3.349 \(\int (d \tan (e+f x))^{7/2} (a+a \tan (e+f x))^3 \, dx\)

Optimal. Leaf size=210 \[ -\frac{4 a^3 d^2 (d \tan (e+f x))^{3/2}}{3 f}+\frac{4 a^3 d^3 \sqrt{d \tan (e+f x)}}{f}-\frac{2 \sqrt{2} a^3 d^{7/2} \tanh ^{-1}\left (\frac{\sqrt{d} \tan (e+f x)+\sqrt{d}}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{f}+\frac{2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{9/2}}{11 d f}+\frac{16 a^3 (d \tan (e+f x))^{9/2}}{33 d f}+\frac{4 a^3 (d \tan (e+f x))^{7/2}}{7 f}-\frac{4 a^3 d (d \tan (e+f x))^{5/2}}{5 f} \]

[Out]

(-2*Sqrt[2]*a^3*d^(7/2)*ArcTanh[(Sqrt[d] + Sqrt[d]*Tan[e + f*x])/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/f + (4*a^3*d
^3*Sqrt[d*Tan[e + f*x]])/f - (4*a^3*d^2*(d*Tan[e + f*x])^(3/2))/(3*f) - (4*a^3*d*(d*Tan[e + f*x])^(5/2))/(5*f)
+ (4*a^3*(d*Tan[e + f*x])^(7/2))/(7*f) + (16*a^3*(d*Tan[e + f*x])^(9/2))/(33*d*f) + (2*(d*Tan[e + f*x])^(9/2)
*(a^3 + a^3*Tan[e + f*x]))/(11*d*f)

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Rubi [A]  time = 0.323781, antiderivative size = 210, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3566, 3630, 3528, 3532, 208} \[ -\frac{4 a^3 d^2 (d \tan (e+f x))^{3/2}}{3 f}+\frac{4 a^3 d^3 \sqrt{d \tan (e+f x)}}{f}-\frac{2 \sqrt{2} a^3 d^{7/2} \tanh ^{-1}\left (\frac{\sqrt{d} \tan (e+f x)+\sqrt{d}}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{f}+\frac{2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{9/2}}{11 d f}+\frac{16 a^3 (d \tan (e+f x))^{9/2}}{33 d f}+\frac{4 a^3 (d \tan (e+f x))^{7/2}}{7 f}-\frac{4 a^3 d (d \tan (e+f x))^{5/2}}{5 f} \]

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^(7/2)*(a + a*Tan[e + f*x])^3,x]

[Out]

(-2*Sqrt[2]*a^3*d^(7/2)*ArcTanh[(Sqrt[d] + Sqrt[d]*Tan[e + f*x])/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/f + (4*a^3*d
^3*Sqrt[d*Tan[e + f*x]])/f - (4*a^3*d^2*(d*Tan[e + f*x])^(3/2))/(3*f) - (4*a^3*d*(d*Tan[e + f*x])^(5/2))/(5*f)
+ (4*a^3*(d*Tan[e + f*x])^(7/2))/(7*f) + (16*a^3*(d*Tan[e + f*x])^(9/2))/(33*d*f) + (2*(d*Tan[e + f*x])^(9/2)
*(a^3 + a^3*Tan[e + f*x]))/(11*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 208

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

Rubi steps

\begin{align*} \int (d \tan (e+f x))^{7/2} (a+a \tan (e+f x))^3 \, dx &=\frac{2 (d \tan (e+f x))^{9/2} \left (a^3+a^3 \tan (e+f x)\right )}{11 d f}+\frac{2 \int (d \tan (e+f x))^{7/2} \left (a^3 d+11 a^3 d \tan (e+f x)+12 a^3 d \tan ^2(e+f x)\right ) \, dx}{11 d}\\ &=\frac{16 a^3 (d \tan (e+f x))^{9/2}}{33 d f}+\frac{2 (d \tan (e+f x))^{9/2} \left (a^3+a^3 \tan (e+f x)\right )}{11 d f}+\frac{2 \int (d \tan (e+f x))^{7/2} \left (-11 a^3 d+11 a^3 d \tan (e+f x)\right ) \, dx}{11 d}\\ &=\frac{4 a^3 (d \tan (e+f x))^{7/2}}{7 f}+\frac{16 a^3 (d \tan (e+f x))^{9/2}}{33 d f}+\frac{2 (d \tan (e+f x))^{9/2} \left (a^3+a^3 \tan (e+f x)\right )}{11 d f}+\frac{2 \int (d \tan (e+f x))^{5/2} \left (-11 a^3 d^2-11 a^3 d^2 \tan (e+f x)\right ) \, dx}{11 d}\\ &=-\frac{4 a^3 d (d \tan (e+f x))^{5/2}}{5 f}+\frac{4 a^3 (d \tan (e+f x))^{7/2}}{7 f}+\frac{16 a^3 (d \tan (e+f x))^{9/2}}{33 d f}+\frac{2 (d \tan (e+f x))^{9/2} \left (a^3+a^3 \tan (e+f x)\right )}{11 d f}+\frac{2 \int (d \tan (e+f x))^{3/2} \left (11 a^3 d^3-11 a^3 d^3 \tan (e+f x)\right ) \, dx}{11 d}\\ &=-\frac{4 a^3 d^2 (d \tan (e+f x))^{3/2}}{3 f}-\frac{4 a^3 d (d \tan (e+f x))^{5/2}}{5 f}+\frac{4 a^3 (d \tan (e+f x))^{7/2}}{7 f}+\frac{16 a^3 (d \tan (e+f x))^{9/2}}{33 d f}+\frac{2 (d \tan (e+f x))^{9/2} \left (a^3+a^3 \tan (e+f x)\right )}{11 d f}+\frac{2 \int \sqrt{d \tan (e+f x)} \left (11 a^3 d^4+11 a^3 d^4 \tan (e+f x)\right ) \, dx}{11 d}\\ &=\frac{4 a^3 d^3 \sqrt{d \tan (e+f x)}}{f}-\frac{4 a^3 d^2 (d \tan (e+f x))^{3/2}}{3 f}-\frac{4 a^3 d (d \tan (e+f x))^{5/2}}{5 f}+\frac{4 a^3 (d \tan (e+f x))^{7/2}}{7 f}+\frac{16 a^3 (d \tan (e+f x))^{9/2}}{33 d f}+\frac{2 (d \tan (e+f x))^{9/2} \left (a^3+a^3 \tan (e+f x)\right )}{11 d f}+\frac{2 \int \frac{-11 a^3 d^5+11 a^3 d^5 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{11 d}\\ &=\frac{4 a^3 d^3 \sqrt{d \tan (e+f x)}}{f}-\frac{4 a^3 d^2 (d \tan (e+f x))^{3/2}}{3 f}-\frac{4 a^3 d (d \tan (e+f x))^{5/2}}{5 f}+\frac{4 a^3 (d \tan (e+f x))^{7/2}}{7 f}+\frac{16 a^3 (d \tan (e+f x))^{9/2}}{33 d f}+\frac{2 (d \tan (e+f x))^{9/2} \left (a^3+a^3 \tan (e+f x)\right )}{11 d f}-\frac{\left (44 a^6 d^9\right ) \operatorname{Subst}\left (\int \frac{1}{-242 a^6 d^{10}+d x^2} \, dx,x,\frac{-11 a^3 d^5-11 a^3 d^5 \tan (e+f x)}{\sqrt{d \tan (e+f x)}}\right )}{f}\\ &=-\frac{2 \sqrt{2} a^3 d^{7/2} \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^3 \sqrt{d \tan (e+f x)}}{f}-\frac{4 a^3 d^2 (d \tan (e+f x))^{3/2}}{3 f}-\frac{4 a^3 d (d \tan (e+f x))^{5/2}}{5 f}+\frac{4 a^3 (d \tan (e+f x))^{7/2}}{7 f}+\frac{16 a^3 (d \tan (e+f x))^{9/2}}{33 d f}+\frac{2 (d \tan (e+f x))^{9/2} \left (a^3+a^3 \tan (e+f x)\right )}{11 d f}\\ \end{align*}

Mathematica [C]  time = 4.4516, size = 375, normalized size = 1.79 \[ \frac{a^3 d^3 \cos (e+f x) (\tan (e+f x)+1)^3 \sqrt{d \tan (e+f x)} \left (3080 \cos ^2(e+f x) \tan ^{\frac{3}{2}}(e+f x) \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};-\tan ^2(e+f x)\right )+420 \sin ^2(e+f x) \tan ^{\frac{7}{2}}(e+f x)+770 \sin (2 (e+f x)) \tan ^{\frac{7}{2}}(e+f x)+1320 \cos ^2(e+f x) \tan ^{\frac{7}{2}}(e+f x)-1848 \cos ^2(e+f x) \tan ^{\frac{5}{2}}(e+f x)-3080 \cos ^2(e+f x) \tan ^{\frac{3}{2}}(e+f x)+2310 \sqrt{2} \cos ^2(e+f x) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (e+f x)}\right )-2310 \sqrt{2} \cos ^2(e+f x) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (e+f x)}+1\right )+9240 \cos ^2(e+f x) \sqrt{\tan (e+f x)}+1155 \sqrt{2} \cos ^2(e+f x) \log \left (\tan (e+f x)-\sqrt{2} \sqrt{\tan (e+f x)}+1\right )-1155 \sqrt{2} \cos ^2(e+f x) \log \left (\tan (e+f x)+\sqrt{2} \sqrt{\tan (e+f x)}+1\right )\right )}{2310 f \sqrt{\tan (e+f x)} (\sin (e+f x)+\cos (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Tan[e + f*x])^(7/2)*(a + a*Tan[e + f*x])^3,x]

[Out]

(a^3*d^3*Cos[e + f*x]*Sqrt[d*Tan[e + f*x]]*(1 + Tan[e + f*x])^3*(2310*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e +
f*x]]]*Cos[e + f*x]^2 - 2310*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]]*Cos[e + f*x]^2 + 1155*Sqrt[2]*Cos[
e + f*x]^2*Log[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]] - 1155*Sqrt[2]*Cos[e + f*x]^2*Log[1 + Sqrt[2]*Sq
rt[Tan[e + f*x]] + Tan[e + f*x]] + 9240*Cos[e + f*x]^2*Sqrt[Tan[e + f*x]] - 3080*Cos[e + f*x]^2*Tan[e + f*x]^(
3/2) + 3080*Cos[e + f*x]^2*Hypergeometric2F1[3/4, 1, 7/4, -Tan[e + f*x]^2]*Tan[e + f*x]^(3/2) - 1848*Cos[e + f
*x]^2*Tan[e + f*x]^(5/2) + 1320*Cos[e + f*x]^2*Tan[e + f*x]^(7/2) + 420*Sin[e + f*x]^2*Tan[e + f*x]^(7/2) + 77
0*Sin[2*(e + f*x)]*Tan[e + f*x]^(7/2)))/(2310*f*(Cos[e + f*x] + Sin[e + f*x])^3*Sqrt[Tan[e + f*x]])

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Maple [B]  time = 0.021, size = 467, normalized size = 2.2 \begin{align*}{\frac{2\,{a}^{3}}{11\,f{d}^{2}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{11}{2}}}}+{\frac{2\,{a}^{3}}{3\,df} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{9}{2}}}}+{\frac{4\,{a}^{3}}{7\,f} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{7}{2}}}}-{\frac{4\,{a}^{3}d}{5\,f} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{4\,{a}^{3}{d}^{2}}{3\,f} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}+4\,{\frac{{a}^{3}{d}^{3}\sqrt{d\tan \left ( fx+e \right ) }}{f}}-{\frac{{a}^{3}{d}^{3}\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}^{3}\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}}{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}^{4}\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}^{4}\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}^{4}\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))^(7/2)*(a+a*tan(f*x+e))^3,x)

[Out]

2/11/f*a^3/d^2*(d*tan(f*x+e))^(11/2)+2/3*a^3*(d*tan(f*x+e))^(9/2)/d/f+4/7*a^3*(d*tan(f*x+e))^(7/2)/f-4/5*a^3*d
*(d*tan(f*x+e))^(5/2)/f-4/3*a^3*d^2*(d*tan(f*x+e))^(3/2)/f+4*a^3*d^3*(d*tan(f*x+e))^(1/2)/f-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*ta
n(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)+1/2/f*a
^3*d^4/(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^4/(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^4/(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))^(7/2)*(a+a*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.70125, size = 892, normalized size = 4.25 \begin{align*} \left [\frac{1155 \, \sqrt{2} a^{3} d^{\frac{7}{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 (105 \, a^{3} d^{3} \tan \left (f x + e\right )^{5} + 385 \, a^{3} d^{3} \tan \left (f x + e\right )^{4} + 330 \, a^{3} d^{3} \tan \left (f x + e\right )^{3} - 462 \, a^{3} d^{3} \tan \left (f x + e\right )^{2} - 770 \, a^{3} d^{3} \tan \left (f x + e\right ) + 2310 \, a^{3} d^{3}\right )} \sqrt{d \tan \left (f x + e\right )}}{1155 \, f}, \frac{2 \,{\left (1155 \, \sqrt{2} a^{3} \sqrt{-d} d^{3} \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 (105 \, a^{3} d^{3} \tan \left (f x + e\right )^{5} + 385 \, a^{3} d^{3} \tan \left (f x + e\right )^{4} + 330 \, a^{3} d^{3} \tan \left (f x + e\right )^{3} - 462 \, a^{3} d^{3} \tan \left (f x + e\right )^{2} - 770 \, a^{3} d^{3} \tan \left (f x + e\right ) + 2310 \, a^{3} d^{3}\right )} \sqrt{d \tan \left (f x + e\right )}\right )}}{1155 \, f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(7/2)*(a+a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

[1/1155*(1155*sqrt(2)*a^3*d^(7/2)*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*(105*a^3*d^3*tan(f*x + e)^5 + 385*a^3*d^3*tan(f*x + e)
^4 + 330*a^3*d^3*tan(f*x + e)^3 - 462*a^3*d^3*tan(f*x + e)^2 - 770*a^3*d^3*tan(f*x + e) + 2310*a^3*d^3)*sqrt(d
*tan(f*x + e)))/f, 2/1155*(1155*sqrt(2)*a^3*sqrt(-d)*d^3*arctan(1/2*sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(-d)*(tan
(f*x + e) + 1)/(d*tan(f*x + e))) + (105*a^3*d^3*tan(f*x + e)^5 + 385*a^3*d^3*tan(f*x + e)^4 + 330*a^3*d^3*tan(
f*x + e)^3 - 462*a^3*d^3*tan(f*x + e)^2 - 770*a^3*d^3*tan(f*x + e) + 2310*a^3*d^3)*sqrt(d*tan(f*x + e)))/f]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**(7/2)*(a+a*tan(f*x+e))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.39553, size = 602, normalized size = 2.87 \begin{align*} -\frac{\sqrt{2}{\left (a^{3} d^{3} \sqrt{{\left | d \right |}} + a^{3} d^{2}{\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^{3} \sqrt{{\left | d \right |}} + a^{3} d^{2}{\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^{3} \sqrt{{\left | d \right |}} - \sqrt{2} a^{3} d^{2}{\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^{3} \sqrt{{\left | d \right |}} - \sqrt{2} a^{3} d^{2}{\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 (105 \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{25} f^{10} \tan \left (f x + e\right )^{5} + 385 \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{25} f^{10} \tan \left (f x + e\right )^{4} + 330 \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{25} f^{10} \tan \left (f x + e\right )^{3} - 462 \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{25} f^{10} \tan \left (f x + e\right )^{2} - 770 \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{25} f^{10} \tan \left (f x + e\right ) + 2310 \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{25} f^{10}\right )}}{1155 \, d^{22} f^{11}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(7/2)*(a+a*tan(f*x+e))^3,x, algorithm="giac")

[Out]

-1/2*sqrt(2)*(a^3*d^3*sqrt(abs(d)) + a^3*d^2*abs(d)^(3/2))*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*s
qrt(abs(d)) + abs(d))/f + 1/2*sqrt(2)*(a^3*d^3*sqrt(abs(d)) + a^3*d^2*abs(d)^(3/2))*log(d*tan(f*x + e) - sqrt(
2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) + abs(d))/f - (sqrt(2)*a^3*d^3*sqrt(abs(d)) - sqrt(2)*a^3*d^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)))/f - (sqrt(2)*a^3*d^3*sqrt(
abs(d)) - sqrt(2)*a^3*d^2*abs(d)^(3/2))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sq
rt(abs(d)))/f + 2/1155*(105*sqrt(d*tan(f*x + e))*a^3*d^25*f^10*tan(f*x + e)^5 + 385*sqrt(d*tan(f*x + e))*a^3*d
^25*f^10*tan(f*x + e)^4 + 330*sqrt(d*tan(f*x + e))*a^3*d^25*f^10*tan(f*x + e)^3 - 462*sqrt(d*tan(f*x + e))*a^3
*d^25*f^10*tan(f*x + e)^2 - 770*sqrt(d*tan(f*x + e))*a^3*d^25*f^10*tan(f*x + e) + 2310*sqrt(d*tan(f*x + e))*a^
3*d^25*f^10)/(d^22*f^11)