3.11 \(\int (b \tan (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=210 \[ \frac{b^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b \tan (c+d x)}}{\sqrt{b}}\right )}{\sqrt{2} d}-\frac{b^{3/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b \tan (c+d x)}}{\sqrt{b}}+1\right )}{\sqrt{2} d}+\frac{b^{3/2} \log \left (\sqrt{b} \tan (c+d x)-\sqrt{2} \sqrt{b \tan (c+d x)}+\sqrt{b}\right )}{2 \sqrt{2} d}-\frac{b^{3/2} \log \left (\sqrt{b} \tan (c+d x)+\sqrt{2} \sqrt{b \tan (c+d x)}+\sqrt{b}\right )}{2 \sqrt{2} d}+\frac{2 b \sqrt{b \tan (c+d x)}}{d} \]

[Out]

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

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Rubi [A]  time = 0.151164, antiderivative size = 210, 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, 211, 1165, 628, 1162, 617, 204} \[ \frac{b^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b \tan (c+d x)}}{\sqrt{b}}\right )}{\sqrt{2} d}-\frac{b^{3/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b \tan (c+d x)}}{\sqrt{b}}+1\right )}{\sqrt{2} d}+\frac{b^{3/2} \log \left (\sqrt{b} \tan (c+d x)-\sqrt{2} \sqrt{b \tan (c+d x)}+\sqrt{b}\right )}{2 \sqrt{2} d}-\frac{b^{3/2} \log \left (\sqrt{b} \tan (c+d x)+\sqrt{2} \sqrt{b \tan (c+d x)}+\sqrt{b}\right )}{2 \sqrt{2} d}+\frac{2 b \sqrt{b \tan (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[c + d*x])^(3/2),x]

[Out]

(b^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[b*Tan[c + d*x]])/Sqrt[b]])/(Sqrt[2]*d) - (b^(3/2)*ArcTan[1 + (Sqrt[2]*Sqrt[b
*Tan[c + d*x]])/Sqrt[b]])/(Sqrt[2]*d) + (b^(3/2)*Log[Sqrt[b] + Sqrt[b]*Tan[c + d*x] - Sqrt[2]*Sqrt[b*Tan[c + d
*x]]])/(2*Sqrt[2]*d) - (b^(3/2)*Log[Sqrt[b] + Sqrt[b]*Tan[c + d*x] + Sqrt[2]*Sqrt[b*Tan[c + d*x]]])/(2*Sqrt[2]
*d) + (2*b*Sqrt[b*Tan[c + d*x]])/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 211

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

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]

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])

Rubi steps

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

Mathematica [A]  time = 0.159505, size = 159, normalized size = 0.76 \[ \frac{(b \tan (c+d x))^{3/2} \left (2 \sqrt{2} \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )-2 \sqrt{2} \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )+8 \sqrt{\tan (c+d x)}+\sqrt{2} \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )-\sqrt{2} \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )\right )}{4 d \tan ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[c + d*x])^(3/2),x]

[Out]

((2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] + Sqrt[2
]*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*
x]] + 8*Sqrt[Tan[c + d*x]])*(b*Tan[c + d*x])^(3/2))/(4*d*Tan[c + d*x]^(3/2))

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Maple [A]  time = 0.015, size = 176, normalized size = 0.8 \begin{align*} 2\,{\frac{b\sqrt{b\tan \left ( dx+c \right ) }}{d}}-{\frac{b\sqrt{2}}{4\,d}\sqrt [4]{{b}^{2}}\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{b\sqrt{2}}{2\,d}\sqrt [4]{{b}^{2}}\arctan \left ({\sqrt{2}\sqrt{b\tan \left ( dx+c \right ) }{\frac{1}{\sqrt [4]{{b}^{2}}}}}+1 \right ) }+{\frac{b\sqrt{2}}{2\,d}\sqrt [4]{{b}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{b\tan \left ( dx+c \right ) }{\frac{1}{\sqrt [4]{{b}^{2}}}}}+1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(d*x+c))^(3/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.696, size = 1328, normalized size = 6.32 \begin{align*} \frac{4 \, \sqrt{2} \left (\frac{b^{6}}{d^{4}}\right )^{\frac{1}{4}} d \arctan \left (-\frac{b^{6} + \sqrt{2} \left (\frac{b^{6}}{d^{4}}\right )^{\frac{3}{4}} b d^{3} \sqrt{\frac{b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}} - \sqrt{2} \left (\frac{b^{6}}{d^{4}}\right )^{\frac{3}{4}} d^{3} \sqrt{\frac{\sqrt{2} \left (\frac{b^{6}}{d^{4}}\right )^{\frac{1}{4}} b d \sqrt{\frac{b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) + b^{3} \sin \left (d x + c\right ) + \sqrt{\frac{b^{6}}{d^{4}}} d^{2} \cos \left (d x + c\right )}{\cos \left (d x + c\right )}}}{b^{6}}\right ) + 4 \, \sqrt{2} \left (\frac{b^{6}}{d^{4}}\right )^{\frac{1}{4}} d \arctan \left (\frac{b^{6} - \sqrt{2} \left (\frac{b^{6}}{d^{4}}\right )^{\frac{3}{4}} b d^{3} \sqrt{\frac{b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}} + \sqrt{2} \left (\frac{b^{6}}{d^{4}}\right )^{\frac{3}{4}} d^{3} \sqrt{-\frac{\sqrt{2} \left (\frac{b^{6}}{d^{4}}\right )^{\frac{1}{4}} b d \sqrt{\frac{b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - b^{3} \sin \left (d x + c\right ) - \sqrt{\frac{b^{6}}{d^{4}}} d^{2} \cos \left (d x + c\right )}{\cos \left (d x + c\right )}}}{b^{6}}\right ) - \sqrt{2} \left (\frac{b^{6}}{d^{4}}\right )^{\frac{1}{4}} d \log \left (\frac{\sqrt{2} \left (\frac{b^{6}}{d^{4}}\right )^{\frac{1}{4}} b d \sqrt{\frac{b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) + b^{3} \sin \left (d x + c\right ) + \sqrt{\frac{b^{6}}{d^{4}}} d^{2} \cos \left (d x + c\right )}{\cos \left (d x + c\right )}\right ) + \sqrt{2} \left (\frac{b^{6}}{d^{4}}\right )^{\frac{1}{4}} d \log \left (-\frac{\sqrt{2} \left (\frac{b^{6}}{d^{4}}\right )^{\frac{1}{4}} b d \sqrt{\frac{b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - b^{3} \sin \left (d x + c\right ) - \sqrt{\frac{b^{6}}{d^{4}}} d^{2} \cos \left (d x + c\right )}{\cos \left (d x + c\right )}\right ) + 8 \, b \sqrt{\frac{b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/4*(4*sqrt(2)*(b^6/d^4)^(1/4)*d*arctan(-(b^6 + sqrt(2)*(b^6/d^4)^(3/4)*b*d^3*sqrt(b*sin(d*x + c)/cos(d*x + c)
) - sqrt(2)*(b^6/d^4)^(3/4)*d^3*sqrt((sqrt(2)*(b^6/d^4)^(1/4)*b*d*sqrt(b*sin(d*x + c)/cos(d*x + c))*cos(d*x +
c) + b^3*sin(d*x + c) + sqrt(b^6/d^4)*d^2*cos(d*x + c))/cos(d*x + c)))/b^6) + 4*sqrt(2)*(b^6/d^4)^(1/4)*d*arct
an((b^6 - sqrt(2)*(b^6/d^4)^(3/4)*b*d^3*sqrt(b*sin(d*x + c)/cos(d*x + c)) + sqrt(2)*(b^6/d^4)^(3/4)*d^3*sqrt(-
(sqrt(2)*(b^6/d^4)^(1/4)*b*d*sqrt(b*sin(d*x + c)/cos(d*x + c))*cos(d*x + c) - b^3*sin(d*x + c) - sqrt(b^6/d^4)
*d^2*cos(d*x + c))/cos(d*x + c)))/b^6) - sqrt(2)*(b^6/d^4)^(1/4)*d*log((sqrt(2)*(b^6/d^4)^(1/4)*b*d*sqrt(b*sin
(d*x + c)/cos(d*x + c))*cos(d*x + c) + b^3*sin(d*x + c) + sqrt(b^6/d^4)*d^2*cos(d*x + c))/cos(d*x + c)) + sqrt
(2)*(b^6/d^4)^(1/4)*d*log(-(sqrt(2)*(b^6/d^4)^(1/4)*b*d*sqrt(b*sin(d*x + c)/cos(d*x + c))*cos(d*x + c) - b^3*s
in(d*x + c) - sqrt(b^6/d^4)*d^2*cos(d*x + c))/cos(d*x + c)) + 8*b*sqrt(b*sin(d*x + c)/cos(d*x + c)))/d

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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{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(d*x+c))**(3/2),x)

[Out]

Integral((b*tan(c + d*x))**(3/2), 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))^(3/2),x, algorithm="giac")

[Out]

Timed out