3.250 \(\int (d \tan (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=212 \[ \frac{d^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} f}-\frac{d^{5/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} f}-\frac{d^{5/2} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{2 \sqrt{2} f}+\frac{d^{5/2} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{2 \sqrt{2} f}+\frac{2 d (d \tan (e+f x))^{3/2}}{3 f} \]

[Out]

(d^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*f) - (d^(5/2)*ArcTan[1 + (Sqrt[2]*Sqrt[d
*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*f) - (d^(5/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f
*x]]])/(2*Sqrt[2]*f) + (d^(5/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(2*Sqrt[2]
*f) + (2*d*(d*Tan[e + f*x])^(3/2))/(3*f)

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Rubi [A]  time = 0.14225, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {3473, 3476, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac{d^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} f}-\frac{d^{5/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} f}-\frac{d^{5/2} \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{2 \sqrt{2} f}+\frac{d^{5/2} \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{2 \sqrt{2} f}+\frac{2 d (d \tan (e+f x))^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^(5/2),x]

[Out]

(d^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*f) - (d^(5/2)*ArcTan[1 + (Sqrt[2]*Sqrt[d
*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*f) - (d^(5/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f
*x]]])/(2*Sqrt[2]*f) + (d^(5/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(2*Sqrt[2]
*f) + (2*d*(d*Tan[e + f*x])^(3/2))/(3*f)

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3476

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

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

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 204

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

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int (d \tan (e+f x))^{5/2} \, dx &=\frac{2 d (d \tan (e+f x))^{3/2}}{3 f}-d^2 \int \sqrt{d \tan (e+f x)} \, dx\\ &=\frac{2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac{d^3 \operatorname{Subst}\left (\int \frac{\sqrt{x}}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{f}\\ &=\frac{2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac{\left (2 d^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{f}\\ &=\frac{2 d (d \tan (e+f x))^{3/2}}{3 f}+\frac{d^3 \operatorname{Subst}\left (\int \frac{d-x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{f}-\frac{d^3 \operatorname{Subst}\left (\int \frac{d+x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{f}\\ &=\frac{2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac{d^{5/2} \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{d}+2 x}{-d-\sqrt{2} \sqrt{d} x-x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{2 \sqrt{2} f}-\frac{d^{5/2} \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{d}-2 x}{-d+\sqrt{2} \sqrt{d} x-x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{2 \sqrt{2} f}-\frac{d^3 \operatorname{Subst}\left (\int \frac{1}{d-\sqrt{2} \sqrt{d} x+x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{2 f}-\frac{d^3 \operatorname{Subst}\left (\int \frac{1}{d+\sqrt{2} \sqrt{d} x+x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{2 f}\\ &=-\frac{d^{5/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{2 \sqrt{2} f}+\frac{d^{5/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{2 \sqrt{2} f}+\frac{2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac{d^{5/2} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} f}+\frac{d^{5/2} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} f}\\ &=\frac{d^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} f}-\frac{d^{5/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} f}-\frac{d^{5/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{2 \sqrt{2} f}+\frac{d^{5/2} \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{2 \sqrt{2} f}+\frac{2 d (d \tan (e+f x))^{3/2}}{3 f}\\ \end{align*}

Mathematica [C]  time = 0.0393839, size = 40, normalized size = 0.19 \[ -\frac{2 d (d \tan (e+f x))^{3/2} \left (\, _2F_1\left (\frac{3}{4},1;\frac{7}{4};-\tan ^2(e+f x)\right )-1\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Tan[e + f*x])^(5/2),x]

[Out]

(-2*d*(-1 + Hypergeometric2F1[3/4, 1, 7/4, -Tan[e + f*x]^2])*(d*Tan[e + f*x])^(3/2))/(3*f)

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Maple [A]  time = 0.011, size = 182, normalized size = 0.9 \begin{align*}{\frac{2\,d}{3\,f} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{{d}^{3}\sqrt{2}}{4\,f}\ln \left ({ \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}-{\frac{{d}^{3}\sqrt{2}}{2\,f}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}+{\frac{{d}^{3}\sqrt{2}}{2\,f}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^(5/2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.76278, size = 1521, normalized size = 7.17 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/12*(12*sqrt(2)*(d^10/f^4)^(1/4)*f*arctan(-(d^10 + sqrt(2)*(d^10/f^4)^(1/4)*d^7*f*sqrt(d*sin(f*x + e)/cos(f*x
+ e)) - sqrt(2)*(d^10/f^4)^(1/4)*f*sqrt((d^15*sin(f*x + e) + sqrt(d^10/f^4)*d^10*f^2*cos(f*x + e) + sqrt(2)*(
d^10/f^4)^(3/4)*d^7*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*cos(f*x + e))/cos(f*x + e)))/d^10)*cos(f*x + e) + 12
*sqrt(2)*(d^10/f^4)^(1/4)*f*arctan((d^10 - sqrt(2)*(d^10/f^4)^(1/4)*d^7*f*sqrt(d*sin(f*x + e)/cos(f*x + e)) +
sqrt(2)*(d^10/f^4)^(1/4)*f*sqrt((d^15*sin(f*x + e) + sqrt(d^10/f^4)*d^10*f^2*cos(f*x + e) - sqrt(2)*(d^10/f^4)
^(3/4)*d^7*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*cos(f*x + e))/cos(f*x + e)))/d^10)*cos(f*x + e) + 3*sqrt(2)*(
d^10/f^4)^(1/4)*f*cos(f*x + e)*log((d^15*sin(f*x + e) + sqrt(d^10/f^4)*d^10*f^2*cos(f*x + e) + sqrt(2)*(d^10/f
^4)^(3/4)*d^7*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*cos(f*x + e))/cos(f*x + e)) - 3*sqrt(2)*(d^10/f^4)^(1/4)*f
*cos(f*x + e)*log((d^15*sin(f*x + e) + sqrt(d^10/f^4)*d^10*f^2*cos(f*x + e) - sqrt(2)*(d^10/f^4)^(3/4)*d^7*f^3
*sqrt(d*sin(f*x + e)/cos(f*x + e))*cos(f*x + e))/cos(f*x + e)) + 8*d^2*sqrt(d*sin(f*x + e)/cos(f*x + e))*sin(f
*x + e))/(f*cos(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \tan{\left (e + f x \right )}\right )^{\frac{5}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**(5/2),x)

[Out]

Integral((d*tan(e + f*x))**(5/2), x)

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Giac [A]  time = 1.28874, size = 277, normalized size = 1.31 \begin{align*} -\frac{1}{12} \,{\left (\frac{6 \, \sqrt{2}{\left | d \right |}^{\frac{3}{2}} \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 )}{f} + \frac{6 \, \sqrt{2}{\left | d \right |}^{\frac{3}{2}} \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 )}{f} - \frac{3 \, \sqrt{2}{\left | d \right |}^{\frac{3}{2}} \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 )}{f} + \frac{3 \, \sqrt{2}{\left | d \right |}^{\frac{3}{2}} \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 )}{f} - \frac{8 \, \sqrt{d \tan \left (f x + e\right )} d \tan \left (f x + e\right )}{f}\right )} d \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

-1/12*(6*sqrt(2)*abs(d)^(3/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))
/f + 6*sqrt(2)*abs(d)^(3/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/
f - 3*sqrt(2)*abs(d)^(3/2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) + abs(d))/f + 3*sqrt
(2)*abs(d)^(3/2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) + abs(d))/f - 8*sqrt(d*tan(f*x
+ e))*d*tan(f*x + e)/f)*d