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

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

[Out]

1/2*d^(3/2)*arctan(1-2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2))/b*2^(1/2)-1/2*d^(3/2)*arctan(1+2^(1/2)*(d*tan(b*x+a
))^(1/2)/d^(1/2))/b*2^(1/2)+1/4*d^(3/2)*ln(d^(1/2)-2^(1/2)*(d*tan(b*x+a))^(1/2)+d^(1/2)*tan(b*x+a))/b*2^(1/2)-
1/4*d^(3/2)*ln(d^(1/2)+2^(1/2)*(d*tan(b*x+a))^(1/2)+d^(1/2)*tan(b*x+a))/b*2^(1/2)+2*d*(d*tan(b*x+a))^(1/2)/b

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3473, 3476, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac {d^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{\sqrt {2} b}-\frac {d^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{\sqrt {2} b}+\frac {d^{3/2} \log \left (\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{2 \sqrt {2} b}-\frac {d^{3/2} \log \left (\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{2 \sqrt {2} b}+\frac {2 d \sqrt {d \tan (a+b x)}}{b} \]

Antiderivative was successfully verified.

[In]

Int[(d*Tan[a + b*x])^(3/2),x]

[Out]

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

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 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 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 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 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 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 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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.27, size = 159, normalized size = 0.76 \[ \frac {(d \tan (a+b x))^{3/2} \left (2 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right )-2 \sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (a+b x)}+1\right )+8 \sqrt {\tan (a+b x)}+\sqrt {2} \log \left (\tan (a+b x)-\sqrt {2} \sqrt {\tan (a+b x)}+1\right )-\sqrt {2} \log \left (\tan (a+b x)+\sqrt {2} \sqrt {\tan (a+b x)}+1\right )\right )}{4 b \tan ^{\frac {3}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Tan[a + b*x])^(3/2),x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.83, size = 533, normalized size = 2.54 \[ \frac {4 \, \sqrt {2} \left (\frac {d^{6}}{b^{4}}\right )^{\frac {1}{4}} b \arctan \left (-\frac {d^{6} + \sqrt {2} \left (\frac {d^{6}}{b^{4}}\right )^{\frac {3}{4}} b^{3} d \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} - \sqrt {2} \left (\frac {d^{6}}{b^{4}}\right )^{\frac {3}{4}} b^{3} \sqrt {\frac {\sqrt {2} \left (\frac {d^{6}}{b^{4}}\right )^{\frac {1}{4}} b d \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} \cos \left (b x + a\right ) + d^{3} \sin \left (b x + a\right ) + \sqrt {\frac {d^{6}}{b^{4}}} b^{2} \cos \left (b x + a\right )}{\cos \left (b x + a\right )}}}{d^{6}}\right ) + 4 \, \sqrt {2} \left (\frac {d^{6}}{b^{4}}\right )^{\frac {1}{4}} b \arctan \left (\frac {d^{6} - \sqrt {2} \left (\frac {d^{6}}{b^{4}}\right )^{\frac {3}{4}} b^{3} d \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} + \sqrt {2} \left (\frac {d^{6}}{b^{4}}\right )^{\frac {3}{4}} b^{3} \sqrt {-\frac {\sqrt {2} \left (\frac {d^{6}}{b^{4}}\right )^{\frac {1}{4}} b d \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} \cos \left (b x + a\right ) - d^{3} \sin \left (b x + a\right ) - \sqrt {\frac {d^{6}}{b^{4}}} b^{2} \cos \left (b x + a\right )}{\cos \left (b x + a\right )}}}{d^{6}}\right ) - \sqrt {2} \left (\frac {d^{6}}{b^{4}}\right )^{\frac {1}{4}} b \log \left (\frac {\sqrt {2} \left (\frac {d^{6}}{b^{4}}\right )^{\frac {1}{4}} b d \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} \cos \left (b x + a\right ) + d^{3} \sin \left (b x + a\right ) + \sqrt {\frac {d^{6}}{b^{4}}} b^{2} \cos \left (b x + a\right )}{\cos \left (b x + a\right )}\right ) + \sqrt {2} \left (\frac {d^{6}}{b^{4}}\right )^{\frac {1}{4}} b \log \left (-\frac {\sqrt {2} \left (\frac {d^{6}}{b^{4}}\right )^{\frac {1}{4}} b d \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} \cos \left (b x + a\right ) - d^{3} \sin \left (b x + a\right ) - \sqrt {\frac {d^{6}}{b^{4}}} b^{2} \cos \left (b x + a\right )}{\cos \left (b x + a\right )}\right ) + 8 \, d \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

maple [A]  time = 0.09, size = 176, normalized size = 0.84 \[ \frac {2 d \sqrt {d \tan \left (b x +a \right )}}{b}+\frac {d \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (b x +a \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 b}-\frac {d \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (b x +a \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (b x +a \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (b x +a \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (b x +a \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{4 b}-\frac {d \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (b x +a \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*d*(d*tan(b*x+a))^(1/2)/b+1/2/b*d*(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(b*x+a))^(1/2)+1)-1/4
/b*d*(d^2)^(1/4)*2^(1/2)*ln((d*tan(b*x+a)+(d^2)^(1/4)*(d*tan(b*x+a))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(b*x+a)-
(d^2)^(1/4)*(d*tan(b*x+a))^(1/2)*2^(1/2)+(d^2)^(1/2)))-1/2/b*d*(d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2)^(1/4)*
(d*tan(b*x+a))^(1/2)+1)

________________________________________________________________________________________

maxima [A]  time = 0.84, size = 170, normalized size = 0.81 \[ -\frac {2 \, \sqrt {2} d^{\frac {5}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + 2 \, \sqrt {2} d^{\frac {5}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + \sqrt {2} d^{\frac {5}{2}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) - \sqrt {2} d^{\frac {5}{2}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) - 8 \, \sqrt {d \tan \left (b x + a\right )} d^{2}}{4 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*sqrt(2)*d^(5/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(b*x + a)))/sqrt(d)) + 2*sqrt(2)*d^(
5/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(b*x + a)))/sqrt(d)) + sqrt(2)*d^(5/2)*log(d*tan(b*x +
a) + sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(d) + d) - sqrt(2)*d^(5/2)*log(d*tan(b*x + a) - sqrt(2)*sqrt(d*tan(b*x
+ a))*sqrt(d) + d) - 8*sqrt(d*tan(b*x + a))*d^2)/(b*d)

________________________________________________________________________________________

mupad [B]  time = 2.65, size = 73, normalized size = 0.35 \[ \frac {2\,d\,\sqrt {d\,\mathrm {tan}\left (a+b\,x\right )}}{b}+\frac {{\left (-1\right )}^{1/4}\,d^{3/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (a+b\,x\right )}}{\sqrt {d}}\right )\,1{}\mathrm {i}}{b}+\frac {{\left (-1\right )}^{1/4}\,d^{3/2}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (a+b\,x\right )}}{\sqrt {d}}\right )\,1{}\mathrm {i}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(a + b*x))^(3/2),x)

[Out]

(2*d*(d*tan(a + b*x))^(1/2))/b + ((-1)^(1/4)*d^(3/2)*atan(((-1)^(1/4)*(d*tan(a + b*x))^(1/2))/d^(1/2))*1i)/b +
((-1)^(1/4)*d^(3/2)*atanh(((-1)^(1/4)*(d*tan(a + b*x))^(1/2))/d^(1/2))*1i)/b

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d*tan(a + b*x))**(3/2), x)

________________________________________________________________________________________