3.4.51 \(\int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^3 \, dx\) [351]

Optimal. Leaf size=160 \[ \frac {2 \sqrt {2} a^3 d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{f}-\frac {4 a^3 d \sqrt {d \tan (e+f x)}}{f}+\frac {4 a^3 (d \tan (e+f x))^{3/2}}{3 f}+\frac {32 a^3 (d \tan (e+f x))^{5/2}}{35 d f}+\frac {2 (d \tan (e+f x))^{5/2} \left (a^3+a^3 \tan (e+f x)\right )}{7 d f} \]

[Out]

2*a^3*d^(3/2)*arctanh(1/2*(d^(1/2)+d^(1/2)*tan(f*x+e))*2^(1/2)/(d*tan(f*x+e))^(1/2))*2^(1/2)/f-4*a^3*d*(d*tan(
f*x+e))^(1/2)/f+4/3*a^3*(d*tan(f*x+e))^(3/2)/f+32/35*a^3*(d*tan(f*x+e))^(5/2)/d/f+2/7*(d*tan(f*x+e))^(5/2)*(a^
3+a^3*tan(f*x+e))/d/f

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Rubi [A]
time = 0.17, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3647, 3711, 3609, 3613, 214} \begin {gather*} \frac {2 \sqrt {2} a^3 d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \tan (e+f x)+\sqrt {d}}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{f}+\frac {32 a^3 (d \tan (e+f x))^{5/2}}{35 d f}+\frac {4 a^3 (d \tan (e+f x))^{3/2}}{3 f}-\frac {4 a^3 d \sqrt {d \tan (e+f x)}}{f}+\frac {2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{5/2}}{7 d f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[2]*a^3*d^(3/2)*ArcTanh[(Sqrt[d] + Sqrt[d]*Tan[e + f*x])/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/f - (4*a^3*d*
Sqrt[d*Tan[e + f*x]])/f + (4*a^3*(d*Tan[e + f*x])^(3/2))/(3*f) + (32*a^3*(d*Tan[e + f*x])^(5/2))/(35*d*f) + (2
*(d*Tan[e + f*x])^(5/2)*(a^3 + a^3*Tan[e + f*x]))/(7*d*f)

Rule 214

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

Rule 3609

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

Rule 3613

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2*(d^2/f),
Subst[Int[1/(2*c*d + b*x^2), x], x, (c - d*Tan[e + f*x])/Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x
] && EqQ[c^2 - d^2, 0]

Rule 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3711

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])
^m*Simp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

Rubi steps

\begin {align*} \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^3 \, dx &=\frac {2 (d \tan (e+f x))^{5/2} \left (a^3+a^3 \tan (e+f x)\right )}{7 d f}+\frac {2 \int (d \tan (e+f x))^{3/2} \left (a^3 d+7 a^3 d \tan (e+f x)+8 a^3 d \tan ^2(e+f x)\right ) \, dx}{7 d}\\ &=\frac {32 a^3 (d \tan (e+f x))^{5/2}}{35 d f}+\frac {2 (d \tan (e+f x))^{5/2} \left (a^3+a^3 \tan (e+f x)\right )}{7 d f}+\frac {2 \int (d \tan (e+f x))^{3/2} \left (-7 a^3 d+7 a^3 d \tan (e+f x)\right ) \, dx}{7 d}\\ &=\frac {4 a^3 (d \tan (e+f x))^{3/2}}{3 f}+\frac {32 a^3 (d \tan (e+f x))^{5/2}}{35 d f}+\frac {2 (d \tan (e+f x))^{5/2} \left (a^3+a^3 \tan (e+f x)\right )}{7 d f}+\frac {2 \int \sqrt {d \tan (e+f x)} \left (-7 a^3 d^2-7 a^3 d^2 \tan (e+f x)\right ) \, dx}{7 d}\\ &=-\frac {4 a^3 d \sqrt {d \tan (e+f x)}}{f}+\frac {4 a^3 (d \tan (e+f x))^{3/2}}{3 f}+\frac {32 a^3 (d \tan (e+f x))^{5/2}}{35 d f}+\frac {2 (d \tan (e+f x))^{5/2} \left (a^3+a^3 \tan (e+f x)\right )}{7 d f}+\frac {2 \int \frac {7 a^3 d^3-7 a^3 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{7 d}\\ &=-\frac {4 a^3 d \sqrt {d \tan (e+f x)}}{f}+\frac {4 a^3 (d \tan (e+f x))^{3/2}}{3 f}+\frac {32 a^3 (d \tan (e+f x))^{5/2}}{35 d f}+\frac {2 (d \tan (e+f x))^{5/2} \left (a^3+a^3 \tan (e+f x)\right )}{7 d f}-\frac {\left (28 a^6 d^5\right ) \text {Subst}\left (\int \frac {1}{-98 a^6 d^6+d x^2} \, dx,x,\frac {7 a^3 d^3+7 a^3 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}\right )}{f}\\ &=\frac {2 \sqrt {2} a^3 d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{f}-\frac {4 a^3 d \sqrt {d \tan (e+f x)}}{f}+\frac {4 a^3 (d \tan (e+f x))^{3/2}}{3 f}+\frac {32 a^3 (d \tan (e+f x))^{5/2}}{35 d f}+\frac {2 (d \tan (e+f x))^{5/2} \left (a^3+a^3 \tan (e+f x)\right )}{7 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 = 2.78, size = 332, normalized size = 2.08 \begin {gather*} -\frac {a^3 \cos (e+f x) (d \tan (e+f x))^{3/2} (1+\tan (e+f x))^3 \left (210 \sqrt {2} \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right ) \cos ^2(e+f x)-210 \sqrt {2} \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (e+f x)}\right ) \cos ^2(e+f x)+105 \sqrt {2} \cos ^2(e+f x) \log \left (1-\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right )-105 \sqrt {2} \cos ^2(e+f x) \log \left (1+\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right )+840 \cos ^2(e+f x) \sqrt {\tan (e+f x)}-280 \cos ^2(e+f x) \tan ^{\frac {3}{2}}(e+f x)+280 \cos ^2(e+f x) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\tan ^2(e+f x)\right ) \tan ^{\frac {3}{2}}(e+f x)-60 \sin ^2(e+f x) \tan ^{\frac {3}{2}}(e+f x)-126 \sin (2 (e+f x)) \tan ^{\frac {3}{2}}(e+f x)\right )}{210 f (\cos (e+f x)+\sin (e+f x))^3 \tan ^{\frac {3}{2}}(e+f x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/210*(a^3*Cos[e + f*x]*(d*Tan[e + f*x])^(3/2)*(1 + Tan[e + f*x])^3*(210*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[
e + f*x]]]*Cos[e + f*x]^2 - 210*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]]*Cos[e + f*x]^2 + 105*Sqrt[2]*Co
s[e + f*x]^2*Log[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]] - 105*Sqrt[2]*Cos[e + f*x]^2*Log[1 + Sqrt[2]*S
qrt[Tan[e + f*x]] + Tan[e + f*x]] + 840*Cos[e + f*x]^2*Sqrt[Tan[e + f*x]] - 280*Cos[e + f*x]^2*Tan[e + f*x]^(3
/2) + 280*Cos[e + f*x]^2*Hypergeometric2F1[3/4, 1, 7/4, -Tan[e + f*x]^2]*Tan[e + f*x]^(3/2) - 60*Sin[e + f*x]^
2*Tan[e + f*x]^(3/2) - 126*Sin[2*(e + f*x)]*Tan[e + f*x]^(3/2)))/(f*(Cos[e + f*x] + Sin[e + f*x])^3*Tan[e + f*
x]^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(338\) vs. \(2(135)=270\).
time = 0.36, size = 339, normalized size = 2.12

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {3 d \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 d^{3} \sqrt {d \tan \left (f x +e \right )}+2 d^{4} \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f \,d^{2}}\)	\(339\)
default	\(\frac {2 a^{3} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {3 d \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 d^{3} \sqrt {d \tan \left (f x +e \right )}+2 d^{4} \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f \,d^{2}}\)	\(339\)

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))^3,x,method=_RETURNVERBOSE)

[Out]

2/f*a^3/d^2*(1/7*(d*tan(f*x+e))^(7/2)+3/5*d*(d*tan(f*x+e))^(5/2)+2/3*d^2*(d*tan(f*x+e))^(3/2)-2*d^3*(d*tan(f*x
+e))^(1/2)+2*d^4*(1/8/d*(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))-1/8/(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 = 171, normalized size = 1.07 \begin {gather*} \frac {105 \, a^{3} d^{3} {\left (\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 )} + \frac {2 \, {\left (15 \, \left (d \tan \left (f x + e\right )\right )^{\frac {7}{2}} a^{3} + 63 \, \left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a^{3} d + 70 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{3} d^{2} - 210 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{3}\right )}}{d}}{105 \, 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))^3,x, algorithm="maxima")

[Out]

1/105*(105*a^3*d^3*(sqrt(2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d) - sqrt(2)*l
og(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d)) + 2*(15*(d*tan(f*x + e))^(7/2)*a^3 + 63
*(d*tan(f*x + e))^(5/2)*a^3*d + 70*(d*tan(f*x + e))^(3/2)*a^3*d^2 - 210*sqrt(d*tan(f*x + e))*a^3*d^3)/d)/(d*f)

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Fricas [A]
time = 1.35, size = 273, normalized size = 1.71 \begin {gather*} \left [\frac {105 \, \sqrt {2} a^{3} d^{\frac {3}{2}} \log \left (\frac {d \tan \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} {\left (\tan \left (f x + e\right ) + 1\right )} + 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, {\left (15 \, a^{3} d \tan \left (f x + e\right )^{3} + 63 \, a^{3} d \tan \left (f x + e\right )^{2} + 70 \, a^{3} d \tan \left (f x + e\right ) - 210 \, a^{3} d\right )} \sqrt {d \tan \left (f x + e\right )}}{105 \, f}, -\frac {2 \, {\left (105 \, \sqrt {2} a^{3} \sqrt {-d} d \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) + 1\right )}}{2 \, d \tan \left (f x + e\right )}\right ) - {\left (15 \, a^{3} d \tan \left (f x + e\right )^{3} + 63 \, a^{3} d \tan \left (f x + e\right )^{2} + 70 \, a^{3} d \tan \left (f x + e\right ) - 210 \, a^{3} d\right )} \sqrt {d \tan \left (f x + e\right )}\right )}}{105 \, f}\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))^3,x, algorithm="fricas")

[Out]

[1/105*(105*sqrt(2)*a^3*d^(3/2)*log((d*tan(f*x + e)^2 + 2*sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d)*(tan(f*x + e) +
1) + 4*d*tan(f*x + e) + d)/(tan(f*x + e)^2 + 1)) + 2*(15*a^3*d*tan(f*x + e)^3 + 63*a^3*d*tan(f*x + e)^2 + 70*
a^3*d*tan(f*x + e) - 210*a^3*d)*sqrt(d*tan(f*x + e)))/f, -2/105*(105*sqrt(2)*a^3*sqrt(-d)*d*arctan(1/2*sqrt(2)
*sqrt(d*tan(f*x + e))*sqrt(-d)*(tan(f*x + e) + 1)/(d*tan(f*x + e))) - (15*a^3*d*tan(f*x + e)^3 + 63*a^3*d*tan(
f*x + e)^2 + 70*a^3*d*tan(f*x + e) - 210*a^3*d)*sqrt(d*tan(f*x + e)))/f]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (142) = 284\).
time = 0.86, size = 379, normalized size = 2.37 \begin {gather*} \frac {1}{210} \, d {\left (\frac {105 \, \sqrt {2} {\left (a^{3} d \sqrt {{\left | d \right |}} + a^{3} {\left | d \right |}^{\frac {3}{2}}\right )} \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 {105 \, \sqrt {2} {\left (a^{3} d \sqrt {{\left | d \right |}} + a^{3} {\left | d \right |}^{\frac {3}{2}}\right )} \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 {210 \, {\left (\sqrt {2} a^{3} d \sqrt {{\left | d \right |}} - \sqrt {2} a^{3} {\left | d \right |}^{\frac {3}{2}}\right )} \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 {210 \, {\left (\sqrt {2} a^{3} d \sqrt {{\left | d \right |}} - \sqrt {2} a^{3} {\left | d \right |}^{\frac {3}{2}}\right )} \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 {4 \, {\left (15 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{21} f^{6} \tan \left (f x + e\right )^{3} + 63 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{21} f^{6} \tan \left (f x + e\right )^{2} + 70 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{21} f^{6} \tan \left (f x + e\right ) - 210 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{21} f^{6}\right )}}{d^{21} f^{7}}\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))^3,x, algorithm="giac")

[Out]

1/210*d*(105*sqrt(2)*(a^3*d*sqrt(abs(d)) + a^3*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) - 105*sqrt(2)*(a^3*d*sqrt(abs(d)) + a^3*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) + 210*(sqrt(2)*a^3*d*sqrt(abs(d)) - sqrt(2)*a^3*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) + 210*(sqrt(2)*a^3
*d*sqrt(abs(d)) - sqrt(2)*a^3*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) + 4*(15*sqrt(d*tan(f*x + e))*a^3*d^21*f^6*tan(f*x + e)^3 + 63*sqrt(d*tan(f*x + e))*a^3*d
^21*f^6*tan(f*x + e)^2 + 70*sqrt(d*tan(f*x + e))*a^3*d^21*f^6*tan(f*x + e) - 210*sqrt(d*tan(f*x + e))*a^3*d^21
*f^6)/(d^21*f^7))

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Mupad [B]
time = 5.15, size = 143, normalized size = 0.89 \begin {gather*} \frac {4\,a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3\,f}+\frac {6\,a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{5\,d\,f}+\frac {2\,a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{7/2}}{7\,d^2\,f}-\frac {4\,a^3\,d\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{f}-\frac {\sqrt {2}\,a^3\,d^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,a^6\,d^{9/2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,32{}\mathrm {i}}{32\,a^6\,d^5+32\,a^6\,d^5\,\mathrm {tan}\left (e+f\,x\right )}\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))^3,x)

[Out]

(4*a^3*(d*tan(e + f*x))^(3/2))/(3*f) + (6*a^3*(d*tan(e + f*x))^(5/2))/(5*d*f) + (2*a^3*(d*tan(e + f*x))^(7/2))
/(7*d^2*f) - (4*a^3*d*(d*tan(e + f*x))^(1/2))/f - (2^(1/2)*a^3*d^(3/2)*atan((2^(1/2)*a^6*d^(9/2)*(d*tan(e + f*
x))^(1/2)*32i)/(32*a^6*d^5 + 32*a^6*d^5*tan(e + f*x)))*2i)/f

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