3.4.57 \(\int \frac {(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{9/2}} \, dx\) [357]

Optimal. Leaf size=165 \[ -\frac {2 \sqrt {2} a^3 \tanh ^{-1}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{d^{9/2} f}-\frac {32 a^3}{35 d^2 f (d \tan (e+f x))^{5/2}}-\frac {4 a^3}{3 d^3 f (d \tan (e+f x))^{3/2}}+\frac {4 a^3}{d^4 f \sqrt {d \tan (e+f x)}}-\frac {2 \left (a^3+a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}} \]

[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)/d^(9/2)/f+4*a^3/d^4/f/(d
*tan(f*x+e))^(1/2)-32/35*a^3/d^2/f/(d*tan(f*x+e))^(5/2)-4/3*a^3/d^3/f/(d*tan(f*x+e))^(3/2)-2/7*(a^3+a^3*tan(f*
x+e))/d/f/(d*tan(f*x+e))^(7/2)

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Rubi [A]
time = 0.19, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3646, 3709, 3610, 3613, 214} \begin {gather*} -\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 )}{d^{9/2} f}+\frac {4 a^3}{d^4 f \sqrt {d \tan (e+f x)}}-\frac {4 a^3}{3 d^3 f (d \tan (e+f x))^{3/2}}-\frac {32 a^3}{35 d^2 f (d \tan (e+f x))^{5/2}}-\frac {2 \left (a^3 \tan (e+f x)+a^3\right )}{7 d f (d \tan (e+f x))^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(9/2),x]

[Out]

(-2*Sqrt[2]*a^3*ArcTanh[(Sqrt[d] + Sqrt[d]*Tan[e + f*x])/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/(d^(9/2)*f) - (32*a^
3)/(35*d^2*f*(d*Tan[e + f*x])^(5/2)) - (4*a^3)/(3*d^3*f*(d*Tan[e + f*x])^(3/2)) + (4*a^3)/(d^4*f*Sqrt[d*Tan[e
+ f*x]]) - (2*(a^3 + a^3*Tan[e + f*x]))/(7*d*f*(d*Tan[e + f*x])^(7/2))

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 3610

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

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 3646

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 3709

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 + b^2)
)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[b*B + a*(A - C) - (A*b - a*B - b*C)*Tan[e +
f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && LtQ[m, -1] && NeQ[a^2
+ b^2, 0]

Rubi steps

\begin {align*} \int \frac {(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{9/2}} \, dx &=-\frac {2 \left (a^3+a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}+\frac {2 \int \frac {8 a^3 d^2+7 a^3 d^2 \tan (e+f x)+a^3 d^2 \tan ^2(e+f x)}{(d \tan (e+f x))^{7/2}} \, dx}{7 d^3}\\ &=-\frac {32 a^3}{35 d^2 f (d \tan (e+f x))^{5/2}}-\frac {2 \left (a^3+a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}+\frac {2 \int \frac {7 a^3 d^3-7 a^3 d^3 \tan (e+f x)}{(d \tan (e+f x))^{5/2}} \, dx}{7 d^5}\\ &=-\frac {32 a^3}{35 d^2 f (d \tan (e+f x))^{5/2}}-\frac {4 a^3}{3 d^3 f (d \tan (e+f x))^{3/2}}-\frac {2 \left (a^3+a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}+\frac {2 \int \frac {-7 a^3 d^4-7 a^3 d^4 \tan (e+f x)}{(d \tan (e+f x))^{3/2}} \, dx}{7 d^7}\\ &=-\frac {32 a^3}{35 d^2 f (d \tan (e+f x))^{5/2}}-\frac {4 a^3}{3 d^3 f (d \tan (e+f x))^{3/2}}+\frac {4 a^3}{d^4 f \sqrt {d \tan (e+f x)}}-\frac {2 \left (a^3+a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}+\frac {2 \int \frac {-7 a^3 d^5+7 a^3 d^5 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{7 d^9}\\ &=-\frac {32 a^3}{35 d^2 f (d \tan (e+f x))^{5/2}}-\frac {4 a^3}{3 d^3 f (d \tan (e+f x))^{3/2}}+\frac {4 a^3}{d^4 f \sqrt {d \tan (e+f x)}}-\frac {2 \left (a^3+a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}-\frac {\left (28 a^6 d\right ) \text {Subst}\left (\int \frac {1}{-98 a^6 d^{10}+d x^2} \, dx,x,\frac {-7 a^3 d^5-7 a^3 d^5 \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 )}{d^{9/2} f}-\frac {32 a^3}{35 d^2 f (d \tan (e+f x))^{5/2}}-\frac {4 a^3}{3 d^3 f (d \tan (e+f x))^{3/2}}+\frac {4 a^3}{d^4 f \sqrt {d \tan (e+f x)}}-\frac {2 \left (a^3+a^3 \tan (e+f x)\right )}{7 d f (d \tan (e+f x))^{7/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.90, size = 175, normalized size = 1.06 \begin {gather*} -\frac {a^3 \cos (e+f x) (1+\cot (e+f x))^3 \left (10 \cos ^2(e+f x) \cot (e+f x) \, _2F_1\left (-\frac {7}{4},1;-\frac {3}{4};-\tan ^2(e+f x)\right )+42 \cos ^2(e+f x) \, _2F_1\left (-\frac {5}{4},1;-\frac {1}{4};-\tan ^2(e+f x)\right )+70 \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};-\tan ^2(e+f x)\right ) \sin ^2(e+f x)+35 \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};-\tan ^2(e+f x)\right ) \sin (2 (e+f x))\right ) \sqrt {d \tan (e+f x)}}{35 d^5 f (\cos (e+f x)+\sin (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(9/2),x]

[Out]

-1/35*(a^3*Cos[e + f*x]*(1 + Cot[e + f*x])^3*(10*Cos[e + f*x]^2*Cot[e + f*x]*Hypergeometric2F1[-7/4, 1, -3/4,
-Tan[e + f*x]^2] + 42*Cos[e + f*x]^2*Hypergeometric2F1[-5/4, 1, -1/4, -Tan[e + f*x]^2] + 70*Hypergeometric2F1[
-1/4, 1, 3/4, -Tan[e + f*x]^2]*Sin[e + f*x]^2 + 35*Hypergeometric2F1[-3/4, 1, 1/4, -Tan[e + f*x]^2]*Sin[2*(e +
f*x)])*Sqrt[d*Tan[e + f*x]])/(d^5*f*(Cos[e + f*x] + Sin[e + f*x])^3)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(337\) vs. \(2(140)=280\).
time = 0.20, size = 338, normalized size = 2.05

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (-\frac {d}{7 \left (d \tan \left (f x +e \right )\right )^{\frac {7}{2}}}-\frac {3}{5 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {2}{d^{2} \sqrt {d \tan \left (f x +e \right )}}-\frac {2}{3 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}}{d^{2}}\right )}{f \,d^{2}}\)	\(338\)
default	\(\frac {2 a^{3} \left (-\frac {d}{7 \left (d \tan \left (f x +e \right )\right )^{\frac {7}{2}}}-\frac {3}{5 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {2}{d^{2} \sqrt {d \tan \left (f x +e \right )}}-\frac {2}{3 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}}{d^{2}}\right )}{f \,d^{2}}\)	\(338\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*tan(f*x+e))^3/(d*tan(f*x+e))^(9/2),x,method=_RETURNVERBOSE)

[Out]

2/f*a^3/d^2*(-1/7*d/(d*tan(f*x+e))^(7/2)-3/5/(d*tan(f*x+e))^(5/2)+2/d^2/(d*tan(f*x+e))^(1/2)-2/3/d/(d*tan(f*x+
e))^(3/2)+1/d^2*(-1/4/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))+1/4/(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)))+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 = 0.53, size = 168, normalized size = 1.02 \begin {gather*} -\frac {\frac {105 \, a^{3} {\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 )}}{d^{3}} - \frac {2 \, {\left (210 \, a^{3} d^{3} \tan \left (f x + e\right )^{3} - 70 \, a^{3} d^{3} \tan \left (f x + e\right )^{2} - 63 \, a^{3} d^{3} \tan \left (f x + e\right ) - 15 \, a^{3} d^{3}\right )}}{\left (d \tan \left (f x + e\right )\right )^{\frac {7}{2}} d^{3}}}{105 \, 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))^(9/2),x, algorithm="maxima")

[Out]

-1/105*(105*a^3*(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))/d^3 - 2*(210*a^3*d^3*tan(f*x + e)^3 - 70*a
^3*d^3*tan(f*x + e)^2 - 63*a^3*d^3*tan(f*x + e) - 15*a^3*d^3)/((d*tan(f*x + e))^(7/2)*d^3))/(d*f)

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Fricas [A]
time = 1.01, size = 304, normalized size = 1.84 \begin {gather*} \left [\frac {105 \, \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 ) \tan \left (f x + e\right )^{4} + 2 \, {\left (210 \, a^{3} \tan \left (f x + e\right )^{3} - 70 \, a^{3} \tan \left (f x + e\right )^{2} - 63 \, a^{3} \tan \left (f x + e\right ) - 15 \, a^{3}\right )} \sqrt {d \tan \left (f x + e\right )}}{105 \, d^{5} f \tan \left (f x + e\right )^{4}}, \frac {2 \, {\left (105 \, \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 ) \tan \left (f x + e\right )^{4} + {\left (210 \, a^{3} \tan \left (f x + e\right )^{3} - 70 \, a^{3} \tan \left (f x + e\right )^{2} - 63 \, a^{3} \tan \left (f x + e\right ) - 15 \, a^{3}\right )} \sqrt {d \tan \left (f x + e\right )}\right )}}{105 \, d^{5} f \tan \left (f x + e\right )^{4}}\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))^(9/2),x, algorithm="fricas")

[Out]

[1/105*(105*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))*tan(f*x + e)^4 + 2*(210*a^3*tan(f*x + e)^3 - 70*a^3*tan(f*x + e)
^2 - 63*a^3*tan(f*x + e) - 15*a^3)*sqrt(d*tan(f*x + e)))/(d^5*f*tan(f*x + e)^4), 2/105*(105*sqrt(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))*tan(f*x + e)^4 + (2
10*a^3*tan(f*x + e)^3 - 70*a^3*tan(f*x + e)^2 - 63*a^3*tan(f*x + e) - 15*a^3)*sqrt(d*tan(f*x + e)))/(d^5*f*tan
(f*x + e)^4)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{3} \left (\int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {9}{2}}}\, dx + \int \frac {3 \tan {\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {9}{2}}}\, dx + \int \frac {3 \tan ^{2}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {9}{2}}}\, dx + \int \frac {\tan ^{3}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {9}{2}}}\, 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))**(9/2),x)

[Out]

a**3*(Integral((d*tan(e + f*x))**(-9/2), x) + Integral(3*tan(e + f*x)/(d*tan(e + f*x))**(9/2), x) + Integral(3
*tan(e + f*x)**2/(d*tan(e + f*x))**(9/2), x) + Integral(tan(e + f*x)**3/(d*tan(e + f*x))**(9/2), x))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 340 vs. \(2 (147) = 294\).
time = 0.97, size = 340, normalized size = 2.06 \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^{6} 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^{6} 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^{6} 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^{6} f} + \frac {2 \, {\left (210 \, a^{3} d^{3} \tan \left (f x + e\right )^{3} - 70 \, a^{3} d^{3} \tan \left (f x + e\right )^{2} - 63 \, a^{3} d^{3} \tan \left (f x + e\right ) - 15 \, a^{3} d^{3}\right )}}{105 \, \sqrt {d \tan \left (f x + e\right )} d^{7} f \tan \left (f x + e\right )^{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))^(9/2),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^6*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^6*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^6*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^6*f) + 2/105*(210*a^3*d^3*tan(f*x + e)^3 - 70*a^3*d^3*tan(f*x + e)^2 - 63*a^3*d^3*tan(f*x + e) - 15*a
^3*d^3)/(sqrt(d*tan(f*x + e))*d^7*f*tan(f*x + e)^3)

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Mupad [B]
time = 5.75, size = 130, normalized size = 0.79 \begin {gather*} -\frac {-4\,d\,a^3\,{\mathrm {tan}\left (e+f\,x\right )}^3+\frac {4\,d\,a^3\,{\mathrm {tan}\left (e+f\,x\right )}^2}{3}+\frac {6\,d\,a^3\,\mathrm {tan}\left (e+f\,x\right )}{5}+\frac {2\,d\,a^3}{7}}{d^2\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{7/2}}-\frac {2\,\sqrt {2}\,a^3\,\mathrm {atanh}\left (\frac {32\,\sqrt {2}\,a^6\,d^{9/2}\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{32\,a^6\,d^5\,f+32\,a^6\,d^5\,f\,\mathrm {tan}\left (e+f\,x\right )}\right )}{d^{9/2}\,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))^(9/2),x)

[Out]

- ((2*a^3*d)/7 + (4*a^3*d*tan(e + f*x)^2)/3 - 4*a^3*d*tan(e + f*x)^3 + (6*a^3*d*tan(e + f*x))/5)/(d^2*f*(d*tan
(e + f*x))^(7/2)) - (2*2^(1/2)*a^3*atanh((32*2^(1/2)*a^6*d^(9/2)*f*(d*tan(e + f*x))^(1/2))/(32*a^6*d^5*f + 32*
a^6*d^5*f*tan(e + f*x))))/(d^(9/2)*f)

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