3.13.41 \(\int (a+b \tan (e+f x))^3 (c+d \tan (e+f x))^{5/2} \, dx\) [1241]

Optimal. Leaf size=322 \[ \frac {(i a+b)^3 (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}-\frac {(i a-b)^3 (c+i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 \left (2 a^3 c d-6 a b^2 c d+3 a^2 b \left (c^2-d^2\right )-b^3 \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}+\frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f} \]

[Out]

(I*a+b)^3*(c-I*d)^(5/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f-(I*a-b)^3*(c+I*d)^(5/2)*arctanh((c+d*t
an(f*x+e))^(1/2)/(c+I*d)^(1/2))/f+2*(2*a^3*c*d-6*a*b^2*c*d+3*a^2*b*(c^2-d^2)-b^3*(c^2-d^2))*(c+d*tan(f*x+e))^(
1/2)/f+2/3*(a^3*d+3*a^2*b*c-3*a*b^2*d-b^3*c)*(c+d*tan(f*x+e))^(3/2)/f+2/5*b*(3*a^2-b^2)*(c+d*tan(f*x+e))^(5/2)
/f-4/63*b^2*(-10*a*d+b*c)*(c+d*tan(f*x+e))^(7/2)/d^2/f+2/9*b^2*(a+b*tan(f*x+e))*(c+d*tan(f*x+e))^(7/2)/d/f

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Rubi [A]
time = 0.63, antiderivative size = 322, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3647, 3711, 3609, 3620, 3618, 65, 214} \begin {gather*} \frac {2 b \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 \left (2 a^3 c d+3 a^2 b \left (c^2-d^2\right )-6 a b^2 c d-b^3 \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 \left (a^3 d+3 a^2 b c-3 a b^2 d-b^3 c\right ) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}+\frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac {(b+i a)^3 (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}-\frac {(-b+i a)^3 (c+i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^3*(c + d*Tan[e + f*x])^(5/2),x]

[Out]

((I*a + b)^3*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f - ((I*a - b)^3*(c + I*d)^(5/2)
*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/f + (2*(2*a^3*c*d - 6*a*b^2*c*d + 3*a^2*b*(c^2 - d^2) - b^3*
(c^2 - d^2))*Sqrt[c + d*Tan[e + f*x]])/f + (2*(3*a^2*b*c - b^3*c + a^3*d - 3*a*b^2*d)*(c + d*Tan[e + f*x])^(3/
2))/(3*f) + (2*b*(3*a^2 - b^2)*(c + d*Tan[e + f*x])^(5/2))/(5*f) - (4*b^2*(b*c - 10*a*d)*(c + d*Tan[e + f*x])^
(7/2))/(63*d^2*f) + (2*b^2*(a + b*Tan[e + f*x])*(c + d*Tan[e + f*x])^(7/2))/(9*d*f)

Rule 65

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 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 3609

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 3618

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

Rule 3620

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

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 (a+b \tan (e+f x))^3 (c+d \tan (e+f x))^{5/2} \, dx &=\frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac {2 \int (c+d \tan (e+f x))^{5/2} \left (\frac {1}{2} \left (9 a^3 d-2 b^2 \left (b c+\frac {7 a d}{2}\right )\right )+\frac {9}{2} b \left (3 a^2-b^2\right ) d \tan (e+f x)-b^2 (b c-10 a d) \tan ^2(e+f x)\right ) \, dx}{9 d}\\ &=-\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}+\frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac {2 \int (c+d \tan (e+f x))^{5/2} \left (\frac {9}{2} a \left (a^2-3 b^2\right ) d+\frac {9}{2} b \left (3 a^2-b^2\right ) d \tan (e+f x)\right ) \, dx}{9 d}\\ &=\frac {2 b \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}+\frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac {2 \int (c+d \tan (e+f x))^{3/2} \left (\frac {9}{2} d \left (a^3 c-3 a b^2 c-3 a^2 b d+b^3 d\right )+\frac {9}{2} d \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right ) \tan (e+f x)\right ) \, dx}{9 d}\\ &=\frac {2 \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}+\frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac {2 \int \sqrt {c+d \tan (e+f x)} \left (-\frac {9}{2} d \left (6 a^2 b c d-2 b^3 c d-a^3 \left (c^2-d^2\right )+3 a b^2 \left (c^2-d^2\right )\right )+\frac {9}{2} d \left (2 a^3 c d-6 a b^2 c d+3 a^2 b \left (c^2-d^2\right )-b^3 \left (c^2-d^2\right )\right ) \tan (e+f x)\right ) \, dx}{9 d}\\ &=\frac {2 \left (2 a^3 c d-6 a b^2 c d+3 a^2 b \left (c^2-d^2\right )-b^3 \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}+\frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac {2 \int \frac {\frac {9}{2} d (a c-b d) \left (a^2 c^2-3 b^2 c^2-8 a b c d-3 a^2 d^2+b^2 d^2\right )+\frac {9}{2} d (b c+a d) \left (3 a^2 c^2-b^2 c^2-8 a b c d-a^2 d^2+3 b^2 d^2\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{9 d}\\ &=\frac {2 \left (2 a^3 c d-6 a b^2 c d+3 a^2 b \left (c^2-d^2\right )-b^3 \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}+\frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac {1}{2} \left ((a-i b)^3 (c-i d)^3\right ) \int \frac {1+i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx+\frac {1}{2} \left ((a+i b)^3 (c+i d)^3\right ) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx\\ &=\frac {2 \left (2 a^3 c d-6 a b^2 c d+3 a^2 b \left (c^2-d^2\right )-b^3 \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}+\frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\left ((i a+b)^3 (c-i d)^3\right ) \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 f}+\frac {\left ((i a-b)^3 (c+i d)^3\right ) \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 f}\\ &=\frac {2 \left (2 a^3 c d-6 a b^2 c d+3 a^2 b \left (c^2-d^2\right )-b^3 \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}+\frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\left ((a-i b)^3 (c-i d)^3\right ) \text {Subst}\left (\int \frac {1}{-1-\frac {i c}{d}+\frac {i x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{d f}-\frac {\left ((a+i b)^3 (c+i d)^3\right ) \text {Subst}\left (\int \frac {1}{-1+\frac {i c}{d}-\frac {i x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{d f}\\ &=\frac {(i a+b)^3 (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}-\frac {(i a-b)^3 (c+i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 \left (2 a^3 c d-6 a b^2 c d+3 a^2 b \left (c^2-d^2\right )-b^3 \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 b \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}+\frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}\\ \end {align*}

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Mathematica [A]
time = 6.21, size = 413, normalized size = 1.28 \begin {gather*} \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac {2 \left (-\frac {2 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{7 d f}+\frac {i \left (\frac {9}{2} a \left (a^2-3 b^2\right ) d-\frac {9}{2} i b \left (3 a^2-b^2\right ) d\right ) \left (\frac {2}{5} (c+d \tan (e+f x))^{5/2}+(c-i d) \left (\frac {2}{3} (c+d \tan (e+f x))^{3/2}+(c-i d) \left (\frac {2 (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{-c+i d}+2 \sqrt {c+d \tan (e+f x)}\right )\right )\right )}{2 f}-\frac {i \left (\frac {9}{2} a \left (a^2-3 b^2\right ) d+\frac {9}{2} i b \left (3 a^2-b^2\right ) d\right ) \left (\frac {2}{5} (c+d \tan (e+f x))^{5/2}+(c+i d) \left (\frac {2}{3} (c+d \tan (e+f x))^{3/2}+(c+i d) \left (\frac {2 (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{-c-i d}+2 \sqrt {c+d \tan (e+f x)}\right )\right )\right )}{2 f}\right )}{9 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])^3*(c + d*Tan[e + f*x])^(5/2),x]

[Out]

(2*b^2*(a + b*Tan[e + f*x])*(c + d*Tan[e + f*x])^(7/2))/(9*d*f) + (2*((-2*b^2*(b*c - 10*a*d)*(c + d*Tan[e + f*
x])^(7/2))/(7*d*f) + ((I/2)*((9*a*(a^2 - 3*b^2)*d)/2 - ((9*I)/2)*b*(3*a^2 - b^2)*d)*((2*(c + d*Tan[e + f*x])^(
5/2))/5 + (c - I*d)*((2*(c + d*Tan[e + f*x])^(3/2))/3 + (c - I*d)*((2*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e
+ f*x]]/Sqrt[c - I*d]])/(-c + I*d) + 2*Sqrt[c + d*Tan[e + f*x]]))))/f - ((I/2)*((9*a*(a^2 - 3*b^2)*d)/2 + ((9
*I)/2)*b*(3*a^2 - b^2)*d)*((2*(c + d*Tan[e + f*x])^(5/2))/5 + (c + I*d)*((2*(c + d*Tan[e + f*x])^(3/2))/3 + (c
+ I*d)*((2*(c + I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/(-c - I*d) + 2*Sqrt[c + d*Tan[e +
f*x]]))))/f))/(9*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(2490\) vs. \(2(286)=572\).
time = 0.50, size = 2491, normalized size = 7.74

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2491\)
default	\(\text {Expression too large to display}\)	\(2491\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(f*x+e))^3*(c+d*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/f/d^2*(1/9*b^3*(c+d*tan(f*x+e))^(9/2)+3/7*a*b^2*d*(c+d*tan(f*x+e))^(7/2)-1/7*b^3*c*(c+d*tan(f*x+e))^(7/2)+3/
5*a^2*b*d^2*(c+d*tan(f*x+e))^(5/2)-1/5*b^3*d^2*(c+d*tan(f*x+e))^(5/2)+1/3*a^3*d^3*(c+d*tan(f*x+e))^(3/2)+a^2*b
*c*d^2*(c+d*tan(f*x+e))^(3/2)-a*b^2*d^3*(c+d*tan(f*x+e))^(3/2)-1/3*b^3*c*d^2*(c+d*tan(f*x+e))^(3/2)+2*a^3*c*d^
3*(c+d*tan(f*x+e))^(1/2)+3*a^2*b*c^2*d^2*(c+d*tan(f*x+e))^(1/2)-3*a^2*b*d^4*(c+d*tan(f*x+e))^(1/2)-6*a*b^2*c*d
^3*(c+d*tan(f*x+e))^(1/2)-b^3*c^2*d^2*(c+d*tan(f*x+e))^(1/2)+b^3*d^4*(c+d*tan(f*x+e))^(1/2)-d^2*(1/4/d*(1/2*((
2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a^3*c^2-(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a^3*d^2-6*(
2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a^2*b*c*d-3*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a*b^2*c
^2+3*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a*b^2*d^2+2*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*b
^3*c*d-(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^3*c^3+3*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^3*c*d^2+9*(2*(c^2+d^2)^(1/2)+2*
c)^(1/2)*a^2*b*c^2*d-3*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2*b*d^3+3*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*b^2*c^3-9*(2*
(c^2+d^2)^(1/2)+2*c)^(1/2)*a*b^2*c*d^2-3*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^3*c^2*d+(2*(c^2+d^2)^(1/2)+2*c)^(1/2)
*b^3*d^3)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))+2*(4*(c^2+d^
2)^(1/2)*a^3*c*d^2+6*(c^2+d^2)^(1/2)*a^2*b*c^2*d-6*(c^2+d^2)^(1/2)*a^2*b*d^3-12*(c^2+d^2)^(1/2)*a*b^2*c*d^2-2*
(c^2+d^2)^(1/2)*b^3*c^2*d+2*(c^2+d^2)^(1/2)*b^3*d^3-1/2*((2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a^3*c^2
-(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a^3*d^2-6*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a^2*b*c
*d-3*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a*b^2*c^2+3*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a
*b^2*d^2+2*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*b^3*c*d-(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^3*c^3+3*(2*(c
^2+d^2)^(1/2)+2*c)^(1/2)*a^3*c*d^2+9*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2*b*c^2*d-3*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)
*a^2*b*d^3+3*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*b^2*c^3-9*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*b^2*c*d^2-3*(2*(c^2+d^2
)^(1/2)+2*c)^(1/2)*b^3*c^2*d+(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^3*d^3)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2
)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/
2)))+1/4/d*(-1/2*((2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a^3*c^2-(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2
)^(1/2)*a^3*d^2-6*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a^2*b*c*d-3*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2
+d^2)^(1/2)*a*b^2*c^2+3*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a*b^2*d^2+2*(2*(c^2+d^2)^(1/2)+2*c)^(1/2
)*(c^2+d^2)^(1/2)*b^3*c*d-(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^3*c^3+3*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^3*c*d^2+9*(2
*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2*b*c^2*d-3*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2*b*d^3+3*(2*(c^2+d^2)^(1/2)+2*c)^(1
/2)*a*b^2*c^3-9*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*b^2*c*d^2-3*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^3*c^2*d+(2*(c^2+d^
2)^(1/2)+2*c)^(1/2)*b^3*d^3)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^
(1/2))+2*(-4*(c^2+d^2)^(1/2)*a^3*c*d^2-6*(c^2+d^2)^(1/2)*a^2*b*c^2*d+6*(c^2+d^2)^(1/2)*a^2*b*d^3+12*(c^2+d^2)^
(1/2)*a*b^2*c*d^2+2*(c^2+d^2)^(1/2)*b^3*c^2*d-2*(c^2+d^2)^(1/2)*b^3*d^3+1/2*((2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^
2+d^2)^(1/2)*a^3*c^2-(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a^3*d^2-6*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^
2+d^2)^(1/2)*a^2*b*c*d-3*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a*b^2*c^2+3*(2*(c^2+d^2)^(1/2)+2*c)^(1/
2)*(c^2+d^2)^(1/2)*a*b^2*d^2+2*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*b^3*c*d-(2*(c^2+d^2)^(1/2)+2*c)^(
1/2)*a^3*c^3+3*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^3*c*d^2+9*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2*b*c^2*d-3*(2*(c^2+d
^2)^(1/2)+2*c)^(1/2)*a^2*b*d^3+3*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*b^2*c^3-9*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*b^2
*c*d^2-3*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^3*c^2*d+(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^3*d^3)*(2*(c^2+d^2)^(1/2)+2*c
)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2
+d^2)^(1/2)-2*c)^(1/2)))))

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^3*(c+d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^3*(c+d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \tan {\left (e + f x \right )}\right )^{3} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))**3*(c+d*tan(f*x+e))**(5/2),x)

[Out]

Integral((a + b*tan(e + f*x))**3*(c + d*tan(e + f*x))**(5/2), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^3*(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choi
ce was done

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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