3.3.83 \(\int (a+i a \tan (c+d x))^{4/3} \, dx\) [283]

Optimal. Leaf size=175 \[ -\frac {a^{4/3} x}{2^{2/3}}-\frac {i \sqrt [3]{2} \sqrt {3} a^{4/3} \text {ArcTan}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{d}+\frac {i a^{4/3} \log (\cos (c+d x))}{2^{2/3} d}+\frac {3 i a^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2^{2/3} d}+\frac {3 i a \sqrt [3]{a+i a \tan (c+d x)}}{d} \]

[Out]

-1/2*a^(4/3)*x*2^(1/3)+1/2*I*a^(4/3)*ln(cos(d*x+c))*2^(1/3)/d+3/2*I*a^(4/3)*ln(2^(1/3)*a^(1/3)-(a+I*a*tan(d*x+
c))^(1/3))*2^(1/3)/d-I*2^(1/3)*a^(4/3)*arctan(1/3*(a^(1/3)+2^(2/3)*(a+I*a*tan(d*x+c))^(1/3))/a^(1/3)*3^(1/2))*
3^(1/2)/d+3*I*a*(a+I*a*tan(d*x+c))^(1/3)/d

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Rubi [A]
time = 0.08, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {3559, 3562, 59, 631, 210, 31} \begin {gather*} -\frac {i \sqrt [3]{2} \sqrt {3} a^{4/3} \text {ArcTan}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{d}+\frac {3 i a^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2^{2/3} d}+\frac {i a^{4/3} \log (\cos (c+d x))}{2^{2/3} d}-\frac {a^{4/3} x}{2^{2/3}}+\frac {3 i a \sqrt [3]{a+i a \tan (c+d x)}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(4/3),x]

[Out]

-((a^(4/3)*x)/2^(2/3)) - (I*2^(1/3)*Sqrt[3]*a^(4/3)*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/(S
qrt[3]*a^(1/3))])/d + (I*a^(4/3)*Log[Cos[c + d*x]])/(2^(2/3)*d) + ((3*I)*a^(4/3)*Log[2^(1/3)*a^(1/3) - (a + I*
a*Tan[c + d*x])^(1/3)])/(2^(2/3)*d) + ((3*I)*a*(a + I*a*Tan[c + d*x])^(1/3))/d

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 59

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, Simp[-L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[3/(2*b*q^2), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x]
&& PosQ[(b*c - a*d)/b]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 3559

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + b*Tan[c + d*x])^(n - 1)/(d*(n - 1))
), x] + Dist[2*a, Int[(a + b*Tan[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && G
tQ[n, 1]

Rule 3562

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[-b/d, Subst[Int[(a + x)^(n - 1)/(a - x), x]
, x, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 + b^2, 0]

Rubi steps

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

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Mathematica [A]
time = 1.26, size = 294, normalized size = 1.68 \begin {gather*} \frac {i a e^{\frac {1}{3} i (c+d x)} \cos (c+d x) \left (6 e^{\frac {2}{3} i (c+d x)}-2 \sqrt {3} \sqrt [3]{1+e^{2 i (c+d x)}} \text {ArcTan}\left (\frac {1+\frac {2 e^{\frac {2}{3} i (c+d x)}}{\sqrt [3]{1+e^{2 i (c+d x)}}}}{\sqrt {3}}\right )+2 \sqrt [3]{1+e^{2 i (c+d x)}} \log \left (1-\frac {e^{\frac {2}{3} i (c+d x)}}{\sqrt [3]{1+e^{2 i (c+d x)}}}\right )-\sqrt [3]{1+e^{2 i (c+d x)}} \log \left (\frac {e^{\frac {4}{3} i (c+d x)}+e^{\frac {2}{3} i (c+d x)} \sqrt [3]{1+e^{2 i (c+d x)}}+\left (1+e^{2 i (c+d x)}\right )^{2/3}}{\left (1+e^{2 i (c+d x)}\right )^{2/3}}\right )\right ) \sqrt [3]{a+i a \tan (c+d x)}}{d \left (1+e^{2 i (c+d x)}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(I*a*E^((I/3)*(c + d*x))*Cos[c + d*x]*(6*E^(((2*I)/3)*(c + d*x)) - 2*Sqrt[3]*(1 + E^((2*I)*(c + d*x)))^(1/3)*A
rcTan[(1 + (2*E^(((2*I)/3)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))^(1/3))/Sqrt[3]] + 2*(1 + E^((2*I)*(c + d*x)))
^(1/3)*Log[1 - E^(((2*I)/3)*(c + d*x))/(1 + E^((2*I)*(c + d*x)))^(1/3)] - (1 + E^((2*I)*(c + d*x)))^(1/3)*Log[
(E^(((4*I)/3)*(c + d*x)) + E^(((2*I)/3)*(c + d*x))*(1 + E^((2*I)*(c + d*x)))^(1/3) + (1 + E^((2*I)*(c + d*x)))
^(2/3))/(1 + E^((2*I)*(c + d*x)))^(2/3)])*(a + I*a*Tan[c + d*x])^(1/3))/(d*(1 + E^((2*I)*(c + d*x))))

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Maple [A]
time = 0.10, size = 151, normalized size = 0.86

method	result	size
derivativedivides	\(\frac {3 i a \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2 \left (\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {2}{3}}}\right ) a \right )}{d}\)	\(151\)
default	\(\frac {3 i a \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2 \left (\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {2}{3}}}\right ) a \right )}{d}\)	\(151\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*I/d*a*((a+I*a*tan(d*x+c))^(1/3)+2*(1/6*2^(1/3)/a^(2/3)*ln((a+I*a*tan(d*x+c))^(1/3)-2^(1/3)*a^(1/3))-1/12*2^(
1/3)/a^(2/3)*ln((a+I*a*tan(d*x+c))^(2/3)+2^(1/3)*a^(1/3)*(a+I*a*tan(d*x+c))^(1/3)+2^(2/3)*a^(2/3))-1/6*2^(1/3)
/a^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2^(2/3)/a^(1/3)*(a+I*a*tan(d*x+c))^(1/3)+1)))*a)

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Maxima [A]
time = 0.52, size = 153, normalized size = 0.87 \begin {gather*} -\frac {i \, {\left (2 \, \sqrt {3} 2^{\frac {1}{3}} a^{\frac {7}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) + 2^{\frac {1}{3}} a^{\frac {7}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) - 2 \cdot 2^{\frac {1}{3}} a^{\frac {7}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) - 6 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{2}\right )}}{2 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*(2*sqrt(3)*2^(1/3)*a^(7/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3)*a^(1/3) + 2*(I*a*tan(d*x + c) + a)^(1/3)
)/a^(1/3)) + 2^(1/3)*a^(7/3)*log(2^(2/3)*a^(2/3) + 2^(1/3)*(I*a*tan(d*x + c) + a)^(1/3)*a^(1/3) + (I*a*tan(d*x
+ c) + a)^(2/3)) - 2*2^(1/3)*a^(7/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1/3)) - 6*(I*a*tan(d*x +
c) + a)^(1/3)*a^2)/(a*d)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 262 vs. \(2 (126) = 252\).
time = 0.80, size = 262, normalized size = 1.50 \begin {gather*} \frac {6 i \cdot 2^{\frac {1}{3}} a \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} + {\left (i \, \sqrt {3} d - d\right )} \left (-\frac {2 i \, a^{4}}{d^{3}}\right )^{\frac {1}{3}} \log \left (\frac {2 \cdot 2^{\frac {1}{3}} a \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} - {\left (\sqrt {3} d + i \, d\right )} \left (-\frac {2 i \, a^{4}}{d^{3}}\right )^{\frac {1}{3}}}{2 \, a}\right ) + {\left (-i \, \sqrt {3} d - d\right )} \left (-\frac {2 i \, a^{4}}{d^{3}}\right )^{\frac {1}{3}} \log \left (\frac {2 \cdot 2^{\frac {1}{3}} a \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} + {\left (\sqrt {3} d - i \, d\right )} \left (-\frac {2 i \, a^{4}}{d^{3}}\right )^{\frac {1}{3}}}{2 \, a}\right ) + 2 \, \left (-\frac {2 i \, a^{4}}{d^{3}}\right )^{\frac {1}{3}} d \log \left (\frac {2^{\frac {1}{3}} a \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} + i \, \left (-\frac {2 i \, a^{4}}{d^{3}}\right )^{\frac {1}{3}} d}{a}\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(6*I*2^(1/3)*a*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) + (I*sqrt(3)*d - d)*(-2*I*a^4/d
^3)^(1/3)*log(1/2*(2*2^(1/3)*a*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) - (sqrt(3)*d + I*d)
*(-2*I*a^4/d^3)^(1/3))/a) + (-I*sqrt(3)*d - d)*(-2*I*a^4/d^3)^(1/3)*log(1/2*(2*2^(1/3)*a*(a/(e^(2*I*d*x + 2*I*
c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) + (sqrt(3)*d - I*d)*(-2*I*a^4/d^3)^(1/3))/a) + 2*(-2*I*a^4/d^3)^(1/3)*d
*log((2^(1/3)*a*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) + I*(-2*I*a^4/d^3)^(1/3)*d)/a))/d

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (i a \tan {\left (c + d x \right )} + a\right )^{\frac {4}{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.99, size = 195, normalized size = 1.11 \begin {gather*} \frac {a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,3{}\mathrm {i}}{d}-\frac {{\left (2{}\mathrm {i}\right )}^{1/3}\,a^{4/3}\,\ln \left (a^2\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,18{}\mathrm {i}+18\,{\left (2{}\mathrm {i}\right )}^{1/3}\,a^{7/3}\,d^2\right )}{d}-\frac {{\left (2{}\mathrm {i}\right )}^{1/3}\,a^{4/3}\,\ln \left (a^2\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,18{}\mathrm {i}+18\,{\left (2{}\mathrm {i}\right )}^{1/3}\,a^{7/3}\,d^2\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{d}+\frac {{\left (2{}\mathrm {i}\right )}^{1/3}\,a^{4/3}\,\ln \left (a^2\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,18{}\mathrm {i}-18\,{\left (2{}\mathrm {i}\right )}^{1/3}\,a^{7/3}\,d^2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^(4/3),x)

[Out]

(a*(a + a*tan(c + d*x)*1i)^(1/3)*3i)/d - (2i^(1/3)*a^(4/3)*log(a^2*d^2*(a + a*tan(c + d*x)*1i)^(1/3)*18i + 18*
2i^(1/3)*a^(7/3)*d^2))/d - (2i^(1/3)*a^(4/3)*log(a^2*d^2*(a + a*tan(c + d*x)*1i)^(1/3)*18i + 18*2i^(1/3)*a^(7/
3)*d^2*((3^(1/2)*1i)/2 - 1/2))*((3^(1/2)*1i)/2 - 1/2))/d + (2i^(1/3)*a^(4/3)*log(a^2*d^2*(a + a*tan(c + d*x)*1
i)^(1/3)*18i - 18*2i^(1/3)*a^(7/3)*d^2*((3^(1/2)*1i)/2 + 1/2))*((3^(1/2)*1i)/2 + 1/2))/d

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