaf size=82 \[ \frac {3 a \tan ^{\frac {2}{3}}(c+d x) \sqrt {a+i a \tan (c+d x)} F_1\left (\frac {2}{3};-\frac {1}{2},1;\frac {5}{3};-i \tan (c+d x),i \tan (c+d x)\right )}{2 d \sqrt {1+i \tan (c+d x)}} \]

[Out]

3/2*a*AppellF1(2/3,-1/2,1,5/3,-I*tan(d*x+c),I*tan(d*x+c))*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^(2/3)/d/(1+I*tan
(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3564, 130, 511, 510} \[ \frac {3 a \tan ^{\frac {2}{3}}(c+d x) \sqrt {a+i a \tan (c+d x)} F_1\left (\frac {2}{3};-\frac {1}{2},1;\frac {5}{3};-i \tan (c+d x),i \tan (c+d x)\right )}{2 d \sqrt {1+i \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*AppellF1[2/3, -1/2, 1, 5/3, (-I)*Tan[c + d*x], I*Tan[c + d*x]]*Tan[c + d*x]^(2/3)*Sqrt[a + I*a*Tan[c + d*
x]])/(2*d*Sqrt[1 + I*Tan[c + d*x]])

Rule 130

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 510

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)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), 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 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[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 3564

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*x)/b)^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*} \int \frac {(a+i a \tan (c+d x))^{3/2}}{\sqrt [3]{\tan (c+d x)}} \, dx &=\frac {\left (i a^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+x}}{\sqrt [3]{-\frac {i x}{a}} \left (-a^2+a x\right )} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int \frac {x \sqrt {a+i a x^3}}{-a^2+i a^2 x^3} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{d}\\ &=-\frac {\left (3 a^3 \sqrt {a+i a \tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {x \sqrt {1+i x^3}}{-a^2+i a^2 x^3} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{d \sqrt {1+i \tan (c+d x)}}\\ &=\frac {3 a F_1\left (\frac {2}{3};-\frac {1}{2},1;\frac {5}{3};-i \tan (c+d x),i \tan (c+d x)\right ) \tan ^{\frac {2}{3}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d \sqrt {1+i \tan (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 16.88, size = 0, normalized size = 0.00 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\sqrt [3]{\tan (c+d x)}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)sym2poly/r2sym(const gen & e,const index_m & i,c
onst vecteur & l) Error: Bad Argument ValueUnable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to
check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Una
ble to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep
/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_
nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2
*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Algebraic extensions not allowed in a
rootofAlgebraic extensions not allowed in a rootofEvaluation time: 8.49int()  Error: Bad Argument Value

________________________________________________________________________________________

maple [F]  time = 1.44, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{\tan \left (d x +c \right )^{\frac {1}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\tan \left (d x + c\right )^{\frac {1}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{1/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}{\sqrt [3]{\tan {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((I*a*(tan(c + d*x) - I))**(3/2)/tan(c + d*x)**(1/3), x)

________________________________________________________________________________________