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*a^3*d^(5/2)*arctan(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^2*(d*ta
n(f*x+e))^(1/2)/f-4/3*a^3*d*(d*tan(f*x+e))^(3/2)/f+4/5*a^3*(d*tan(f*x+e))^(5/2)/f+40/63*a^3*(d*tan(f*x+e))^(7/
2)/d/f+2/9*(d*tan(f*x+e))^(7/2)*(a^3+a^3*tan(f*x+e))/d/f

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Rubi [A]  time = 0.28, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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 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 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]

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*}

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Mathematica [C]  time = 6.11, 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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fricas [A]  time = 0.52, size = 303, normalized size = 1.63 \[ \left [\frac {315 \, \sqrt {2} a^{3} \sqrt {-d} d^{2} \log \left (\frac {d \tan \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) - 1\right )} - 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, {\left (35 \, a^{3} d^{2} \tan \left (f x + e\right )^{4} + 135 \, a^{3} d^{2} \tan \left (f x + e\right )^{3} + 126 \, a^{3} d^{2} \tan \left (f x + e\right )^{2} - 210 \, a^{3} d^{2} \tan \left (f x + e\right ) - 630 \, a^{3} d^{2}\right )} \sqrt {d \tan \left (f x + e\right )}}{315 \, f}, \frac {2 \, {\left (315 \, \sqrt {2} a^{3} d^{\frac {5}{2}} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} {\left (\tan \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {d} \tan \left (f x + e\right )}\right ) + {\left (35 \, a^{3} d^{2} \tan \left (f x + e\right )^{4} + 135 \, a^{3} d^{2} \tan \left (f x + e\right )^{3} + 126 \, a^{3} d^{2} \tan \left (f x + e\right )^{2} - 210 \, a^{3} d^{2} \tan \left (f x + e\right ) - 630 \, a^{3} d^{2}\right )} \sqrt {d \tan \left (f x + e\right )}\right )}}{315 \, f}\right ] \]

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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giac [B]  time = 2.79, size = 405, normalized size = 2.18 \[ \frac {\sqrt {2} {\left (a^{3} d^{2} \sqrt {{\left | d \right |}} - a^{3} d {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{2 \, f} - \frac {\sqrt {2} {\left (a^{3} d^{2} \sqrt {{\left | d \right |}} - a^{3} d {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{2 \, f} + \frac {{\left (\sqrt {2} a^{3} d^{2} \sqrt {{\left | d \right |}} + \sqrt {2} a^{3} d {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{f} + \frac {{\left (\sqrt {2} a^{3} d^{2} \sqrt {{\left | d \right |}} + \sqrt {2} a^{3} d {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{f} + \frac {2 \, {\left (35 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{20} f^{8} \tan \left (f x + e\right )^{4} + 135 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{20} f^{8} \tan \left (f x + e\right )^{3} + 126 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{20} f^{8} \tan \left (f x + e\right )^{2} - 210 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{20} f^{8} \tan \left (f x + e\right ) - 630 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{20} f^{8}\right )}}{315 \, d^{18} f^{9}} \]

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)

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maple [B]  time = 0.26, size = 446, normalized size = 2.40 \[ \frac {2 a^{3} \left (d \tan \left (f x +e \right )\right )^{\frac {9}{2}}}{9 f \,d^{2}}+\frac {6 a^{3} \left (d \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7 d f}+\frac {4 a^{3} \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f}-\frac {4 a^{3} d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3 f}-\frac {4 a^{3} d^{2} \sqrt {d \tan \left (f x +e \right )}}{f}+\frac {a^{3} d^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{2 f}+\frac {a^{3} d^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f}-\frac {a^{3} d^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f}+\frac {a^{3} d^{3} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{2 f \left (d^{2}\right )^{\frac {1}{4}}}+\frac {a^{3} d^{3} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f \left (d^{2}\right )^{\frac {1}{4}}}-\frac {a^{3} d^{3} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f \left (d^{2}\right )^{\frac {1}{4}}} \]

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

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maxima [A]  time = 0.71, size = 180, normalized size = 0.97 \[ \frac {2 \, {\left (315 \, a^{3} d^{4} {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}}\right )} + \frac {35 \, \left (d \tan \left (f x + e\right )\right )^{\frac {9}{2}} a^{3} + 135 \, \left (d \tan \left (f x + e\right )\right )^{\frac {7}{2}} a^{3} d + 126 \, \left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a^{3} d^{2} - 210 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{3} d^{3} - 630 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{4}}{d}\right )}}{315 \, d f} \]

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]

2/315*(315*a^3*d^4*(sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) + s
qrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d)) + (35*(d*tan(f*x + e))
^(9/2)*a^3 + 135*(d*tan(f*x + e))^(7/2)*a^3*d + 126*(d*tan(f*x + e))^(5/2)*a^3*d^2 - 210*(d*tan(f*x + e))^(3/2
)*a^3*d^3 - 630*sqrt(d*tan(f*x + e))*a^3*d^4)/d)/(d*f)

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mupad [B]  time = 5.96, size = 176, normalized size = 0.95 \[ \frac {4\,a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{5\,f}-\frac {4\,a^3\,d^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{f}+\frac {6\,a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{7/2}}{7\,d\,f}+\frac {2\,a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{9/2}}{9\,d^2\,f}-\frac {4\,a^3\,d\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3\,f}+\frac {\sqrt {2}\,a^3\,d^{5/2}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}+\frac {\sqrt {2}\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{2\,d^{3/2}}\right )\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{3} \left (\int \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}\, dx + \int 3 \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan {\left (e + f x \right )}\, dx + \int 3 \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan ^{2}{\left (e + f x \right )}\, dx + \int \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan ^{3}{\left (e + f x \right )}\, dx\right ) \]

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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