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

Optimal. Leaf size=165 \[ \frac{11 d^{7/2} \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{8 a^3 f}+\frac{d^{7/2} \tan ^{-1}\left (\frac{\sqrt{d}-\sqrt{d} \tan (e+f x)}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{2 \sqrt{2} a^3 f}-\frac{7 d^3 \sqrt{d \tan (e+f x)}}{8 a^3 f (\tan (e+f x)+1)}-\frac{d^2 (d \tan (e+f x))^{3/2}}{4 a f (a \tan (e+f x)+a)^2} \]

[Out]

(11*d^(7/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(8*a^3*f) + (d^(7/2)*ArcTan[(Sqrt[d] - Sqrt[d]*Tan[e + f*x])
/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/(2*Sqrt[2]*a^3*f) - (7*d^3*Sqrt[d*Tan[e + f*x]])/(8*a^3*f*(1 + Tan[e + f*x])
) - (d^2*(d*Tan[e + f*x])^(3/2))/(4*a*f*(a + a*Tan[e + f*x])^2)

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Rubi [A]  time = 0.636871, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {3565, 3645, 3653, 3532, 205, 3634, 63} \[ \frac{11 d^{7/2} \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{8 a^3 f}+\frac{d^{7/2} \tan ^{-1}\left (\frac{\sqrt{d}-\sqrt{d} \tan (e+f x)}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{2 \sqrt{2} a^3 f}-\frac{7 d^3 \sqrt{d \tan (e+f x)}}{8 a^3 f (\tan (e+f x)+1)}-\frac{d^2 (d \tan (e+f x))^{3/2}}{4 a f (a \tan (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(11*d^(7/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(8*a^3*f) + (d^(7/2)*ArcTan[(Sqrt[d] - Sqrt[d]*Tan[e + f*x])
/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/(2*Sqrt[2]*a^3*f) - (7*d^3*Sqrt[d*Tan[e + f*x]])/(8*a^3*f*(1 + Tan[e + f*x])
) - (d^2*(d*Tan[e + f*x])^(3/2))/(4*a*f*(a + a*Tan[e + f*x])^2)

Rule 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3645

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*d^2 + c*(c*C - B*d))*(a + b*T
an[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), I
nt[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c
*m + a*d*(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b*(d*(B*c - A*d)*(m + n + 1
) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3653

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

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]

Rule 3634

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rubi steps

\begin{align*} \int \frac{(d \tan (e+f x))^{7/2}}{(a+a \tan (e+f x))^3} \, dx &=-\frac{d^2 (d \tan (e+f x))^{3/2}}{4 a f (a+a \tan (e+f x))^2}+\frac{\int \frac{\sqrt{d \tan (e+f x)} \left (\frac{3 a^2 d^3}{2}-2 a^2 d^3 \tan (e+f x)+\frac{7}{2} a^2 d^3 \tan ^2(e+f x)\right )}{(a+a \tan (e+f x))^2} \, dx}{4 a^3}\\ &=-\frac{7 d^3 \sqrt{d \tan (e+f x)}}{8 a^3 f (1+\tan (e+f x))}-\frac{d^2 (d \tan (e+f x))^{3/2}}{4 a f (a+a \tan (e+f x))^2}+\frac{\int \frac{\frac{7 a^4 d^4}{2}-4 a^4 d^4 \tan (e+f x)+\frac{7}{2} a^4 d^4 \tan ^2(e+f x)}{\sqrt{d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{8 a^6}\\ &=-\frac{7 d^3 \sqrt{d \tan (e+f x)}}{8 a^3 f (1+\tan (e+f x))}-\frac{d^2 (d \tan (e+f x))^{3/2}}{4 a f (a+a \tan (e+f x))^2}+\frac{\int \frac{-4 a^5 d^4-4 a^5 d^4 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{16 a^8}+\frac{\left (11 d^4\right ) \int \frac{1+\tan ^2(e+f x)}{\sqrt{d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{16 a^2}\\ &=-\frac{7 d^3 \sqrt{d \tan (e+f x)}}{8 a^3 f (1+\tan (e+f x))}-\frac{d^2 (d \tan (e+f x))^{3/2}}{4 a f (a+a \tan (e+f x))^2}+\frac{\left (11 d^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{d x} (a+a x)} \, dx,x,\tan (e+f x)\right )}{16 a^2 f}-\frac{\left (2 a^2 d^8\right ) \operatorname{Subst}\left (\int \frac{1}{32 a^{10} d^8+d x^2} \, dx,x,\frac{-4 a^5 d^4+4 a^5 d^4 \tan (e+f x)}{\sqrt{d \tan (e+f x)}}\right )}{f}\\ &=\frac{d^{7/2} \tan ^{-1}\left (\frac{\sqrt{d}-\sqrt{d} \tan (e+f x)}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{2 \sqrt{2} a^3 f}-\frac{7 d^3 \sqrt{d \tan (e+f x)}}{8 a^3 f (1+\tan (e+f x))}-\frac{d^2 (d \tan (e+f x))^{3/2}}{4 a f (a+a \tan (e+f x))^2}+\frac{\left (11 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{a+\frac{a x^2}{d}} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{8 a^2 f}\\ &=\frac{11 d^{7/2} \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{8 a^3 f}+\frac{d^{7/2} \tan ^{-1}\left (\frac{\sqrt{d}-\sqrt{d} \tan (e+f x)}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{2 \sqrt{2} a^3 f}-\frac{7 d^3 \sqrt{d \tan (e+f x)}}{8 a^3 f (1+\tan (e+f x))}-\frac{d^2 (d \tan (e+f x))^{3/2}}{4 a f (a+a \tan (e+f x))^2}\\ \end{align*}

Mathematica [A]  time = 2.77351, size = 183, normalized size = 1.11 \[ \frac{(d \tan (e+f x))^{7/2} (\sin (e+f x)+\cos (e+f x))^3 \left (\frac{2 \left (2 \sqrt{2} \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (e+f x)}\right )-2 \sqrt{2} \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (e+f x)}+1\right )+11 \tan ^{-1}\left (\sqrt{\tan (e+f x)}\right )\right ) \csc (e+f x) \sec ^2(e+f x)}{\tan ^{\frac{5}{2}}(e+f x)}-\frac{\csc ^5(e+f x) (9 \sin (2 (e+f x))+7 \cos (2 (e+f x))+7)}{(\cot (e+f x)+1)^2}\right )}{16 a^3 f (\tan (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[e + f*x] + Sin[e + f*x])^3*(-((Csc[e + f*x]^5*(7 + 7*Cos[2*(e + f*x)] + 9*Sin[2*(e + f*x)]))/(1 + Cot[e
+ f*x])^2) + (2*(2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]] - 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e +
f*x]]] + 11*ArcTan[Sqrt[Tan[e + f*x]]])*Csc[e + f*x]*Sec[e + f*x]^2)/Tan[e + f*x]^(5/2))*(d*Tan[e + f*x])^(7/2
))/(16*a^3*f*(1 + Tan[e + f*x])^3)

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Maple [B]  time = 0.035, size = 440, normalized size = 2.7 \begin{align*} -{\frac{{d}^{3}\sqrt{2}}{16\,f{a}^{3}}\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{{d}^{3}\sqrt{2}}{8\,f{a}^{3}}\sqrt [4]{{d}^{2}}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }+{\frac{{d}^{3}\sqrt{2}}{8\,f{a}^{3}}\sqrt [4]{{d}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }-{\frac{{d}^{4}\sqrt{2}}{16\,f{a}^{3}}\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{{d}^{4}\sqrt{2}}{8\,f{a}^{3}}\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{{d}^{4}\sqrt{2}}{8\,f{a}^{3}}\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{9\,{d}^{4}}{8\,f{a}^{3} \left ( d\tan \left ( fx+e \right ) +d \right ) ^{2}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{7\,{d}^{5}}{8\,f{a}^{3} \left ( d\tan \left ( fx+e \right ) +d \right ) ^{2}}\sqrt{d\tan \left ( fx+e \right ) }}+{\frac{11}{8\,f{a}^{3}}{d}^{{\frac{7}{2}}}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{d}}}} \right ) } \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]

-1/16/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)))-1/8/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)+1/8/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)-1/16/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)))-1/8/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)+1/8/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)-9/8/f/a^3*d^4/(d*tan(f*x+e)+d)^2*(d*tan(f*x+e))^(3/2)-7/8/f/a^3*d^5/(d*tan(f*
x+e)+d)^2*(d*tan(f*x+e))^(1/2)+11/8*d^(7/2)*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/f

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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 = 1.84833, size = 1162, normalized size = 7.04 \begin{align*} \left [\frac{2 \,{\left (\sqrt{2} d^{3} \tan \left (f x + e\right )^{2} + 2 \, \sqrt{2} d^{3} \tan \left (f x + e\right ) + \sqrt{2} d^{3}\right )} \sqrt{-d} \log \left (\frac{d \tan \left (f x + e\right )^{2} - 2 \, \sqrt{d \tan \left (f x + e\right )}{\left (\sqrt{2} \tan \left (f x + e\right ) - \sqrt{2}\right )} \sqrt{-d} - 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 11 \,{\left (d^{3} \tan \left (f x + e\right )^{2} + 2 \, d^{3} \tan \left (f x + e\right ) + d^{3}\right )} \sqrt{-d} \log \left (\frac{d \tan \left (f x + e\right ) + 2 \, \sqrt{d \tan \left (f x + e\right )} \sqrt{-d} - d}{\tan \left (f x + e\right ) + 1}\right ) - 2 \,{\left (9 \, d^{3} \tan \left (f x + e\right ) + 7 \, d^{3}\right )} \sqrt{d \tan \left (f x + e\right )}}{16 \,{\left (a^{3} f \tan \left (f x + e\right )^{2} + 2 \, a^{3} f \tan \left (f x + e\right ) + a^{3} f\right )}}, \frac{11 \,{\left (d^{3} \tan \left (f x + e\right )^{2} + 2 \, d^{3} \tan \left (f x + e\right ) + d^{3}\right )} \sqrt{d} \arctan \left (\frac{\sqrt{d \tan \left (f x + e\right )}}{\sqrt{d}}\right ) - 2 \,{\left (\sqrt{2} d^{3} \tan \left (f x + e\right )^{2} + 2 \, \sqrt{2} d^{3} \tan \left (f x + e\right ) + \sqrt{2} d^{3}\right )} \sqrt{d} \arctan \left (\frac{\sqrt{d \tan \left (f x + e\right )}{\left (\sqrt{2} \tan \left (f x + e\right ) - \sqrt{2}\right )}}{2 \, \sqrt{d} \tan \left (f x + e\right )}\right ) -{\left (9 \, d^{3} \tan \left (f x + e\right ) + 7 \, d^{3}\right )} \sqrt{d \tan \left (f x + e\right )}}{8 \,{\left (a^{3} f \tan \left (f x + e\right )^{2} + 2 \, a^{3} f \tan \left (f x + e\right ) + a^{3} f\right )}}\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]

[1/16*(2*(sqrt(2)*d^3*tan(f*x + e)^2 + 2*sqrt(2)*d^3*tan(f*x + e) + sqrt(2)*d^3)*sqrt(-d)*log((d*tan(f*x + e)^
2 - 2*sqrt(d*tan(f*x + e))*(sqrt(2)*tan(f*x + e) - sqrt(2))*sqrt(-d) - 4*d*tan(f*x + e) + d)/(tan(f*x + e)^2 +
1)) + 11*(d^3*tan(f*x + e)^2 + 2*d^3*tan(f*x + e) + d^3)*sqrt(-d)*log((d*tan(f*x + e) + 2*sqrt(d*tan(f*x + e)
)*sqrt(-d) - d)/(tan(f*x + e) + 1)) - 2*(9*d^3*tan(f*x + e) + 7*d^3)*sqrt(d*tan(f*x + e)))/(a^3*f*tan(f*x + e)
^2 + 2*a^3*f*tan(f*x + e) + a^3*f), 1/8*(11*(d^3*tan(f*x + e)^2 + 2*d^3*tan(f*x + e) + d^3)*sqrt(d)*arctan(sqr
t(d*tan(f*x + e))/sqrt(d)) - 2*(sqrt(2)*d^3*tan(f*x + e)^2 + 2*sqrt(2)*d^3*tan(f*x + e) + sqrt(2)*d^3)*sqrt(d)
*arctan(1/2*sqrt(d*tan(f*x + e))*(sqrt(2)*tan(f*x + e) - sqrt(2))/(sqrt(d)*tan(f*x + e))) - (9*d^3*tan(f*x + e
) + 7*d^3)*sqrt(d*tan(f*x + e)))/(a^3*f*tan(f*x + e)^2 + 2*a^3*f*tan(f*x + e) + a^3*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 = 1.37346, size = 435, normalized size = 2.64 \begin{align*} -\frac{1}{16} \, d^{4}{\left (\frac{2 \, \sqrt{2}{\left (d \sqrt{{\left | d \right |}} +{\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 )}{a^{3} d^{2} f} + \frac{2 \, \sqrt{2}{\left (d \sqrt{{\left | d \right |}} +{\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 )}{a^{3} d^{2} f} - \frac{22 \, \arctan \left (\frac{\sqrt{d \tan \left (f x + e\right )}}{\sqrt{d}}\right )}{a^{3} \sqrt{d} f} + \frac{\sqrt{2}{\left (d \sqrt{{\left | d \right |}} -{\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 )}{a^{3} d^{2} f} - \frac{\sqrt{2}{\left (d \sqrt{{\left | d \right |}} -{\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 )}{a^{3} d^{2} f} + \frac{2 \,{\left (9 \, \sqrt{d \tan \left (f x + e\right )} d \tan \left (f x + e\right ) + 7 \, \sqrt{d \tan \left (f x + e\right )} d\right )}}{{\left (d \tan \left (f x + e\right ) + d\right )}^{2} a^{3} 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="giac")

[Out]

-1/16*d^4*(2*sqrt(2)*(d*sqrt(abs(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)))/(a^3*d^2*f) + 2*sqrt(2)*(d*sqrt(abs(d)) + abs(d)^(3/2))*arctan(-1/2*sqrt(2)*(sqrt(2)*s
qrt(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/(a^3*d^2*f) - 22*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a^3
*sqrt(d)*f) + sqrt(2)*(d*sqrt(abs(d)) - abs(d)^(3/2))*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(a
bs(d)) + abs(d))/(a^3*d^2*f) - sqrt(2)*(d*sqrt(abs(d)) - abs(d)^(3/2))*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan
(f*x + e))*sqrt(abs(d)) + abs(d))/(a^3*d^2*f) + 2*(9*sqrt(d*tan(f*x + e))*d*tan(f*x + e) + 7*sqrt(d*tan(f*x +
e))*d)/((d*tan(f*x + e) + d)^2*a^3*f))