3.2.51 \(\int \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2 \, dx\) [151]

Optimal. Leaf size=90 \[ \frac {4 (-1)^{3/4} a^2 \sqrt {d} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {4 i a^2 \sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2 (d \tan (e+f x))^{3/2}}{3 d f} \]

[Out]

4*(-1)^(3/4)*a^2*arctan((-1)^(3/4)*(d*tan(f*x+e))^(1/2)/d^(1/2))*d^(1/2)/f+4*I*a^2*(d*tan(f*x+e))^(1/2)/f-2/3*
a^2*(d*tan(f*x+e))^(3/2)/d/f

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Rubi [A]
time = 0.09, antiderivative size = 90, 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 = {3624, 3609, 3614, 211} \begin {gather*} \frac {4 (-1)^{3/4} a^2 \sqrt {d} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {2 a^2 (d \tan (e+f x))^{3/2}}{3 d f}+\frac {4 i a^2 \sqrt {d \tan (e+f x)}}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[d*Tan[e + f*x]]*(a + I*a*Tan[e + f*x])^2,x]

[Out]

(4*(-1)^(3/4)*a^2*Sqrt[d]*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/f + ((4*I)*a^2*Sqrt[d*Tan[e + f*x
]])/f - (2*a^2*(d*Tan[e + f*x])^(3/2))/(3*d*f)

Rule 211

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

Rule 3609

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*
((a + b*Tan[e + f*x])^m/(f*m)), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3614

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2*(c^2/f), S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]

Rule 3624

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
d^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e
+ f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[m, -1] &&  !(EqQ[m, 2] && EqQ
[a, 0])

Rubi steps

\begin {align*} \int \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2 \, dx &=-\frac {2 a^2 (d \tan (e+f x))^{3/2}}{3 d f}+\int \sqrt {d \tan (e+f x)} \left (2 a^2+2 i a^2 \tan (e+f x)\right ) \, dx\\ &=\frac {4 i a^2 \sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2 (d \tan (e+f x))^{3/2}}{3 d f}+\int \frac {-2 i a^2 d+2 a^2 d \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx\\ &=\frac {4 i a^2 \sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2 (d \tan (e+f x))^{3/2}}{3 d f}-\frac {\left (8 a^4 d^2\right ) \text {Subst}\left (\int \frac {1}{-2 i a^2 d^2-2 a^2 d x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{f}\\ &=\frac {4 (-1)^{3/4} a^2 \sqrt {d} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {4 i a^2 \sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2 (d \tan (e+f x))^{3/2}}{3 d f}\\ \end {align*}

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Mathematica [A]
time = 1.63, size = 102, normalized size = 1.13 \begin {gather*} \frac {2 i a^2 \left (-6 \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )+(6+i \tan (e+f x)) \sqrt {i \tan (e+f x)}\right ) \sqrt {d \tan (e+f x)}}{3 f \sqrt {i \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d*Tan[e + f*x]]*(a + I*a*Tan[e + f*x])^2,x]

[Out]

(((2*I)/3)*a^2*(-6*ArcTanh[Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x)))]] + (6 + I*Tan[e + f*x])*
Sqrt[I*Tan[e + f*x]])*Sqrt[d*Tan[e + f*x]])/(f*Sqrt[I*Tan[e + f*x]])

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 310 vs. \(2 (73 ) = 146\).
time = 0.12, size = 311, normalized size = 3.46

method	result	size
derivativedivides	\(\frac {2 a^{2} \left (-\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 i d \sqrt {d \tan \left (f x +e \right )}-2 d^{2} \left (\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f d}\)	\(311\)
default	\(\frac {2 a^{2} \left (-\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 i d \sqrt {d \tan \left (f x +e \right )}-2 d^{2} \left (\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f d}\)	\(311\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

2/f*a^2/d*(-1/3*(d*tan(f*x+e))^(3/2)+2*I*d*(d*tan(f*x+e))^(1/2)-2*d^2*(1/8*I/d*(d^2)^(1/4)*2^(1/2)*(ln((d*tan(
f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^
(1/2)+(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(
f*x+e))^(1/2)+1))-1/8/(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/
2))/(d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f
*x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))))

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (75) = 150\).
time = 0.51, size = 204, normalized size = 2.27 \begin {gather*} \frac {3 \, a^{2} d^{2} {\left (-\frac {\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (i + 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\left (i + 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} - 4 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{2} + 24 i \, \sqrt {d \tan \left (f x + e\right )} a^{2} d}{6 \, d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*a^2*d^2*(-(2*I - 2)*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt
(d) - (2*I - 2)*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) - (I +
1)*sqrt(2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d) + (I + 1)*sqrt(2)*log(d*tan
(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d)) - 4*(d*tan(f*x + e))^(3/2)*a^2 + 24*I*sqrt(d*ta
n(f*x + e))*a^2*d)/(d*f)

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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. 322 vs. \(2 (75) = 150\).
time = 0.40, size = 322, normalized size = 3.58 \begin {gather*} \frac {3 \, \sqrt {\frac {16 i \, a^{4} d}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {{\left (-4 i \, a^{2} d e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt {\frac {16 i \, a^{4} d}{f^{2}}} {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\right ) - 3 \, \sqrt {\frac {16 i \, a^{4} d}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {{\left (-4 i \, a^{2} d e^{\left (2 i \, f x + 2 i \, e\right )} + \sqrt {\frac {16 i \, a^{4} d}{f^{2}}} {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\right ) - 8 \, {\left (-7 i \, a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 5 i \, a^{2}\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*sqrt(16*I*a^4*d/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*log(1/2*(-4*I*a^2*d*e^(2*I*f*x + 2*I*e) + sqrt(16*I*a
^4*d/f^2)*(I*f*e^(2*I*f*x + 2*I*e) + I*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^
(-2*I*f*x - 2*I*e)/a^2) - 3*sqrt(16*I*a^4*d/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*log(1/2*(-4*I*a^2*d*e^(2*I*f*x +
2*I*e) + sqrt(16*I*a^4*d/f^2)*(-I*f*e^(2*I*f*x + 2*I*e) - I*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f
*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/a^2) - 8*(-7*I*a^2*e^(2*I*f*x + 2*I*e) - 5*I*a^2)*sqrt((-I*d*e^(2*I*f*
x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/(f*e^(2*I*f*x + 2*I*e) + f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - a^{2} \left (\int \left (- \sqrt {d \tan {\left (e + f x \right )}}\right )\, dx + \int \sqrt {d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\, dx + \int \left (- 2 i \sqrt {d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}\right )\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**(1/2)*(a+I*a*tan(f*x+e))**2,x)

[Out]

-a**2*(Integral(-sqrt(d*tan(e + f*x)), x) + Integral(sqrt(d*tan(e + f*x))*tan(e + f*x)**2, x) + Integral(-2*I*
sqrt(d*tan(e + f*x))*tan(e + f*x), x))

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Giac [A]
time = 0.59, size = 129, normalized size = 1.43 \begin {gather*} \frac {4 \, \sqrt {2} a^{2} \sqrt {d} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right )} a^{2} d^{3} f^{2} \tan \left (f x + e\right ) - 6 i \, \sqrt {d \tan \left (f x + e\right )} a^{2} d^{3} f^{2}\right )}}{3 \, d^{3} f^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*sqrt(2)*a^2*sqrt(d)*arctan(8*sqrt(d^2)*sqrt(d*tan(f*x + e))/(4*I*sqrt(2)*d^(3/2) + 4*sqrt(2)*sqrt(d^2)*sqrt(
d)))/(f*(I*d/sqrt(d^2) + 1)) - 2/3*(sqrt(d*tan(f*x + e))*a^2*d^3*f^2*tan(f*x + e) - 6*I*sqrt(d*tan(f*x + e))*a
^2*d^3*f^2)/(d^3*f^3)

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Mupad [B]
time = 4.28, size = 74, normalized size = 0.82 \begin {gather*} \frac {a^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,4{}\mathrm {i}}{f}-\frac {2\,a^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3\,d\,f}-\frac {2\,\sqrt {4{}\mathrm {i}}\,a^2\,\sqrt {d}\,\mathrm {atanh}\left (\frac {\sqrt {4{}\mathrm {i}}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}\right )}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(e + f*x))^(1/2)*(a + a*tan(e + f*x)*1i)^2,x)

[Out]

(a^2*(d*tan(e + f*x))^(1/2)*4i)/f - (2*a^2*(d*tan(e + f*x))^(3/2))/(3*d*f) - (2*4i^(1/2)*a^2*d^(1/2)*atanh((4i
^(1/2)*(d*tan(e + f*x))^(1/2))/(2*d^(1/2))))/f

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