3.4.53 \(\int \frac {(a+a \tan (e+f x))^3}{\sqrt {d \tan (e+f x)}} \, dx\) [353]

Optimal. Leaf size=117 \[ -\frac {2 \sqrt {2} a^3 \tanh ^{-1}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {d} f}+\frac {16 a^3 \sqrt {d \tan (e+f x)}}{3 d f}+\frac {2 \sqrt {d \tan (e+f x)} \left (a^3+a^3 \tan (e+f x)\right )}{3 d f} \]

[Out]

-2*a^3*arctanh(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/d^(1/2)+16/3*a^3*(d*ta
n(f*x+e))^(1/2)/d/f+2/3*(d*tan(f*x+e))^(1/2)*(a^3+a^3*tan(f*x+e))/d/f

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Rubi [A]
time = 0.11, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3647, 3711, 3613, 214} \begin {gather*} \frac {16 a^3 \sqrt {d \tan (e+f x)}}{3 d f}+\frac {2 \left (a^3 \tan (e+f x)+a^3\right ) \sqrt {d \tan (e+f x)}}{3 d f}-\frac {2 \sqrt {2} a^3 \tanh ^{-1}\left (\frac {\sqrt {d} \tan (e+f x)+\sqrt {d}}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {d} f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[2]*a^3*ArcTanh[(Sqrt[d] + Sqrt[d]*Tan[e + f*x])/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/(Sqrt[d]*f) + (16*a^
3*Sqrt[d*Tan[e + f*x]])/(3*d*f) + (2*Sqrt[d*Tan[e + f*x]]*(a^3 + a^3*Tan[e + f*x]))/(3*d*f)

Rule 214

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

Rule 3613

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 3647

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 3711

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 \frac {(a+a \tan (e+f x))^3}{\sqrt {d \tan (e+f x)}} \, dx &=\frac {2 \sqrt {d \tan (e+f x)} \left (a^3+a^3 \tan (e+f x)\right )}{3 d f}+\frac {2 \int \frac {a^3 d+3 a^3 d \tan (e+f x)+4 a^3 d \tan ^2(e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{3 d}\\ &=\frac {16 a^3 \sqrt {d \tan (e+f x)}}{3 d f}+\frac {2 \sqrt {d \tan (e+f x)} \left (a^3+a^3 \tan (e+f x)\right )}{3 d f}+\frac {2 \int \frac {-3 a^3 d+3 a^3 d \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{3 d}\\ &=\frac {16 a^3 \sqrt {d \tan (e+f x)}}{3 d f}+\frac {2 \sqrt {d \tan (e+f x)} \left (a^3+a^3 \tan (e+f x)\right )}{3 d f}-\frac {\left (12 a^6 d\right ) \text {Subst}\left (\int \frac {1}{-18 a^6 d^2+d x^2} \, dx,x,\frac {-3 a^3 d-3 a^3 d \tan (e+f x)}{\sqrt {d \tan (e+f x)}}\right )}{f}\\ &=-\frac {2 \sqrt {2} a^3 \tanh ^{-1}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {d} f}+\frac {16 a^3 \sqrt {d \tan (e+f x)}}{3 d f}+\frac {2 \sqrt {d \tan (e+f x)} \left (a^3+a^3 \tan (e+f x)\right )}{3 d f}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 6.09, size = 553, normalized size = 4.73 \begin {gather*} \frac {6 \cos ^2(e+f x) \sin (e+f x) (a+a \tan (e+f x))^3}{f (\cos (e+f x)+\sin (e+f x))^3 \sqrt {d \tan (e+f x)}}+\frac {2 \cos (e+f x) \sin ^2(e+f x) (a+a \tan (e+f x))^3}{3 f (\cos (e+f x)+\sin (e+f x))^3 \sqrt {d \tan (e+f x)}}+\frac {4 \cos (e+f x) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\tan ^2(e+f x)\right ) \sin ^2(e+f x) (a+a \tan (e+f x))^3}{3 f (\cos (e+f x)+\sin (e+f x))^3 \sqrt {d \tan (e+f x)}}+\frac {\sqrt {2} \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right ) \cos ^3(e+f x) \sqrt {\tan (e+f x)} (a+a \tan (e+f x))^3}{f (\cos (e+f x)+\sin (e+f x))^3 \sqrt {d \tan (e+f x)}}-\frac {\sqrt {2} \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (e+f x)}\right ) \cos ^3(e+f x) \sqrt {\tan (e+f x)} (a+a \tan (e+f x))^3}{f (\cos (e+f x)+\sin (e+f x))^3 \sqrt {d \tan (e+f x)}}+\frac {\cos ^3(e+f x) \log \left (1-\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right ) \sqrt {\tan (e+f x)} (a+a \tan (e+f x))^3}{\sqrt {2} f (\cos (e+f x)+\sin (e+f x))^3 \sqrt {d \tan (e+f x)}}-\frac {\cos ^3(e+f x) \log \left (1+\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right ) \sqrt {\tan (e+f x)} (a+a \tan (e+f x))^3}{\sqrt {2} f (\cos (e+f x)+\sin (e+f x))^3 \sqrt {d \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6*Cos[e + f*x]^2*Sin[e + f*x]*(a + a*Tan[e + f*x])^3)/(f*(Cos[e + f*x] + Sin[e + f*x])^3*Sqrt[d*Tan[e + f*x]]
) + (2*Cos[e + f*x]*Sin[e + f*x]^2*(a + a*Tan[e + f*x])^3)/(3*f*(Cos[e + f*x] + Sin[e + f*x])^3*Sqrt[d*Tan[e +
f*x]]) + (4*Cos[e + f*x]*Hypergeometric2F1[3/4, 1, 7/4, -Tan[e + f*x]^2]*Sin[e + f*x]^2*(a + a*Tan[e + f*x])^
3)/(3*f*(Cos[e + f*x] + Sin[e + f*x])^3*Sqrt[d*Tan[e + f*x]]) + (Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]
]*Cos[e + f*x]^3*Sqrt[Tan[e + f*x]]*(a + a*Tan[e + f*x])^3)/(f*(Cos[e + f*x] + Sin[e + f*x])^3*Sqrt[d*Tan[e +
f*x]]) - (Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]]*Cos[e + f*x]^3*Sqrt[Tan[e + f*x]]*(a + a*Tan[e + f*x]
)^3)/(f*(Cos[e + f*x] + Sin[e + f*x])^3*Sqrt[d*Tan[e + f*x]]) + (Cos[e + f*x]^3*Log[1 - Sqrt[2]*Sqrt[Tan[e + f
*x]] + Tan[e + f*x]]*Sqrt[Tan[e + f*x]]*(a + a*Tan[e + f*x])^3)/(Sqrt[2]*f*(Cos[e + f*x] + Sin[e + f*x])^3*Sqr
t[d*Tan[e + f*x]]) - (Cos[e + f*x]^3*Log[1 + Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]]*Sqrt[Tan[e + f*x]]*(a
+ a*Tan[e + f*x])^3)/(Sqrt[2]*f*(Cos[e + f*x] + Sin[e + f*x])^3*Sqrt[d*Tan[e + f*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(308\) vs. \(2(98)=196\).
time = 0.27, size = 309, normalized size = 2.64

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+3 d \sqrt {d \tan \left (f x +e \right )}-2 d^{2} \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f \,d^{2}}\)	\(309\)
default	\(\frac {2 a^{3} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+3 d \sqrt {d \tan \left (f x +e \right )}-2 d^{2} \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f \,d^{2}}\)	\(309\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*tan(f*x+e))^3/(d*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.50, size = 130, normalized size = 1.11 \begin {gather*} -\frac {3 \, a^{3} d {\left (\frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} - \frac {2 \, {\left (\left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{3} + 9 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d\right )}}{d}}{3 \, d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*tan(f*x+e))^3/(d*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

-1/3*(3*a^3*d*(sqrt(2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d) - sqrt(2)*log(d*
tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d)) - 2*((d*tan(f*x + e))^(3/2)*a^3 + 9*sqrt(d*t
an(f*x + e))*a^3*d)/d)/(d*f)

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Fricas [A]
time = 0.87, size = 210, normalized size = 1.79 \begin {gather*} \left [\frac {3 \, \sqrt {2} a^{3} \sqrt {d} \log \left (\frac {\tan \left (f x + e\right )^{2} - \frac {2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} {\left (\tan \left (f x + e\right ) + 1\right )}}{\sqrt {d}} + 4 \, \tan \left (f x + e\right ) + 1}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, {\left (a^{3} \tan \left (f x + e\right ) + 9 \, a^{3}\right )} \sqrt {d \tan \left (f x + e\right )}}{3 \, d f}, \frac {2 \, {\left (3 \, \sqrt {2} a^{3} d \sqrt {-\frac {1}{d}} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-\frac {1}{d}} {\left (\tan \left (f x + e\right ) + 1\right )}}{2 \, \tan \left (f x + e\right )}\right ) + {\left (a^{3} \tan \left (f x + e\right ) + 9 \, a^{3}\right )} \sqrt {d \tan \left (f x + e\right )}\right )}}{3 \, d f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*tan(f*x+e))^3/(d*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[1/3*(3*sqrt(2)*a^3*sqrt(d)*log((tan(f*x + e)^2 - 2*sqrt(2)*sqrt(d*tan(f*x + e))*(tan(f*x + e) + 1)/sqrt(d) +
4*tan(f*x + e) + 1)/(tan(f*x + e)^2 + 1)) + 2*(a^3*tan(f*x + e) + 9*a^3)*sqrt(d*tan(f*x + e)))/(d*f), 2/3*(3*s
qrt(2)*a^3*d*sqrt(-1/d)*arctan(1/2*sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(-1/d)*(tan(f*x + e) + 1)/tan(f*x + e)) +
(a^3*tan(f*x + e) + 9*a^3)*sqrt(d*tan(f*x + e)))/(d*f)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{3} \left (\int \frac {1}{\sqrt {d \tan {\left (e + f x \right )}}}\, dx + \int \frac {3 \tan {\left (e + f x \right )}}{\sqrt {d \tan {\left (e + f x \right )}}}\, dx + \int \frac {3 \tan ^{2}{\left (e + f x \right )}}{\sqrt {d \tan {\left (e + f x \right )}}}\, dx + \int \frac {\tan ^{3}{\left (e + f x \right )}}{\sqrt {d \tan {\left (e + f x \right )}}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 313 vs. \(2 (103) = 206\).
time = 0.71, size = 313, normalized size = 2.68 \begin {gather*} -\frac {\sqrt {2} {\left (a^{3} d \sqrt {{\left | d \right |}} + a^{3} {\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 \, d^{2} f} + \frac {\sqrt {2} {\left (a^{3} d \sqrt {{\left | d \right |}} + a^{3} {\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 \, d^{2} f} - \frac {{\left (\sqrt {2} a^{3} d \sqrt {{\left | d \right |}} - \sqrt {2} a^{3} {\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 )}{d^{2} f} - \frac {{\left (\sqrt {2} a^{3} d \sqrt {{\left | d \right |}} - \sqrt {2} a^{3} {\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 )}{d^{2} f} + \frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right )} a^{3} d^{5} f^{2} \tan \left (f x + e\right ) + 9 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{5} f^{2}\right )}}{3 \, d^{6} f^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*tan(f*x+e))^3/(d*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

-1/2*sqrt(2)*(a^3*d*sqrt(abs(d)) + a^3*abs(d)^(3/2))*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(ab
s(d)) + abs(d))/(d^2*f) + 1/2*sqrt(2)*(a^3*d*sqrt(abs(d)) + a^3*abs(d)^(3/2))*log(d*tan(f*x + e) - sqrt(2)*sqr
t(d*tan(f*x + e))*sqrt(abs(d)) + abs(d))/(d^2*f) - (sqrt(2)*a^3*d*sqrt(abs(d)) - sqrt(2)*a^3*abs(d)^(3/2))*arc
tan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/(d^2*f) - (sqrt(2)*a^3*d*sqrt(ab
s(d)) - sqrt(2)*a^3*abs(d)^(3/2))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(abs
(d)))/(d^2*f) + 2/3*(sqrt(d*tan(f*x + e))*a^3*d^5*f^2*tan(f*x + e) + 9*sqrt(d*tan(f*x + e))*a^3*d^5*f^2)/(d^6*
f^3)

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Mupad [B]
time = 4.26, size = 100, normalized size = 0.85 \begin {gather*} \frac {6\,a^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{d\,f}+\frac {2\,a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3\,d^2\,f}-\frac {2\,\sqrt {2}\,a^3\,\mathrm {atanh}\left (\frac {32\,\sqrt {2}\,a^6\,\sqrt {d}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{32\,a^6\,d+32\,a^6\,d\,\mathrm {tan}\left (e+f\,x\right )}\right )}{\sqrt {d}\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x))^3/(d*tan(e + f*x))^(1/2),x)

[Out]

(6*a^3*(d*tan(e + f*x))^(1/2))/(d*f) + (2*a^3*(d*tan(e + f*x))^(3/2))/(3*d^2*f) - (2*2^(1/2)*a^3*atanh((32*2^(
1/2)*a^6*d^(1/2)*(d*tan(e + f*x))^(1/2))/(32*a^6*d + 32*a^6*d*tan(e + f*x))))/(d^(1/2)*f)

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