∫ tan2 _ 3 (c + dx)(a + ia tan(c + dx))3∕2 dx

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

Optimal. Leaf size=82 \[ \frac {3 a \tan ^{\frac {5}{3}}(c+d x) \sqrt {a+i a \tan (c+d x)} F_1\left (\frac {5}{3};-\frac {1}{2},1;\frac {8}{3};-i \tan (c+d x),i \tan (c+d x)\right )}{5 d \sqrt {1+i \tan (c+d x)}} \]

[Out]

3/5*a*AppellF1(5/3,-1/2,1,8/3,-I*tan(d*x+c),I*tan(d*x+c))*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^(5/3)/d/(1+I*tan

(d*x+c))^(1/2)

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Rubi [A] time = 0.15, 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} \[ \frac {3 a \tan ^{\frac {5}{3}}(c+d x) \sqrt {a+i a \tan (c+d x)} F_1\left (\frac {5}{3};-\frac {1}{2},1;\frac {8}{3};-i \tan (c+d x),i \tan (c+d x)\right )}{5 d \sqrt {1+i \tan (c+d x)}} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(3*a*AppellF1[5/3, -1/2, 1, 8/3, (-I)*Tan[c + d*x], I*Tan[c + d*x]]*Tan[c + d*x]^(5/3)*Sqrt[a + I*a*Tan[c + d*

x]])/(5*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 \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx &=\frac {\left (i a^2\right ) \operatorname {Subst}\left (\int \frac {\left (-\frac {i x}{a}\right )^{2/3} \sqrt {a+x}}{-a^2+a x} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int \frac {x^4 \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^4 \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 {5}{3};-\frac {1}{2},1;\frac {8}{3};-i \tan (c+d x),i \tan (c+d x)\right ) \tan ^{\frac {5}{3}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d \sqrt {1+i \tan (c+d x)}}\\ \end {align*}

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Mathematica [F] time = 14.23, size = 0, normalized size = 0.00 \[ \int \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx \]

Veriﬁcation is Not applicable to the result.

[In]

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

[Out]

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

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

[Out]

Timed out

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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(tan(d*x+c)^(2/3)*(a+I*a*tan(d*x+c))^(3/2),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)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)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)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)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)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)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)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab

le 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_n

ostep/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_no

step/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*p

i/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 chec

k 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 t

o 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)Un

able 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_noste

p/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)Simplification assuming c near 0Unab

le to check sign: (pi/x/2)>(-pi/x/2)Unable to check sign: (pi/x/2)>(-pi/x/2)Unable to check sign: (pi/x/2)>(-p

i/x/2)Unable to check sign: (pi/x/2)>(-pi/x/2)Unable to check sign: (pi/x/2)>(-pi/x/2)Unable to check sign: (p

i/x/2)>(-pi/x/2)Unable to check sign: (pi/x/2)>(-pi/x/2)Unable to check sign: (pi/x/2)>(-pi/x/2)Unable to chec

k sign: (pi/x/2)>(-pi/x/2)Unable to check sign: (pi/x/2)>(-pi/x/2)Simplification assuming a near 0Simplificati

on assuming c near 0Simplification assuming d near 0Simplification assuming x near 0Warning, choosing root of

[1,0,0,0,0,0,%%%{-2,[1]%%%}] at parameters values [-82]Simplification assuming c near 0Unable to check sign: (

pi/x/2)>(-pi/x/2)Unable to check sign: (pi/x/2)>(-pi/x/2)Unable to check sign: (pi/x/2)>(-pi/x/2)Unable to che

ck sign: (pi/x/2)>(-pi/x/2)Unable to check sign: (pi/x/2)>(-pi/x/2)Unable to check sign: (pi/x/2)>(-pi/x/2)Una

ble to check sign: (pi/x/2)>(-pi/x/2)Unable to check sign: (pi/x/2)>(-pi/x/2)Unable to check sign: (pi/x/2)>(-

pi/x/2)Unable to check sign: (pi/x/2)>(-pi/x/2)Simplification assuming a near 0Simplification assuming c near

0Simplification assuming d near 0Simplification assuming x near 0sym2poly/r2sym(const gen & e,const index_m &

i,const vecteur & l) Error: Bad Argument ValueSimplification assuming c near 0Unable to check sign: (pi/x/2)>(

-pi/x/2)Unable to check sign: (pi/x/2)>(-pi/x/2)Unable to check sign: (pi/x/2)>(-pi/x/2)Unable to check sign:

(pi/x/2)>(-pi/x/2)Unable to check sign: (pi/x/2)>(-pi/x/2)Unable to check sign: (pi/x/2)>(-pi/x/2)Unable to ch

eck sign: (pi/x/2)>(-pi/x/2)Unable to check sign: (pi/x/2)>(-pi/x/2)Unable to check sign: (pi/x/2)>(-pi/x/2)Un

able to check sign: (pi/x/2)>(-pi/x/2)Simplification assuming a near 0Simplification assuming c near 0Simplifi

cation assuming d near 0Simplification assuming x near 0sym2poly/r2sym(const gen & e,const index_m & i,const v

ecteur & l) Error: Bad Argument ValueEvaluation time: 2.38Done

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maple [F] time = 1.32, size = 0, normalized size = 0.00 \[ \int \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}\, dx \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [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(tan(d*x+c)**(2/3)*(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Timed out

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