3.4.59 \(\int \frac {(d \tan (e+f x))^{3/2}}{a+a \tan (e+f x)} \, dx\) [359]

Optimal. Leaf size=87 \[ \frac {d^{3/2} \text {ArcTan}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a f}-\frac {d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a f} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3654, 3613, 214, 3715, 65, 211} \begin {gather*} \frac {d^{3/2} \text {ArcTan}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a f}-\frac {d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \tan (e+f x)+\sqrt {d}}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 214

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

Rule 3613

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]

Rule 3654

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(3/2)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1
/(c^2 + d^2), Int[Simp[a^2*c - b^2*c + 2*a*b*d + (2*a*b*c - a^2*d + b^2*d)*Tan[e + f*x], x]/Sqrt[a + b*Tan[e +
f*x]], x], x] + Dist[(b*c - a*d)^2/(c^2 + d^2), Int[(1 + Tan[e + f*x]^2)/(Sqrt[a + b*Tan[e + f*x]]*(c + 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] && NeQ[c^2 + d^2
, 0]

Rule 3715

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rubi steps

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

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Mathematica [A]
time = 0.35, size = 107, normalized size = 1.23 \begin {gather*} \frac {\left (4 \text {ArcTan}\left (\sqrt {\tan (e+f x)}\right )+\sqrt {2} \left (\log \left (-1+\sqrt {2} \sqrt {\tan (e+f x)}-\tan (e+f x)\right )-\log \left (1+\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right )\right )\right ) (d \tan (e+f x))^{3/2}}{4 a f \tan ^{\frac {3}{2}}(e+f x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(297\) vs. \(2(70)=140\).
time = 0.29, size = 298, normalized size = 3.43

method	result	size
derivativedivides	\(\frac {2 d^{2} \left (\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2 \sqrt {d}}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f a}\)	\(298\)
default	\(\frac {2 d^{2} \left (\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2 \sqrt {d}}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{16 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f a}\)	\(298\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^(3/2)/(a+a*tan(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

2/f/a*d^2*(1/2/d^(1/2)*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))-1/16/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))+1/16/(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 [A]
time = 0.53, size = 117, normalized size = 1.34 \begin {gather*} -\frac {\frac {d^{3} {\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 )}}{a} - \frac {4 \, d^{\frac {5}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a}}{4 \, d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(d^3*(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*tan(
f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d))/a - 4*d^(5/2)*arctan(sqrt(d*tan(f*x + e))/sqrt(d
))/a)/(d*f)

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Fricas [A]
time = 1.05, size = 223, normalized size = 2.56 \begin {gather*} \left [\frac {\sqrt {2} \sqrt {-d} d \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )} {\left (\sqrt {2} \tan \left (f x + e\right ) + \sqrt {2}\right )} \sqrt {-d}}{2 \, d \tan \left (f x + e\right )}\right ) + \sqrt {-d} d \log \left (\frac {d \tan \left (f x + e\right ) + 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right )}{2 \, a f}, \frac {\sqrt {2} d^{\frac {3}{2}} \log \left (\frac {d \tan \left (f x + e\right )^{2} - 2 \, \sqrt {d \tan \left (f x + e\right )} {\left (\sqrt {2} \tan \left (f x + e\right ) + \sqrt {2}\right )} \sqrt {d} + 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 4 \, d^{\frac {3}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{4 \, a f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(3/2)/(a+a*tan(f*x+e)),x, algorithm="fricas")

[Out]

[1/2*(sqrt(2)*sqrt(-d)*d*arctan(1/2*sqrt(d*tan(f*x + e))*(sqrt(2)*tan(f*x + e) + sqrt(2))*sqrt(-d)/(d*tan(f*x
+ e))) + sqrt(-d)*d*log((d*tan(f*x + e) + 2*sqrt(d*tan(f*x + e))*sqrt(-d) - d)/(tan(f*x + e) + 1)))/(a*f), 1/4
*(sqrt(2)*d^(3/2)*log((d*tan(f*x + e)^2 - 2*sqrt(d*tan(f*x + e))*(sqrt(2)*tan(f*x + e) + sqrt(2))*sqrt(d) + 4*
d*tan(f*x + e) + d)/(tan(f*x + e)^2 + 1)) + 4*d^(3/2)*arctan(sqrt(d*tan(f*x + e))/sqrt(d)))/(a*f)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\tan {\left (e + f x \right )} + 1}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**(3/2)/(a+a*tan(f*x+e)),x)

[Out]

Integral((d*tan(e + f*x))**(3/2)/(tan(e + f*x) + 1), x)/a

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 263 vs. \(2 (73) = 146\).
time = 0.57, size = 263, normalized size = 3.02 \begin {gather*} -\frac {1}{8} \, d {\left (\frac {2 \, \sqrt {2} {\left (d \sqrt {{\left | d \right |}} - {\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 )}{a d f} + \frac {2 \, \sqrt {2} {\left (d \sqrt {{\left | d \right |}} - {\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 )}{a d f} - \frac {8 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a f} + \frac {\sqrt {2} {\left (d \sqrt {{\left | d \right |}} + {\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 )}{a d f} - \frac {\sqrt {2} {\left (d \sqrt {{\left | d \right |}} + {\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 )}{a d f}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(3/2)/(a+a*tan(f*x+e)),x, algorithm="giac")

[Out]

-1/8*d*(2*sqrt(2)*(d*sqrt(abs(d)) - abs(d)^(3/2))*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(f*x
+ e)))/sqrt(abs(d)))/(a*d*f) + 2*sqrt(2)*(d*sqrt(abs(d)) - abs(d)^(3/2))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs
(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/(a*d*f) - 8*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a*f) +
sqrt(2)*(d*sqrt(abs(d)) + abs(d)^(3/2))*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) + abs(d
))/(a*d*f) - sqrt(2)*(d*sqrt(abs(d)) + abs(d)^(3/2))*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(ab
s(d)) + abs(d))/(a*d*f))

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Mupad [B]
time = 4.23, size = 78, normalized size = 0.90 \begin {gather*} \frac {d^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{a\,f}-\frac {\sqrt {2}\,d^{3/2}\,\mathrm {atanh}\left (\frac {12\,\sqrt {2}\,d^{25/2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{12\,d^{13}\,\mathrm {tan}\left (e+f\,x\right )+12\,d^{13}}\right )}{2\,a\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(e + f*x))^(3/2)/(a + a*tan(e + f*x)),x)

[Out]

(d^(3/2)*atan((d*tan(e + f*x))^(1/2)/d^(1/2)))/(a*f) - (2^(1/2)*d^(3/2)*atanh((12*2^(1/2)*d^(25/2)*(d*tan(e +
f*x))^(1/2))/(12*d^13*tan(e + f*x) + 12*d^13)))/(2*a*f)

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