3.299 \(\int \frac {\sqrt {\tan (c+d x)}}{(a+i a \tan (c+d x))^{4/3}} \, dx\)

Optimal. Leaf size=84 \[ \frac {2 \sqrt [3]{1+i \tan (c+d x)} \tan ^{\frac {3}{2}}(c+d x) F_1\left (\frac {3}{2};\frac {7}{3},1;\frac {5}{2};-i \tan (c+d x),i \tan (c+d x)\right )}{3 a d \sqrt [3]{a+i a \tan (c+d x)}} \]

[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]  time = 0.17, antiderivative size = 84, 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} \[ \frac {2 \sqrt [3]{1+i \tan (c+d x)} \tan ^{\frac {3}{2}}(c+d x) F_1\left (\frac {3}{2};\frac {7}{3},1;\frac {5}{2};-i \tan (c+d x),i \tan (c+d x)\right )}{3 a d \sqrt [3]{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[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 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 {\sqrt {\tan (c+d x)}}{(a+i a \tan (c+d x))^{4/3}} \, dx &=\frac {\left (i a^2\right ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 F_1\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]  time = 15.29, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\tan (c+d x)}}{(a+i a \tan (c+d x))^{4/3}} \, dx \]

Verification is Not applicable to the result.

[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]

________________________________________________________________________________________

fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ \frac {2^{\frac {2}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (15 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 12 i \, e^{\left (5 i \, d x + 5 i \, c\right )} + 42 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 24 i \, e^{\left (3 i \, d x + 3 i \, c\right )} + 39 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 12 i \, e^{\left (i \, d x + i \, c\right )} + 12 i\right )} e^{\left (\frac {4}{3} i \, d x + \frac {4}{3} i \, c\right )} + 32 \, {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - 4 \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + 4 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} {\rm integral}\left (\frac {2^{\frac {2}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-4 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 48 i \, e^{\left (5 i \, d x + 5 i \, c\right )} - 47 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 66 i \, e^{\left (3 i \, d x + 3 i \, c\right )} - 47 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 18 i \, e^{\left (i \, d x + i \, c\right )} - 4 i\right )} e^{\left (\frac {4}{3} i \, d x + \frac {4}{3} i \, c\right )}}{16 \, {\left (a^{2} d e^{\left (7 i \, d x + 7 i \, c\right )} - 6 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 11 \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} - 2 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} - 12 \, a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )} + 8 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}}, x\right )}{32 \, {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - 4 \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + 4 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(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))
*(15*I*e^(6*I*d*x + 6*I*c) - 12*I*e^(5*I*d*x + 5*I*c) + 42*I*e^(4*I*d*x + 4*I*c) - 24*I*e^(3*I*d*x + 3*I*c) +
39*I*e^(2*I*d*x + 2*I*c) - 12*I*e^(I*d*x + I*c) + 12*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))

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\tan \left (d x + c\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [F]  time = 1.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\tan }\left (d x +c \right )}{\left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[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.

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{4/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\tan {\left (c + d x \right )}}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________