3.1249 \(\int \frac{(a+b \tan (e+f x))^2}{\sqrt{c+d \tan (e+f x)}} \, dx\)

Optimal. Leaf size=134 \[ -\frac{i (a-i b)^2 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f \sqrt{c-i d}}+\frac{i (a+i b)^2 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f \sqrt{c+i d}}+\frac{2 b^2 \sqrt{c+d \tan (e+f x)}}{d f} \]

[Out]

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

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Rubi [A]  time = 0.246715, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {3543, 3539, 3537, 63, 208} \[ -\frac{i (a-i b)^2 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f \sqrt{c-i d}}+\frac{i (a+i b)^2 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f \sqrt{c+i d}}+\frac{2 b^2 \sqrt{c+d \tan (e+f x)}}{d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3543

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
(d^2*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e
+ f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[m, -1] &&  !(EqQ[m, 2] && EqQ
[a, 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 \frac{(a+b \tan (e+f x))^2}{\sqrt{c+d \tan (e+f x)}} \, dx &=\frac{2 b^2 \sqrt{c+d \tan (e+f x)}}{d f}+\int \frac{a^2-b^2+2 a b \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx\\ &=\frac{2 b^2 \sqrt{c+d \tan (e+f x)}}{d f}+\frac{1}{2} (a-i b)^2 \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx+\frac{1}{2} (a+i b)^2 \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx\\ &=\frac{2 b^2 \sqrt{c+d \tan (e+f x)}}{d f}+\frac{\left (i (a-i b)^2\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+i b)^2\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 b^2 \sqrt{c+d \tan (e+f x)}}{d f}-\frac{(a-i b)^2 \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)^2 \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-i b)^2 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{\sqrt{c-i d} f}+\frac{i (a+i b)^2 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{\sqrt{c+i d} f}+\frac{2 b^2 \sqrt{c+d \tan (e+f x)}}{d f}\\ \end{align*}

Mathematica [A]  time = 0.215893, size = 129, normalized size = 0.96 \[ \frac{-\frac{i (a-i b)^2 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{\sqrt{c-i d}}+\frac{i (a+i b)^2 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{\sqrt{c+i d}}+\frac{2 b^2 \sqrt{c+d \tan (e+f x)}}{d}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.073, size = 5680, normalized size = 42.4 \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))^2/(c+d*tan(f*x+e))^(1/2),x)

[Out]

result too large to display

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

[Out]

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

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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))^2/(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 )^{2}}{\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))**2/(c+d*tan(f*x+e))**(1/2),x)

[Out]

Integral((a + b*tan(e + f*x))**2/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 )}^{2}}{\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))^2/(c+d*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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