3.1.31 \(\int (b \tan ^3(c+d x))^{3/2} \, dx\) [31]

Optimal. Leaf size=286 \[ -\frac {2 b \sqrt {b \tan ^3(c+d x)}}{3 d}-\frac {b \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right ) \sqrt {b \tan ^3(c+d x)}}{\sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)}+\frac {b \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right ) \sqrt {b \tan ^3(c+d x)}}{\sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)}+\frac {b \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right ) \sqrt {b \tan ^3(c+d x)}}{2 \sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)}-\frac {b \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right ) \sqrt {b \tan ^3(c+d x)}}{2 \sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 b \tan ^2(c+d x) \sqrt {b \tan ^3(c+d x)}}{7 d} \]

[Out]

-2/3*b*(b*tan(d*x+c)^3)^(1/2)/d+1/2*b*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*(b*tan(d*x+c)^3)^(1/2)/d*2^(1/2)/tan
(d*x+c)^(3/2)+1/2*b*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*(b*tan(d*x+c)^3)^(1/2)/d*2^(1/2)/tan(d*x+c)^(3/2)+1/4*b
*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))*(b*tan(d*x+c)^3)^(1/2)/d*2^(1/2)/tan(d*x+c)^(3/2)-1/4*b*ln(1+2^(1/2
)*tan(d*x+c)^(1/2)+tan(d*x+c))*(b*tan(d*x+c)^3)^(1/2)/d*2^(1/2)/tan(d*x+c)^(3/2)+2/7*b*(b*tan(d*x+c)^3)^(1/2)*
tan(d*x+c)^2/d

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Rubi [A]
time = 0.09, antiderivative size = 286, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3739, 3554, 3557, 335, 303, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {b \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right ) \sqrt {b \tan ^3(c+d x)}}{\sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)}+\frac {b \text {ArcTan}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right ) \sqrt {b \tan ^3(c+d x)}}{\sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b \sqrt {b \tan ^3(c+d x)}}{3 d}+\frac {2 b \tan ^2(c+d x) \sqrt {b \tan ^3(c+d x)}}{7 d}+\frac {b \sqrt {b \tan ^3(c+d x)} \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)}-\frac {b \sqrt {b \tan ^3(c+d x)} \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Tan[c + d*x]^3)^(3/2),x]

[Out]

(-2*b*Sqrt[b*Tan[c + d*x]^3])/(3*d) - (b*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]*Sqrt[b*Tan[c + d*x]^3])/(Sqrt[
2]*d*Tan[c + d*x]^(3/2)) + (b*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]*Sqrt[b*Tan[c + d*x]^3])/(Sqrt[2]*d*Tan[c
+ d*x]^(3/2)) + (b*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]*Sqrt[b*Tan[c + d*x]^3])/(2*Sqrt[2]*d*Tan
[c + d*x]^(3/2)) - (b*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]*Sqrt[b*Tan[c + d*x]^3])/(2*Sqrt[2]*d*
Tan[c + d*x]^(3/2)) + (2*b*Tan[c + d*x]^2*Sqrt[b*Tan[c + d*x]^3])/(7*d)

Rule 210

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

Rule 303

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 335

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 631

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 642

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 1176

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 1179

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 3554

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 3557

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 3739

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p])), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rubi steps

\begin {align*} \int \left (b \tan ^3(c+d x)\right )^{3/2} \, dx &=\frac {\left (b \sqrt {b \tan ^3(c+d x)}\right ) \int \tan ^{\frac {9}{2}}(c+d x) \, dx}{\tan ^{\frac {3}{2}}(c+d x)}\\ &=\frac {2 b \tan ^2(c+d x) \sqrt {b \tan ^3(c+d x)}}{7 d}-\frac {\left (b \sqrt {b \tan ^3(c+d x)}\right ) \int \tan ^{\frac {5}{2}}(c+d x) \, dx}{\tan ^{\frac {3}{2}}(c+d x)}\\ &=-\frac {2 b \sqrt {b \tan ^3(c+d x)}}{3 d}+\frac {2 b \tan ^2(c+d x) \sqrt {b \tan ^3(c+d x)}}{7 d}+\frac {\left (b \sqrt {b \tan ^3(c+d x)}\right ) \int \sqrt {\tan (c+d x)} \, dx}{\tan ^{\frac {3}{2}}(c+d x)}\\ &=-\frac {2 b \sqrt {b \tan ^3(c+d x)}}{3 d}+\frac {2 b \tan ^2(c+d x) \sqrt {b \tan ^3(c+d x)}}{7 d}+\frac {\left (b \sqrt {b \tan ^3(c+d x)}\right ) \text {Subst}\left (\int \frac {\sqrt {x}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d \tan ^{\frac {3}{2}}(c+d x)}\\ &=-\frac {2 b \sqrt {b \tan ^3(c+d x)}}{3 d}+\frac {2 b \tan ^2(c+d x) \sqrt {b \tan ^3(c+d x)}}{7 d}+\frac {\left (2 b \sqrt {b \tan ^3(c+d x)}\right ) \text {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d \tan ^{\frac {3}{2}}(c+d x)}\\ &=-\frac {2 b \sqrt {b \tan ^3(c+d x)}}{3 d}+\frac {2 b \tan ^2(c+d x) \sqrt {b \tan ^3(c+d x)}}{7 d}-\frac {\left (b \sqrt {b \tan ^3(c+d x)}\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d \tan ^{\frac {3}{2}}(c+d x)}+\frac {\left (b \sqrt {b \tan ^3(c+d x)}\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d \tan ^{\frac {3}{2}}(c+d x)}\\ &=-\frac {2 b \sqrt {b \tan ^3(c+d x)}}{3 d}+\frac {2 b \tan ^2(c+d x) \sqrt {b \tan ^3(c+d x)}}{7 d}+\frac {\left (b \sqrt {b \tan ^3(c+d x)}\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {\left (b \sqrt {b \tan ^3(c+d x)}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {\left (b \sqrt {b \tan ^3(c+d x)}\right ) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)}+\frac {\left (b \sqrt {b \tan ^3(c+d x)}\right ) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)}\\ &=-\frac {2 b \sqrt {b \tan ^3(c+d x)}}{3 d}+\frac {b \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right ) \sqrt {b \tan ^3(c+d x)}}{2 \sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)}-\frac {b \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right ) \sqrt {b \tan ^3(c+d x)}}{2 \sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 b \tan ^2(c+d x) \sqrt {b \tan ^3(c+d x)}}{7 d}+\frac {\left (b \sqrt {b \tan ^3(c+d x)}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)}-\frac {\left (b \sqrt {b \tan ^3(c+d x)}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)}\\ &=-\frac {2 b \sqrt {b \tan ^3(c+d x)}}{3 d}-\frac {b \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right ) \sqrt {b \tan ^3(c+d x)}}{\sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)}+\frac {b \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right ) \sqrt {b \tan ^3(c+d x)}}{\sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)}+\frac {b \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right ) \sqrt {b \tan ^3(c+d x)}}{2 \sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)}-\frac {b \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right ) \sqrt {b \tan ^3(c+d x)}}{2 \sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 b \tan ^2(c+d x) \sqrt {b \tan ^3(c+d x)}}{7 d}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.07, size = 54, normalized size = 0.19 \begin {gather*} \frac {2 b \sqrt {b \tan ^3(c+d x)} \left (-7+7 \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\tan ^2(c+d x)\right )+3 \tan ^2(c+d x)\right )}{21 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[c + d*x]^3)^(3/2),x]

[Out]

(2*b*Sqrt[b*Tan[c + d*x]^3]*(-7 + 7*Hypergeometric2F1[3/4, 1, 7/4, -Tan[c + d*x]^2] + 3*Tan[c + d*x]^2))/(21*d
)

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Maple [A]
time = 0.06, size = 236, normalized size = 0.83

method	result	size
derivativedivides	\(\frac {\left (b \left (\tan ^{3}\left (d x +c \right )\right )\right )^{\frac {3}{2}} \left (24 \left (b \tan \left (d x +c \right )\right )^{\frac {7}{2}} \left (b^{2}\right )^{\frac {1}{4}}+21 b^{4} \sqrt {2}\, \ln \left (-\frac {\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}-b \tan \left (d x +c \right )-\sqrt {b^{2}}}{b \tan \left (d x +c \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+42 b^{4} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}+\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )+42 b^{4} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}-\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )-56 b^{2} \left (b \tan \left (d x +c \right )\right )^{\frac {3}{2}} \left (b^{2}\right )^{\frac {1}{4}}\right )}{84 d \tan \left (d x +c \right )^{3} \left (b \tan \left (d x +c \right )\right )^{\frac {3}{2}} b^{2} \left (b^{2}\right )^{\frac {1}{4}}}\)	\(236\)
default	\(\frac {\left (b \left (\tan ^{3}\left (d x +c \right )\right )\right )^{\frac {3}{2}} \left (24 \left (b \tan \left (d x +c \right )\right )^{\frac {7}{2}} \left (b^{2}\right )^{\frac {1}{4}}+21 b^{4} \sqrt {2}\, \ln \left (-\frac {\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}-b \tan \left (d x +c \right )-\sqrt {b^{2}}}{b \tan \left (d x +c \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+42 b^{4} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}+\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )+42 b^{4} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}-\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )-56 b^{2} \left (b \tan \left (d x +c \right )\right )^{\frac {3}{2}} \left (b^{2}\right )^{\frac {1}{4}}\right )}{84 d \tan \left (d x +c \right )^{3} \left (b \tan \left (d x +c \right )\right )^{\frac {3}{2}} b^{2} \left (b^{2}\right )^{\frac {1}{4}}}\)	\(236\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/84/d*(b*tan(d*x+c)^3)^(3/2)*(24*(b*tan(d*x+c))^(7/2)*(b^2)^(1/4)+21*b^4*2^(1/2)*ln(-((b^2)^(1/4)*(b*tan(d*x+
c))^(1/2)*2^(1/2)-b*tan(d*x+c)-(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)
))+42*b^4*2^(1/2)*arctan((2^(1/2)*(b*tan(d*x+c))^(1/2)+(b^2)^(1/4))/(b^2)^(1/4))+42*b^4*2^(1/2)*arctan((2^(1/2
)*(b*tan(d*x+c))^(1/2)-(b^2)^(1/4))/(b^2)^(1/4))-56*b^2*(b*tan(d*x+c))^(3/2)*(b^2)^(1/4))/tan(d*x+c)^3/(b*tan(
d*x+c))^(3/2)/b^2/(b^2)^(1/4)

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Maxima [A]
time = 0.52, size = 140, normalized size = 0.49 \begin {gather*} \frac {24 \, b^{\frac {3}{2}} \tan \left (d x + c\right )^{\frac {7}{2}} - 56 \, b^{\frac {3}{2}} \tan \left (d x + c\right )^{\frac {3}{2}} + 21 \, {\left (2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} b^{\frac {3}{2}}}{84 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/84*(24*b^(3/2)*tan(d*x + c)^(7/2) - 56*b^(3/2)*tan(d*x + c)^(3/2) + 21*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2
) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - sqrt(2)*log(sqr
t(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + sqrt(2)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))*b^(
3/2))/d

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.53, size = 253, normalized size = 0.88 \begin {gather*} \frac {1}{84} \, b {\left (\frac {42 \, \sqrt {2} {\left | b \right |}^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | b \right |}} + 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {{\left | b \right |}}}\right )}{b d} + \frac {42 \, \sqrt {2} {\left | b \right |}^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | b \right |}} - 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {{\left | b \right |}}}\right )}{b d} - \frac {21 \, \sqrt {2} {\left | b \right |}^{\frac {3}{2}} \log \left (b \tan \left (d x + c\right ) + \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {{\left | b \right |}} + {\left | b \right |}\right )}{b d} + \frac {21 \, \sqrt {2} {\left | b \right |}^{\frac {3}{2}} \log \left (b \tan \left (d x + c\right ) - \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {{\left | b \right |}} + {\left | b \right |}\right )}{b d} + \frac {8 \, {\left (3 \, \sqrt {b \tan \left (d x + c\right )} b^{21} d^{6} \tan \left (d x + c\right )^{3} - 7 \, \sqrt {b \tan \left (d x + c\right )} b^{21} d^{6} \tan \left (d x + c\right )\right )}}{b^{21} d^{7}}\right )} \mathrm {sgn}\left (\tan \left (d x + c\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/84*b*(42*sqrt(2)*abs(b)^(3/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(b)) + 2*sqrt(b*tan(d*x + c)))/sqrt(abs(b)
))/(b*d) + 42*sqrt(2)*abs(b)^(3/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(b)) - 2*sqrt(b*tan(d*x + c)))/sqrt(ab
s(b)))/(b*d) - 21*sqrt(2)*abs(b)^(3/2)*log(b*tan(d*x + c) + sqrt(2)*sqrt(b*tan(d*x + c))*sqrt(abs(b)) + abs(b)
)/(b*d) + 21*sqrt(2)*abs(b)^(3/2)*log(b*tan(d*x + c) - sqrt(2)*sqrt(b*tan(d*x + c))*sqrt(abs(b)) + abs(b))/(b*
d) + 8*(3*sqrt(b*tan(d*x + c))*b^21*d^6*tan(d*x + c)^3 - 7*sqrt(b*tan(d*x + c))*b^21*d^6*tan(d*x + c))/(b^21*d
^7))*sgn(tan(d*x + c))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^3\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(c + d*x)^3)^(3/2),x)

[Out]

int((b*tan(c + d*x)^3)^(3/2), x)

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