3.1241 \(\int (a+b \tan (e+f x))^3 (c+d \tan (e+f x))^{5/2} \, dx\)

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.907082, antiderivative size = 322, normalized size of antiderivative = 1., 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 = {3566, 3630, 3528, 3539, 3537, 63, 208} \[ \frac{2 \left (3 a^2 b \left (c^2-d^2\right )+2 a^3 c d-6 a b^2 c d-b^3 \left (c^2-d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 b \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{5/2}}{5 f}+\frac{2 \left (3 a^2 b c+a^3 d-3 a b^2 d-b^3 c\right ) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac{4 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac{(b+i a)^3 (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f}-\frac{(-b+i a)^3 (c+i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f} \]

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 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]

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 3539

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 3537

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^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 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 208

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

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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 6.20127, size = 413, normalized size = 1.28 \[ \frac{2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac{2 \left (\frac{i \left (\frac{9}{2} a d \left (a^2-3 b^2\right )-\frac{9}{2} i b d \left (3 a^2-b^2\right )\right ) \left (\frac{2}{5} (c+d \tan (e+f x))^{5/2}+(c-i d) \left (\frac{2}{3} (c+d \tan (e+f x))^{3/2}+(c-i d) \left (2 \sqrt{c+d \tan (e+f x)}+\frac{2 (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{-c+i d}\right )\right )\right )}{2 f}-\frac{i \left (\frac{9}{2} a d \left (a^2-3 b^2\right )+\frac{9}{2} i b d \left (3 a^2-b^2\right )\right ) \left (\frac{2}{5} (c+d \tan (e+f x))^{5/2}+(c+i d) \left (\frac{2}{3} (c+d \tan (e+f x))^{3/2}+(c+i d) \left (2 \sqrt{c+d \tan (e+f x)}+\frac{2 (c+i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{-c-i d}\right )\right )\right )}{2 f}-\frac{2 b^2 (b c-10 a d) (c+d \tan (e+f x))^{7/2}}{7 d f}\right )}{9 d} \]

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]  time = 0.079, size = 5053, normalized size = 15.7 \begin{align*} \text{output too large to display} \end{align*}

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)

[Out]

result too large to display

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

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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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

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]

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