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]

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

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Rubi [A]  time = 0.64, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, 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 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]

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 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 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 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 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]

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*}

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Mathematica [A]  time = 2.89, 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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fricas [A]  time = 0.55, size = 447, normalized size = 2.71 \[ \left [\frac {2 \, {\left (\sqrt {2} d^{3} \tan \left (f x + e\right )^{2} + 2 \, \sqrt {2} d^{3} \tan \left (f x + e\right ) + \sqrt {2} d^{3}\right )} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right )^{2} - 2 \, \sqrt {d \tan \left (f x + e\right )} {\left (\sqrt {2} \tan \left (f x + e\right ) - \sqrt {2}\right )} \sqrt {-d} - 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 11 \, {\left (d^{3} \tan \left (f x + e\right )^{2} + 2 \, d^{3} \tan \left (f x + e\right ) + d^{3}\right )} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right ) + 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right ) - 2 \, {\left (9 \, d^{3} \tan \left (f x + e\right ) + 7 \, d^{3}\right )} \sqrt {d \tan \left (f x + e\right )}}{16 \, {\left (a^{3} f \tan \left (f x + e\right )^{2} + 2 \, a^{3} f \tan \left (f x + e\right ) + a^{3} f\right )}}, \frac {11 \, {\left (d^{3} \tan \left (f x + e\right )^{2} + 2 \, d^{3} \tan \left (f x + e\right ) + d^{3}\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right ) - 2 \, {\left (\sqrt {2} d^{3} \tan \left (f x + e\right )^{2} + 2 \, \sqrt {2} d^{3} \tan \left (f x + e\right ) + \sqrt {2} d^{3}\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )} {\left (\sqrt {2} \tan \left (f x + e\right ) - \sqrt {2}\right )}}{2 \, \sqrt {d} \tan \left (f x + e\right )}\right ) - {\left (9 \, d^{3} \tan \left (f x + e\right ) + 7 \, d^{3}\right )} \sqrt {d \tan \left (f x + e\right )}}{8 \, {\left (a^{3} f \tan \left (f x + e\right )^{2} + 2 \, a^{3} f \tan \left (f x + e\right ) + a^{3} f\right )}}\right ] \]

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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giac [B]  time = 3.83, size = 326, normalized size = 1.98 \[ -\frac {1}{16} \, d^{3} {\left (\frac {2 \, \sqrt {2} {\left (d \sqrt {{\left | d \right |}} + {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{a^{3} d f} + \frac {2 \, \sqrt {2} {\left (d \sqrt {{\left | d \right |}} + {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{a^{3} d f} - \frac {22 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{3} f} + \frac {\sqrt {2} {\left (d \sqrt {{\left | d \right |}} - {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{a^{3} d f} - \frac {\sqrt {2} {\left (d \sqrt {{\left | d \right |}} - {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{a^{3} d f} + \frac {2 \, {\left (9 \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right ) + 7 \, \sqrt {d \tan \left (f x + e\right )} d^{2}\right )}}{{\left (d \tan \left (f x + e\right ) + d\right )}^{2} a^{3} f}\right )} \]

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^3*(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*f) + 2*sqrt(2)*(d*sqrt(abs(d)) + abs(d)^(3/2))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqr
t(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/(a^3*d*f) - 22*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/
(a^3*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*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*f) + 2*(9*sqrt(d*tan(f*x + e))*d^2*tan(f*x + e) + 7*sqrt(d*tan(f*x + e))*d^
2)/((d*tan(f*x + e) + d)^2*a^3*f))

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maple [B]  time = 0.33, size = 440, normalized size = 2.67 \[ -\frac {d^{3} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{16 f \,a^{3}}-\frac {d^{3} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{8 f \,a^{3}}+\frac {d^{3} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{8 f \,a^{3}}-\frac {d^{4} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{16 f \,a^{3} \left (d^{2}\right )^{\frac {1}{4}}}-\frac {d^{4} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{8 f \,a^{3} \left (d^{2}\right )^{\frac {1}{4}}}+\frac {d^{4} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{8 f \,a^{3} \left (d^{2}\right )^{\frac {1}{4}}}-\frac {9 d^{4} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{8 f \,a^{3} \left (d \tan \left (f x +e \right )+d \right )^{2}}-\frac {7 d^{5} \sqrt {d \tan \left (f x +e \right )}}{8 f \,a^{3} \left (d \tan \left (f x +e \right )+d \right )^{2}}+\frac {11 d^{\frac {7}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{8 a^{3} f} \]

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*2^(1/2)/(d^2)^(1/4)*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*2^(1/2)/(d^2)^
(1/4)*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)+1/8/f/a^3*d^4*2^(1/2)/(d^2)^(1/4)*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 [A]  time = 0.80, size = 184, normalized size = 1.12 \[ -\frac {\frac {2 \, d^{5} {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}}\right )}}{a^{3}} - \frac {11 \, d^{\frac {9}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{3}} + \frac {9 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d^{5} + 7 \, \sqrt {d \tan \left (f x + e\right )} d^{6}}{a^{3} d^{2} \tan \left (f x + e\right )^{2} + 2 \, a^{3} d^{2} \tan \left (f x + e\right ) + a^{3} d^{2}}}{8 \, d f} \]

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]

-1/8*(2*d^5*(sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) + sqrt(2)*
arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d))/a^3 - 11*d^(9/2)*arctan(sqrt(
d*tan(f*x + e))/sqrt(d))/a^3 + (9*(d*tan(f*x + e))^(3/2)*d^5 + 7*sqrt(d*tan(f*x + e))*d^6)/(a^3*d^2*tan(f*x +
e)^2 + 2*a^3*d^2*tan(f*x + e) + a^3*d^2))/(d*f)

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mupad [B]  time = 4.86, size = 178, normalized size = 1.08 \[ \frac {11\,d^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f}-\frac {\frac {7\,d^5\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{8}+\frac {9\,d^4\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{8}}{f\,a^3\,d^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+2\,f\,a^3\,d^2\,\mathrm {tan}\left (e+f\,x\right )+f\,a^3\,d^2}-\frac {\sqrt {2}\,d^{7/2}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}+\frac {\sqrt {2}\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{2\,d^{3/2}}\right )\right )}{8\,a^3\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}{\tan ^{3}{\left (e + f x \right )} + 3 \tan ^{2}{\left (e + f x \right )} + 3 \tan {\left (e + f x \right )} + 1}\, dx}{a^{3}} \]

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]

Integral((d*tan(e + f*x))**(7/2)/(tan(e + f*x)**3 + 3*tan(e + f*x)**2 + 3*tan(e + f*x) + 1), x)/a**3

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