3.10 \(\int (b \tan (c+d x))^{5/2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.144697, 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} \[ \frac{b^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b \tan (c+d x)}}{\sqrt{b}}\right )}{\sqrt{2} d}-\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b \tan (c+d x)}}{\sqrt{b}}+1\right )}{\sqrt{2} d}-\frac{b^{5/2} \log \left (\sqrt{b} \tan (c+d x)-\sqrt{2} \sqrt{b \tan (c+d x)}+\sqrt{b}\right )}{2 \sqrt{2} d}+\frac{b^{5/2} \log \left (\sqrt{b} \tan (c+d x)+\sqrt{2} \sqrt{b \tan (c+d x)}+\sqrt{b}\right )}{2 \sqrt{2} d}+\frac{2 b (b \tan (c+d x))^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[c + d*x])^(5/2),x]

[Out]

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

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

Mathematica [C]  time = 0.0683589, size = 40, normalized size = 0.19 \[ -\frac{2 b (b \tan (c+d x))^{3/2} \left (\, _2F_1\left (\frac{3}{4},1;\frac{7}{4};-\tan ^2(c+d x)\right )-1\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[c + d*x])^(5/2),x]

[Out]

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

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Maple [A]  time = 0.015, size = 182, normalized size = 0.9 \begin{align*}{\frac{2\,b}{3\,d} \left ( b\tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{{b}^{3}\sqrt{2}}{4\,d}\ln \left ({ \left ( b\tan \left ( dx+c \right ) -\sqrt [4]{{b}^{2}}\sqrt{b\tan \left ( dx+c \right ) }\sqrt{2}+\sqrt{{b}^{2}} \right ) \left ( b\tan \left ( dx+c \right ) +\sqrt [4]{{b}^{2}}\sqrt{b\tan \left ( dx+c \right ) }\sqrt{2}+\sqrt{{b}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{b}^{2}}}}}-{\frac{{b}^{3}\sqrt{2}}{2\,d}\arctan \left ({\sqrt{2}\sqrt{b\tan \left ( dx+c \right ) }{\frac{1}{\sqrt [4]{{b}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{b}^{2}}}}}+{\frac{{b}^{3}\sqrt{2}}{2\,d}\arctan \left ( -{\sqrt{2}\sqrt{b\tan \left ( dx+c \right ) }{\frac{1}{\sqrt [4]{{b}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(d*x+c))^(5/2),x)

[Out]

2/3*b*(b*tan(d*x+c))^(3/2)/d-1/4/d*b^3/(b^2)^(1/4)*2^(1/2)*ln((b*tan(d*x+c)-(b^2)^(1/4)*(b*tan(d*x+c))^(1/2)*2
^(1/2)+(b^2)^(1/2))/(b*tan(d*x+c)+(b^2)^(1/4)*(b*tan(d*x+c))^(1/2)*2^(1/2)+(b^2)^(1/2)))-1/2/d*b^3/(b^2)^(1/4)
*2^(1/2)*arctan(2^(1/2)/(b^2)^(1/4)*(b*tan(d*x+c))^(1/2)+1)+1/2/d*b^3/(b^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(b^2
)^(1/4)*(b*tan(d*x+c))^(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((b*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.71795, 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((b*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(d*x+c))**(5/2),x)

[Out]

Integral((b*tan(c + d*x))**(5/2), x)

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out