3.1254 \(\int \frac{(a+b \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=216 \[ -\frac{2 b \left (a d (2 b c-a d)-b^2 \left (2 c^2+d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{d^2 f \left (c^2+d^2\right )}-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))}{d f \left (c^2+d^2\right ) \sqrt{c+d \tan (e+f x)}}-\frac{(-b+i a)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{3/2}}+\frac{(b+i a)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.503254, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3565, 3630, 3539, 3537, 63, 208} \[ -\frac{2 b \left (a d (2 b c-a d)-b^2 \left (2 c^2+d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{d^2 f \left (c^2+d^2\right )}-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))}{d f \left (c^2+d^2\right ) \sqrt{c+d \tan (e+f x)}}-\frac{(-b+i a)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{3/2}}+\frac{(b+i a)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3565

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

Mathematica [C]  time = 1.80615, size = 287, normalized size = 1.33 \[ \frac{\frac{\left (3 a^2 b c+a^3 (-d)+3 a b^2 d-b^3 c\right ) \left ((d-i c) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c+d \tan (e+f x)}{c-i d}\right )+(d+i c) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c+d \tan (e+f x)}{c+i d}\right )\right )}{\left (c^2+d^2\right ) \sqrt{c+d \tan (e+f x)}}-i b \left (3 a^2-b^2\right ) \left (\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{\sqrt{c-i d}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{\sqrt{c+i d}}\right )+\frac{4 b^2 (b c-2 a d)}{d \sqrt{c+d \tan (e+f x)}}+\frac{2 b^2 (a+b \tan (e+f x))}{\sqrt{c+d \tan (e+f x)}}}{d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*b*(3*a^2 - b^2)*(ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]]/Sqrt[c - I*d] - ArcTanh[Sqrt[c + d*Tan[
e + f*x]]/Sqrt[c + I*d]]/Sqrt[c + I*d]) + (4*b^2*(b*c - 2*a*d))/(d*Sqrt[c + d*Tan[e + f*x]]) + ((3*a^2*b*c - b
^3*c - a^3*d + 3*a*b^2*d)*(((-I)*c + d)*Hypergeometric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c - I*d)] + (I*c
+ d)*Hypergeometric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c + I*d)]))/((c^2 + d^2)*Sqrt[c + d*Tan[e + f*x]])
+ (2*b^2*(a + b*Tan[e + f*x]))/Sqrt[c + d*Tan[e + f*x]])/(d*f)

________________________________________________________________________________________

Maple [B]  time = 0.082, size = 16343, normalized size = 75.7 \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))^(3/2),x)

[Out]

result too large to display

________________________________________________________________________________________

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))^(3/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

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))^(3/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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