3.4.44 \(\int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2 \, dx\) [344]

Optimal. Leaf size=246 \[ \frac {\sqrt {2} a^2 d^{3/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {\sqrt {2} a^2 d^{3/2} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {a^2 d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} f}+\frac {a^2 d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} f}+\frac {4 a^2 (d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f} \]

[Out]

-1/2*a^2*d^(3/2)*ln(d^(1/2)-2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/f*2^(1/2)+1/2*a^2*d^(3/2)*ln(d^(1
/2)+2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/f*2^(1/2)+a^2*d^(3/2)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/
2)/d^(1/2))*2^(1/2)/f-a^2*d^(3/2)*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*2^(1/2)/f+4/3*a^2*(d*tan(f*x+
e))^(3/2)/f+2/5*a^2*(d*tan(f*x+e))^(5/2)/d/f

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Rubi [A]
time = 0.16, antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3624, 12, 16, 3554, 3557, 335, 303, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {\sqrt {2} a^2 d^{3/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {\sqrt {2} a^2 d^{3/2} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{f}-\frac {a^2 d^{3/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} f}+\frac {a^2 d^{3/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} f}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}+\frac {4 a^2 (d \tan (e+f x))^{3/2}}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^(3/2)*(a + a*Tan[e + f*x])^2,x]

[Out]

(Sqrt[2]*a^2*d^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/f - (Sqrt[2]*a^2*d^(3/2)*ArcTan[1 + (
Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/f - (a^2*d^(3/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*T
an[e + f*x]]])/(Sqrt[2]*f) + (a^2*d^(3/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/
(Sqrt[2]*f) + (4*a^2*(d*Tan[e + f*x])^(3/2))/(3*f) + (2*a^2*(d*Tan[e + f*x])^(5/2))/(5*d*f)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 3624

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
d^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e
+ f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[m, -1] &&  !(EqQ[m, 2] && EqQ
[a, 0])

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.40, size = 52, normalized size = 0.21 \begin {gather*} \frac {2 a^2 (d \tan (e+f x))^{3/2} \left (10-10 \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\tan ^2(e+f x)\right )+3 \tan (e+f x)\right )}{15 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d*Tan[e + f*x])^(3/2)*(a + a*Tan[e + f*x])^2,x]

[Out]

(2*a^2*(d*Tan[e + f*x])^(3/2)*(10 - 10*Hypergeometric2F1[3/4, 1, 7/4, -Tan[e + f*x]^2] + 3*Tan[e + f*x]))/(15*
f)

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Maple [A]
time = 0.26, size = 172, normalized size = 0.70

method	result	size
derivativedivides	\(\frac {2 a^{2} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {d^{3} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f d}\)	\(172\)
default	\(\frac {2 a^{2} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {d^{3} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f d}\)	\(172\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^(3/2)*(a+a*tan(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

2/f*a^2/d*(1/5*(d*tan(f*x+e))^(5/2)+2/3*d*(d*tan(f*x+e))^(3/2)-1/4*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)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^
(1/2)+1)))

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Maxima [A]
time = 0.52, size = 203, normalized size = 0.83 \begin {gather*} -\frac {15 \, a^{2} d^{3} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} - 12 \, \left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a^{2} - 40 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{2} d}{30 \, d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(15*a^2*d^3*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) +
2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) - sqrt(2)*log(d*tan(
f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d) + sqrt(2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan
(f*x + e))*sqrt(d) + d)/sqrt(d)) - 12*(d*tan(f*x + e))^(5/2)*a^2 - 40*(d*tan(f*x + e))^(3/2)*a^2*d)/(d*f)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 774 vs. \(2 (203) = 406\).
time = 1.18, size = 774, normalized size = 3.15 \begin {gather*} \frac {60 \, \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \arctan \left (-\frac {a^{8} d^{6} + \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {1}{4}} a^{6} d^{4} f \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} - \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \sqrt {\frac {a^{12} d^{9} \sin \left (f x + e\right ) + \sqrt {\frac {a^{8} d^{6}}{f^{4}}} a^{8} d^{6} f^{2} \cos \left (f x + e\right ) + \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {3}{4}} a^{6} d^{4} f^{3} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\cos \left (f x + e\right )}}}{a^{8} d^{6}}\right ) \cos \left (f x + e\right )^{2} + 60 \, \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \arctan \left (\frac {a^{8} d^{6} - \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {1}{4}} a^{6} d^{4} f \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} + \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \sqrt {\frac {a^{12} d^{9} \sin \left (f x + e\right ) + \sqrt {\frac {a^{8} d^{6}}{f^{4}}} a^{8} d^{6} f^{2} \cos \left (f x + e\right ) - \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {3}{4}} a^{6} d^{4} f^{3} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\cos \left (f x + e\right )}}}{a^{8} d^{6}}\right ) \cos \left (f x + e\right )^{2} + 15 \, \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \cos \left (f x + e\right )^{2} \log \left (\frac {a^{12} d^{9} \sin \left (f x + e\right ) + \sqrt {\frac {a^{8} d^{6}}{f^{4}}} a^{8} d^{6} f^{2} \cos \left (f x + e\right ) + \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {3}{4}} a^{6} d^{4} f^{3} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\cos \left (f x + e\right )}\right ) - 15 \, \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \cos \left (f x + e\right )^{2} \log \left (\frac {a^{12} d^{9} \sin \left (f x + e\right ) + \sqrt {\frac {a^{8} d^{6}}{f^{4}}} a^{8} d^{6} f^{2} \cos \left (f x + e\right ) - \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {3}{4}} a^{6} d^{4} f^{3} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\cos \left (f x + e\right )}\right ) - 4 \, {\left (3 \, a^{2} d \cos \left (f x + e\right )^{2} - 10 \, a^{2} d \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 3 \, a^{2} d\right )} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{30 \, f \cos \left (f x + e\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(60*sqrt(2)*(a^8*d^6/f^4)^(1/4)*f*arctan(-(a^8*d^6 + sqrt(2)*(a^8*d^6/f^4)^(1/4)*a^6*d^4*f*sqrt(d*sin(f*x
+ e)/cos(f*x + e)) - sqrt(2)*(a^8*d^6/f^4)^(1/4)*f*sqrt((a^12*d^9*sin(f*x + e) + sqrt(a^8*d^6/f^4)*a^8*d^6*f^
2*cos(f*x + e) + sqrt(2)*(a^8*d^6/f^4)^(3/4)*a^6*d^4*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*cos(f*x + e))/cos(f
*x + e)))/(a^8*d^6))*cos(f*x + e)^2 + 60*sqrt(2)*(a^8*d^6/f^4)^(1/4)*f*arctan((a^8*d^6 - sqrt(2)*(a^8*d^6/f^4)
^(1/4)*a^6*d^4*f*sqrt(d*sin(f*x + e)/cos(f*x + e)) + sqrt(2)*(a^8*d^6/f^4)^(1/4)*f*sqrt((a^12*d^9*sin(f*x + e)
+ sqrt(a^8*d^6/f^4)*a^8*d^6*f^2*cos(f*x + e) - sqrt(2)*(a^8*d^6/f^4)^(3/4)*a^6*d^4*f^3*sqrt(d*sin(f*x + e)/co
s(f*x + e))*cos(f*x + e))/cos(f*x + e)))/(a^8*d^6))*cos(f*x + e)^2 + 15*sqrt(2)*(a^8*d^6/f^4)^(1/4)*f*cos(f*x
+ e)^2*log((a^12*d^9*sin(f*x + e) + sqrt(a^8*d^6/f^4)*a^8*d^6*f^2*cos(f*x + e) + sqrt(2)*(a^8*d^6/f^4)^(3/4)*a
^6*d^4*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*cos(f*x + e))/cos(f*x + e)) - 15*sqrt(2)*(a^8*d^6/f^4)^(1/4)*f*co
s(f*x + e)^2*log((a^12*d^9*sin(f*x + e) + sqrt(a^8*d^6/f^4)*a^8*d^6*f^2*cos(f*x + e) - sqrt(2)*(a^8*d^6/f^4)^(
3/4)*a^6*d^4*f^3*sqrt(d*sin(f*x + e)/cos(f*x + e))*cos(f*x + e))/cos(f*x + e)) - 4*(3*a^2*d*cos(f*x + e)^2 - 1
0*a^2*d*cos(f*x + e)*sin(f*x + e) - 3*a^2*d)*sqrt(d*sin(f*x + e)/cos(f*x + e)))/(f*cos(f*x + e)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral((d*tan(e + f*x))**(3/2), x) + Integral(2*(d*tan(e + f*x))**(3/2)*tan(e + f*x), x) + Integral((d
*tan(e + f*x))**(3/2)*tan(e + f*x)**2, x))

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Giac [A]
time = 0.68, size = 274, normalized size = 1.11 \begin {gather*} -\frac {1}{30} \, {\left (\frac {30 \, \sqrt {2} a^{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 )}{d f} + \frac {30 \, \sqrt {2} a^{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 )}{d f} - \frac {15 \, \sqrt {2} a^{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 )}{d f} + \frac {15 \, \sqrt {2} a^{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 )}{d f} - \frac {4 \, {\left (3 \, \sqrt {d \tan \left (f x + e\right )} a^{2} d^{10} f^{4} \tan \left (f x + e\right )^{2} + 10 \, \sqrt {d \tan \left (f x + e\right )} a^{2} d^{10} f^{4} \tan \left (f x + e\right )\right )}}{d^{10} f^{5}}\right )} d \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(30*sqrt(2)*a^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)))/(d*f) + 30*sqrt(2)*a^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)))/(d*f) - 15*sqrt(2)*a^2*abs(d)^(3/2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d
)) + abs(d))/(d*f) + 15*sqrt(2)*a^2*abs(d)^(3/2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)
) + abs(d))/(d*f) - 4*(3*sqrt(d*tan(f*x + e))*a^2*d^10*f^4*tan(f*x + e)^2 + 10*sqrt(d*tan(f*x + e))*a^2*d^10*f
^4*tan(f*x + e))/(d^10*f^5))*d

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Mupad [B]
time = 4.64, size = 104, normalized size = 0.42 \begin {gather*} \frac {4\,a^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3\,f}+\frac {2\,a^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{5\,d\,f}-\frac {2\,{\left (-1\right )}^{1/4}\,a^2\,d^{3/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{f}-\frac {{\left (-1\right )}^{1/4}\,a^2\,d^{3/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{\sqrt {d}}\right )\,2{}\mathrm {i}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*a^2*(d*tan(e + f*x))^(3/2))/(3*f) + (2*a^2*(d*tan(e + f*x))^(5/2))/(5*d*f) - (2*(-1)^(1/4)*a^2*d^(3/2)*atan
(((-1)^(1/4)*(d*tan(e + f*x))^(1/2))/d^(1/2)))/f - ((-1)^(1/4)*a^2*d^(3/2)*atan(((-1)^(1/4)*(d*tan(e + f*x))^(
1/2)*1i)/d^(1/2))*2i)/f

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