3.350 \(\int (d \tan (e+f x))^{5/2} (a+a \tan (e+f x))^3 \, dx\)

Optimal. Leaf size=186 \[ -\frac{2 \sqrt{2} a^3 d^{5/2} \tan ^{-1}\left (\frac{\sqrt{d}-\sqrt{d} \tan (e+f x)}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{f}-\frac{4 a^3 d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{7/2}}{9 d f}+\frac{40 a^3 (d \tan (e+f x))^{7/2}}{63 d f}+\frac{4 a^3 (d \tan (e+f x))^{5/2}}{5 f}-\frac{4 a^3 d (d \tan (e+f x))^{3/2}}{3 f} \]

[Out]

(-2*Sqrt[2]*a^3*d^(5/2)*ArcTan[(Sqrt[d] - Sqrt[d]*Tan[e + f*x])/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/f - (4*a^3*d^
2*Sqrt[d*Tan[e + f*x]])/f - (4*a^3*d*(d*Tan[e + f*x])^(3/2))/(3*f) + (4*a^3*(d*Tan[e + f*x])^(5/2))/(5*f) + (4
0*a^3*(d*Tan[e + f*x])^(7/2))/(63*d*f) + (2*(d*Tan[e + f*x])^(7/2)*(a^3 + a^3*Tan[e + f*x]))/(9*d*f)

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Rubi [A]  time = 0.278976, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3566, 3630, 3528, 3532, 205} \[ -\frac{2 \sqrt{2} a^3 d^{5/2} \tan ^{-1}\left (\frac{\sqrt{d}-\sqrt{d} \tan (e+f x)}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{f}-\frac{4 a^3 d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{7/2}}{9 d f}+\frac{40 a^3 (d \tan (e+f x))^{7/2}}{63 d f}+\frac{4 a^3 (d \tan (e+f x))^{5/2}}{5 f}-\frac{4 a^3 d (d \tan (e+f x))^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^(5/2)*(a + a*Tan[e + f*x])^3,x]

[Out]

(-2*Sqrt[2]*a^3*d^(5/2)*ArcTan[(Sqrt[d] - Sqrt[d]*Tan[e + f*x])/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/f - (4*a^3*d^
2*Sqrt[d*Tan[e + f*x]])/f - (4*a^3*d*(d*Tan[e + f*x])^(3/2))/(3*f) + (4*a^3*(d*Tan[e + f*x])^(5/2))/(5*f) + (4
0*a^3*(d*Tan[e + f*x])^(7/2))/(63*d*f) + (2*(d*Tan[e + f*x])^(7/2)*(a^3 + a^3*Tan[e + f*x]))/(9*d*f)

Rule 3566

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 3630

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]

Rule 3528

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 3532

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 205

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

Rubi steps

\begin{align*} \int (d \tan (e+f x))^{5/2} (a+a \tan (e+f x))^3 \, dx &=\frac{2 (d \tan (e+f x))^{7/2} \left (a^3+a^3 \tan (e+f x)\right )}{9 d f}+\frac{2 \int (d \tan (e+f x))^{5/2} \left (a^3 d+9 a^3 d \tan (e+f x)+10 a^3 d \tan ^2(e+f x)\right ) \, dx}{9 d}\\ &=\frac{40 a^3 (d \tan (e+f x))^{7/2}}{63 d f}+\frac{2 (d \tan (e+f x))^{7/2} \left (a^3+a^3 \tan (e+f x)\right )}{9 d f}+\frac{2 \int (d \tan (e+f x))^{5/2} \left (-9 a^3 d+9 a^3 d \tan (e+f x)\right ) \, dx}{9 d}\\ &=\frac{4 a^3 (d \tan (e+f x))^{5/2}}{5 f}+\frac{40 a^3 (d \tan (e+f x))^{7/2}}{63 d f}+\frac{2 (d \tan (e+f x))^{7/2} \left (a^3+a^3 \tan (e+f x)\right )}{9 d f}+\frac{2 \int (d \tan (e+f x))^{3/2} \left (-9 a^3 d^2-9 a^3 d^2 \tan (e+f x)\right ) \, dx}{9 d}\\ &=-\frac{4 a^3 d (d \tan (e+f x))^{3/2}}{3 f}+\frac{4 a^3 (d \tan (e+f x))^{5/2}}{5 f}+\frac{40 a^3 (d \tan (e+f x))^{7/2}}{63 d f}+\frac{2 (d \tan (e+f x))^{7/2} \left (a^3+a^3 \tan (e+f x)\right )}{9 d f}+\frac{2 \int \sqrt{d \tan (e+f x)} \left (9 a^3 d^3-9 a^3 d^3 \tan (e+f x)\right ) \, dx}{9 d}\\ &=-\frac{4 a^3 d^2 \sqrt{d \tan (e+f x)}}{f}-\frac{4 a^3 d (d \tan (e+f x))^{3/2}}{3 f}+\frac{4 a^3 (d \tan (e+f x))^{5/2}}{5 f}+\frac{40 a^3 (d \tan (e+f x))^{7/2}}{63 d f}+\frac{2 (d \tan (e+f x))^{7/2} \left (a^3+a^3 \tan (e+f x)\right )}{9 d f}+\frac{2 \int \frac{9 a^3 d^4+9 a^3 d^4 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{9 d}\\ &=-\frac{4 a^3 d^2 \sqrt{d \tan (e+f x)}}{f}-\frac{4 a^3 d (d \tan (e+f x))^{3/2}}{3 f}+\frac{4 a^3 (d \tan (e+f x))^{5/2}}{5 f}+\frac{40 a^3 (d \tan (e+f x))^{7/2}}{63 d f}+\frac{2 (d \tan (e+f x))^{7/2} \left (a^3+a^3 \tan (e+f x)\right )}{9 d f}-\frac{\left (36 a^6 d^7\right ) \operatorname{Subst}\left (\int \frac{1}{162 a^6 d^8+d x^2} \, dx,x,\frac{9 a^3 d^4-9 a^3 d^4 \tan (e+f x)}{\sqrt{d \tan (e+f x)}}\right )}{f}\\ &=-\frac{2 \sqrt{2} a^3 d^{5/2} \tan ^{-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^2 \sqrt{d \tan (e+f x)}}{f}-\frac{4 a^3 d (d \tan (e+f x))^{3/2}}{3 f}+\frac{4 a^3 (d \tan (e+f x))^{5/2}}{5 f}+\frac{40 a^3 (d \tan (e+f x))^{7/2}}{63 d f}+\frac{2 (d \tan (e+f x))^{7/2} \left (a^3+a^3 \tan (e+f x)\right )}{9 d f}\\ \end{align*}

Mathematica [C]  time = 6.10809, size = 729, normalized size = 3.92 \[ \frac{4 \cos ^3(e+f x) \cot (e+f x) (a \tan (e+f x)+a)^3 (d \tan (e+f x))^{5/2} \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};-\tan ^2(e+f x)\right )}{3 f (\sin (e+f x)+\cos (e+f x))^3}-\frac{\sqrt{2} \cos ^3(e+f x) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (e+f x)}\right ) (a \tan (e+f x)+a)^3 (d \tan (e+f x))^{5/2}}{f \tan ^{\frac{5}{2}}(e+f x) (\sin (e+f x)+\cos (e+f x))^3}+\frac{\sqrt{2} \cos ^3(e+f x) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (e+f x)}+1\right ) (a \tan (e+f x)+a)^3 (d \tan (e+f x))^{5/2}}{f \tan ^{\frac{5}{2}}(e+f x) (\sin (e+f x)+\cos (e+f x))^3}+\frac{4 \cos ^3(e+f x) (a \tan (e+f x)+a)^3 (d \tan (e+f x))^{5/2}}{5 f (\sin (e+f x)+\cos (e+f x))^3}+\frac{6 \sin (e+f x) \cos ^2(e+f x) (a \tan (e+f x)+a)^3 (d \tan (e+f x))^{5/2}}{7 f (\sin (e+f x)+\cos (e+f x))^3}+\frac{2 \sin ^2(e+f x) \cos (e+f x) (a \tan (e+f x)+a)^3 (d \tan (e+f x))^{5/2}}{9 f (\sin (e+f x)+\cos (e+f x))^3}-\frac{\cos ^3(e+f x) (a \tan (e+f x)+a)^3 (d \tan (e+f x))^{5/2} \log \left (\tan (e+f x)-\sqrt{2} \sqrt{\tan (e+f x)}+1\right )}{\sqrt{2} f \tan ^{\frac{5}{2}}(e+f x) (\sin (e+f x)+\cos (e+f x))^3}+\frac{\cos ^3(e+f x) (a \tan (e+f x)+a)^3 (d \tan (e+f x))^{5/2} \log \left (\tan (e+f x)+\sqrt{2} \sqrt{\tan (e+f x)}+1\right )}{\sqrt{2} f \tan ^{\frac{5}{2}}(e+f x) (\sin (e+f x)+\cos (e+f x))^3}-\frac{4 \cos ^3(e+f x) \cot ^2(e+f x) (a \tan (e+f x)+a)^3 (d \tan (e+f x))^{5/2}}{f (\sin (e+f x)+\cos (e+f x))^3}-\frac{4 \cos ^3(e+f x) \cot (e+f x) (a \tan (e+f x)+a)^3 (d \tan (e+f x))^{5/2}}{3 f (\sin (e+f x)+\cos (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Tan[e + f*x])^(5/2)*(a + a*Tan[e + f*x])^3,x]

[Out]

(4*Cos[e + f*x]^3*(d*Tan[e + f*x])^(5/2)*(a + a*Tan[e + f*x])^3)/(5*f*(Cos[e + f*x] + Sin[e + f*x])^3) - (4*Co
s[e + f*x]^3*Cot[e + f*x]*(d*Tan[e + f*x])^(5/2)*(a + a*Tan[e + f*x])^3)/(3*f*(Cos[e + f*x] + Sin[e + f*x])^3)
- (4*Cos[e + f*x]^3*Cot[e + f*x]^2*(d*Tan[e + f*x])^(5/2)*(a + a*Tan[e + f*x])^3)/(f*(Cos[e + f*x] + Sin[e +
f*x])^3) + (4*Cos[e + f*x]^3*Cot[e + f*x]*Hypergeometric2F1[3/4, 1, 7/4, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(5/
2)*(a + a*Tan[e + f*x])^3)/(3*f*(Cos[e + f*x] + Sin[e + f*x])^3) + (6*Cos[e + f*x]^2*Sin[e + f*x]*(d*Tan[e + f
*x])^(5/2)*(a + a*Tan[e + f*x])^3)/(7*f*(Cos[e + f*x] + Sin[e + f*x])^3) + (2*Cos[e + f*x]*Sin[e + f*x]^2*(d*T
an[e + f*x])^(5/2)*(a + a*Tan[e + f*x])^3)/(9*f*(Cos[e + f*x] + Sin[e + f*x])^3) - (Sqrt[2]*ArcTan[1 - Sqrt[2]
*Sqrt[Tan[e + f*x]]]*Cos[e + f*x]^3*(d*Tan[e + f*x])^(5/2)*(a + a*Tan[e + f*x])^3)/(f*(Cos[e + f*x] + Sin[e +
f*x])^3*Tan[e + f*x]^(5/2)) + (Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]]*Cos[e + f*x]^3*(d*Tan[e + f*x])^
(5/2)*(a + a*Tan[e + f*x])^3)/(f*(Cos[e + f*x] + Sin[e + f*x])^3*Tan[e + f*x]^(5/2)) - (Cos[e + f*x]^3*Log[1 -
Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]]*(d*Tan[e + f*x])^(5/2)*(a + a*Tan[e + f*x])^3)/(Sqrt[2]*f*(Cos[e +
f*x] + Sin[e + f*x])^3*Tan[e + f*x]^(5/2)) + (Cos[e + f*x]^3*Log[1 + Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x
]]*(d*Tan[e + f*x])^(5/2)*(a + a*Tan[e + f*x])^3)/(Sqrt[2]*f*(Cos[e + f*x] + Sin[e + f*x])^3*Tan[e + f*x]^(5/2
))

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Maple [B]  time = 0.019, size = 446, normalized size = 2.4 \begin{align*}{\frac{2\,{a}^{3}}{9\,f{d}^{2}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{9}{2}}}}+{\frac{6\,{a}^{3}}{7\,df} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{7}{2}}}}+{\frac{4\,{a}^{3}}{5\,f} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{4\,{a}^{3}d}{3\,f} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-4\,{\frac{{a}^{3}{d}^{2}\sqrt{d\tan \left ( fx+e \right ) }}{f}}+{\frac{{a}^{3}{d}^{2}\sqrt{2}}{2\,f}\sqrt [4]{{d}^{2}}\ln \left ({ \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ) }+{\frac{{a}^{3}{d}^{2}\sqrt{2}}{f}\sqrt [4]{{d}^{2}}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }-{\frac{{a}^{3}{d}^{2}\sqrt{2}}{f}\sqrt [4]{{d}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }+{\frac{{a}^{3}{d}^{3}\sqrt{2}}{2\,f}\ln \left ({ \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}+{\frac{{a}^{3}{d}^{3}\sqrt{2}}{f}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}-{\frac{{a}^{3}{d}^{3}\sqrt{2}}{f}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^(5/2)*(a+a*tan(f*x+e))^3,x)

[Out]

2/9/f*a^3/d^2*(d*tan(f*x+e))^(9/2)+6/7*a^3*(d*tan(f*x+e))^(7/2)/d/f+4/5*a^3*(d*tan(f*x+e))^(5/2)/f-4/3*a^3*d*(
d*tan(f*x+e))^(3/2)/f-4*a^3*d^2*(d*tan(f*x+e))^(1/2)/f+1/2/f*a^3*d^2*(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)))+1/f*a^3*d^2*(d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-1/f*a^3*d^2*(d^2)^
(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)+1/2/f*a^3*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)))+1/f*a^3*d^3/(d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-1/f*a^3
*d^3/(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(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((d*tan(f*x+e))^(5/2)*(a+a*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.71175, size = 801, normalized size = 4.31 \begin{align*} \left [\frac{315 \, \sqrt{2} a^{3} \sqrt{-d} d^{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 (35 \, a^{3} d^{2} \tan \left (f x + e\right )^{4} + 135 \, a^{3} d^{2} \tan \left (f x + e\right )^{3} + 126 \, a^{3} d^{2} \tan \left (f x + e\right )^{2} - 210 \, a^{3} d^{2} \tan \left (f x + e\right ) - 630 \, a^{3} d^{2}\right )} \sqrt{d \tan \left (f x + e\right )}}{315 \, f}, \frac{2 \,{\left (315 \, \sqrt{2} a^{3} d^{\frac{5}{2}} \arctan \left (\frac{\sqrt{2} \sqrt{d \tan \left (f x + e\right )}{\left (\tan \left (f x + e\right ) - 1\right )}}{2 \, \sqrt{d} \tan \left (f x + e\right )}\right ) +{\left (35 \, a^{3} d^{2} \tan \left (f x + e\right )^{4} + 135 \, a^{3} d^{2} \tan \left (f x + e\right )^{3} + 126 \, a^{3} d^{2} \tan \left (f x + e\right )^{2} - 210 \, a^{3} d^{2} \tan \left (f x + e\right ) - 630 \, a^{3} d^{2}\right )} \sqrt{d \tan \left (f x + e\right )}\right )}}{315 \, f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(5/2)*(a+a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

[1/315*(315*sqrt(2)*a^3*sqrt(-d)*d^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*(35*a^3*d^2*tan(f*x + e)^4 + 135*a^3*d^2*tan(f*x +
e)^3 + 126*a^3*d^2*tan(f*x + e)^2 - 210*a^3*d^2*tan(f*x + e) - 630*a^3*d^2)*sqrt(d*tan(f*x + e)))/f, 2/315*(3
15*sqrt(2)*a^3*d^(5/2)*arctan(1/2*sqrt(2)*sqrt(d*tan(f*x + e))*(tan(f*x + e) - 1)/(sqrt(d)*tan(f*x + e))) + (3
5*a^3*d^2*tan(f*x + e)^4 + 135*a^3*d^2*tan(f*x + e)^3 + 126*a^3*d^2*tan(f*x + e)^2 - 210*a^3*d^2*tan(f*x + e)
- 630*a^3*d^2)*sqrt(d*tan(f*x + e)))/f]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int \left (d \tan{\left (e + f x \right )}\right )^{\frac{5}{2}}\, dx + \int 3 \left (d \tan{\left (e + f x \right )}\right )^{\frac{5}{2}} \tan{\left (e + f x \right )}\, dx + \int 3 \left (d \tan{\left (e + f x \right )}\right )^{\frac{5}{2}} \tan ^{2}{\left (e + f x \right )}\, dx + \int \left (d \tan{\left (e + f x \right )}\right )^{\frac{5}{2}} \tan ^{3}{\left (e + f x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**(5/2)*(a+a*tan(f*x+e))**3,x)

[Out]

a**3*(Integral((d*tan(e + f*x))**(5/2), x) + Integral(3*(d*tan(e + f*x))**(5/2)*tan(e + f*x), x) + Integral(3*
(d*tan(e + f*x))**(5/2)*tan(e + f*x)**2, x) + Integral((d*tan(e + f*x))**(5/2)*tan(e + f*x)**3, x))

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Giac [B]  time = 1.49181, size = 547, normalized size = 2.94 \begin{align*} \frac{\sqrt{2}{\left (a^{3} d^{2} \sqrt{{\left | d \right |}} - a^{3} d{\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 )}{2 \, f} - \frac{\sqrt{2}{\left (a^{3} d^{2} \sqrt{{\left | d \right |}} - a^{3} d{\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 )}{2 \, f} + \frac{{\left (\sqrt{2} a^{3} d^{2} \sqrt{{\left | d \right |}} + \sqrt{2} a^{3} d{\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 )}{f} + \frac{{\left (\sqrt{2} a^{3} d^{2} \sqrt{{\left | d \right |}} + \sqrt{2} a^{3} d{\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 )}{f} + \frac{2 \,{\left (35 \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{20} f^{8} \tan \left (f x + e\right )^{4} + 135 \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{20} f^{8} \tan \left (f x + e\right )^{3} + 126 \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{20} f^{8} \tan \left (f x + e\right )^{2} - 210 \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{20} f^{8} \tan \left (f x + e\right ) - 630 \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{20} f^{8}\right )}}{315 \, d^{18} f^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(5/2)*(a+a*tan(f*x+e))^3,x, algorithm="giac")

[Out]

1/2*sqrt(2)*(a^3*d^2*sqrt(abs(d)) - a^3*d*abs(d)^(3/2))*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt
(abs(d)) + abs(d))/f - 1/2*sqrt(2)*(a^3*d^2*sqrt(abs(d)) - a^3*d*abs(d)^(3/2))*log(d*tan(f*x + e) - sqrt(2)*sq
rt(d*tan(f*x + e))*sqrt(abs(d)) + abs(d))/f + (sqrt(2)*a^3*d^2*sqrt(abs(d)) + sqrt(2)*a^3*d*abs(d)^(3/2))*arct
an(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/f + (sqrt(2)*a^3*d^2*sqrt(abs(d))
+ sqrt(2)*a^3*d*abs(d)^(3/2))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)
))/f + 2/315*(35*sqrt(d*tan(f*x + e))*a^3*d^20*f^8*tan(f*x + e)^4 + 135*sqrt(d*tan(f*x + e))*a^3*d^20*f^8*tan(
f*x + e)^3 + 126*sqrt(d*tan(f*x + e))*a^3*d^20*f^8*tan(f*x + e)^2 - 210*sqrt(d*tan(f*x + e))*a^3*d^20*f^8*tan(
f*x + e) - 630*sqrt(d*tan(f*x + e))*a^3*d^20*f^8)/(d^18*f^9)