3.338 \(\int \sqrt {d \tan (e+f x)} (a+a \tan (e+f x)) \, dx\)

Optimal. Leaf size=72 \[ \frac {2 a \sqrt {d \tan (e+f x)}}{f}-\frac {\sqrt {2} a \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d} \tan (e+f x)+\sqrt {d}}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{f} \]

[Out]

-a*arctanh(1/2*(d^(1/2)+d^(1/2)*tan(f*x+e))*2^(1/2)/(d*tan(f*x+e))^(1/2))*2^(1/2)*d^(1/2)/f+2*a*(d*tan(f*x+e))
^(1/2)/f

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Rubi [A]  time = 0.07, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3528, 3532, 208} \[ \frac {2 a \sqrt {d \tan (e+f x)}}{f}-\frac {\sqrt {2} a \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d} \tan (e+f x)+\sqrt {d}}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*a*Sqrt[d]*ArcTanh[(Sqrt[d] + Sqrt[d]*Tan[e + f*x])/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/f) + (2*a*Sqrt[
d*Tan[e + f*x]])/f

Rule 208

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

Rule 3528

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 3532

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

Rubi steps

\begin {align*} \int \sqrt {d \tan (e+f x)} (a+a \tan (e+f x)) \, dx &=\frac {2 a \sqrt {d \tan (e+f x)}}{f}+\int \frac {-a d+a d \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx\\ &=\frac {2 a \sqrt {d \tan (e+f x)}}{f}-\frac {\left (2 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{-2 a^2 d^2+d x^2} \, dx,x,\frac {-a d-a d \tan (e+f x)}{\sqrt {d \tan (e+f x)}}\right )}{f}\\ &=-\frac {\sqrt {2} a \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{f}+\frac {2 a \sqrt {d \tan (e+f x)}}{f}\\ \end {align*}

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Mathematica [C]  time = 0.11, size = 92, normalized size = 1.28 \[ \frac {(1+i) a \sqrt {d \tan (e+f x)} \left (\sqrt [4]{-1} \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (e+f x)}\right )+(1-i) \sqrt {\tan (e+f x)}-(-1)^{3/4} \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\tan (e+f x)}\right )\right )}{f \sqrt {\tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 157, normalized size = 2.18 \[ \left [\frac {\sqrt {2} a \sqrt {d} \log \left (\frac {d \tan \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} {\left (\tan \left (f x + e\right ) + 1\right )} + 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 4 \, \sqrt {d \tan \left (f x + e\right )} a}{2 \, f}, \frac {\sqrt {2} a \sqrt {-d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) + 1\right )}}{2 \, d \tan \left (f x + e\right )}\right ) + 2 \, \sqrt {d \tan \left (f x + e\right )} a}{f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(2)*a*sqrt(d)*log((d*tan(f*x + e)^2 - 2*sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d)*(tan(f*x + e) + 1) + 4*
d*tan(f*x + e) + d)/(tan(f*x + e)^2 + 1)) + 4*sqrt(d*tan(f*x + e))*a)/f, (sqrt(2)*a*sqrt(-d)*arctan(1/2*sqrt(2
)*sqrt(d*tan(f*x + e))*sqrt(-d)*(tan(f*x + e) + 1)/(d*tan(f*x + e))) + 2*sqrt(d*tan(f*x + e))*a)/f]

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giac [B]  time = 0.98, size = 249, normalized size = 3.46 \[ \frac {2 \, \sqrt {d \tan \left (f x + e\right )} a}{f} - \frac {\sqrt {2} {\left (a d \sqrt {{\left | d \right |}} - a {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{2 \, d f} - \frac {\sqrt {2} {\left (a d \sqrt {{\left | d \right |}} - a {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{2 \, d f} - \frac {\sqrt {2} {\left (a d \sqrt {{\left | d \right |}} + a {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{4 \, d f} + \frac {\sqrt {2} {\left (a d \sqrt {{\left | d \right |}} + a {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{4 \, d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(d*tan(f*x + e))*a/f - 1/2*sqrt(2)*(a*d*sqrt(abs(d)) - a*abs(d)^(3/2))*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(
abs(d)) + 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/(d*f) - 1/2*sqrt(2)*(a*d*sqrt(abs(d)) - a*abs(d)^(3/2))*arctan
(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/(d*f) - 1/4*sqrt(2)*(a*d*sqrt(abs(
d)) + a*abs(d)^(3/2))*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) + abs(d))/(d*f) + 1/4*sqr
t(2)*(a*d*sqrt(abs(d)) + a*abs(d)^(3/2))*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) + abs(
d))/(d*f)

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maple [B]  time = 0.20, size = 337, normalized size = 4.68 \[ \frac {2 a \sqrt {d \tan \left (f x +e \right )}}{f}-\frac {a \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{4 f}-\frac {a \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f}+\frac {a \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f}+\frac {a d \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{4 f \left (d^{2}\right )^{\frac {1}{4}}}+\frac {a d \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f \left (d^{2}\right )^{\frac {1}{4}}}-\frac {a d \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f \left (d^{2}\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.82, size = 101, normalized size = 1.40 \[ -\frac {a d^{2} {\left (\frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} - 4 \, \sqrt {d \tan \left (f x + e\right )} a d}{2 \, d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.46, size = 125, normalized size = 1.74 \[ \frac {2\,a\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{f}+\frac {{\left (-1\right )}^{1/4}\,a\,\sqrt {d}\,\left (\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )-\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\right )}{f}+\frac {{\left (-1\right )}^{1/4}\,a\,\sqrt {d}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\,1{}\mathrm {i}}{f}+\frac {{\left (-1\right )}^{1/4}\,a\,\sqrt {d}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\,1{}\mathrm {i}}{f} \]

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)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int \sqrt {d \tan {\left (e + f x \right )}}\, dx + \int \sqrt {d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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