3.1248 \(\int \frac{(a+b \tan (e+f x))^3}{\sqrt{c+d \tan (e+f x)}} \, dx\)

Optimal. Leaf size=178 \[ -\frac{4 b^2 (b c-4 a d) \sqrt{c+d \tan (e+f x)}}{3 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}{3 d f}-\frac{(-b+i a)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f \sqrt{c+i d}}+\frac{(b+i a)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f \sqrt{c-i d}} \]

[Out]

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

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Rubi [A]  time = 0.430224, antiderivative size = 178, 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 = {3566, 3630, 3539, 3537, 63, 208} \[ -\frac{4 b^2 (b c-4 a d) \sqrt{c+d \tan (e+f x)}}{3 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}{3 d f}-\frac{(-b+i a)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f \sqrt{c+i d}}+\frac{(b+i a)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f \sqrt{c-i d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I*a + b)^3*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f) - ((I*a - b)^3*ArcTanh[Sqrt[c
+ d*Tan[e + f*x]]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f) - (4*b^2*(b*c - 4*a*d)*Sqrt[c + d*Tan[e + f*x]])/(3*d^2*f)
+ (2*b^2*(a + b*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]])/(3*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 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}{\sqrt{c+d \tan (e+f x)}} \, dx &=\frac{2 b^2 (a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}{3 d f}+\frac{2 \int \frac{\frac{1}{2} \left (3 a^3 d-b^2 (2 b c+a d)\right )+\frac{3}{2} b \left (3 a^2-b^2\right ) d \tan (e+f x)-b^2 (b c-4 a d) \tan ^2(e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{3 d}\\ &=-\frac{4 b^2 (b c-4 a d) \sqrt{c+d \tan (e+f x)}}{3 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}{3 d f}+\frac{2 \int \frac{\frac{3}{2} a \left (a^2-3 b^2\right ) d+\frac{3}{2} b \left (3 a^2-b^2\right ) d \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{3 d}\\ &=-\frac{4 b^2 (b c-4 a d) \sqrt{c+d \tan (e+f x)}}{3 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}{3 d f}+\frac{1}{2} (a-i b)^3 \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx+\frac{1}{2} (a+i b)^3 \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx\\ &=-\frac{4 b^2 (b c-4 a d) \sqrt{c+d \tan (e+f x)}}{3 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}{3 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 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 f}\\ &=-\frac{4 b^2 (b c-4 a d) \sqrt{c+d \tan (e+f x)}}{3 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}{3 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 )}{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 )}{d f}\\ &=\frac{(i a+b)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{\sqrt{c-i d} f}-\frac{(i a-b)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{\sqrt{c+i d} f}-\frac{4 b^2 (b c-4 a d) \sqrt{c+d \tan (e+f x)}}{3 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}{3 d f}\\ \end{align*}

Mathematica [A]  time = 1.01059, size = 178, normalized size = 1. \[ \frac{2 \left (\frac{2 b^2 (4 a d-b c) \sqrt{c+d \tan (e+f x)}}{d}+b^2 (a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}-\frac{3 i d (a-i b)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{2 \sqrt{c-i d}}+\frac{3 i d (a+i b)^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{2 \sqrt{c+i d}}\right )}{3 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.084, size = 8262, normalized size = 46.4 \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))^(1/2),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))^(1/2),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))^(1/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \tan{\left (e + f x \right )}\right )^{3}}{\sqrt{c + d \tan{\left (e + f x \right )}}}\, 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))**(1/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\sqrt{d \tan \left (f x + e\right ) + c}}\,{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))^(1/2),x, algorithm="giac")

[Out]

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