3.1252 \(\int \frac{1}{(a+b \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}} \, dx\)

Optimal. Leaf size=244 \[ -\frac{b^2 \sqrt{c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))}-\frac{b^{3/2} \left (-5 a^2 d+4 a b c-b^2 d\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{f \left (a^2+b^2\right )^2 (b c-a d)^{3/2}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (a-i b)^2 \sqrt{c-i d}}+\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (a+i b)^2 \sqrt{c+i d}} \]

[Out]

((-I)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((a - I*b)^2*Sqrt[c - I*d]*f) + (I*ArcTanh[Sqrt[c + d*T
an[e + f*x]]/Sqrt[c + I*d]])/((a + I*b)^2*Sqrt[c + I*d]*f) - (b^(3/2)*(4*a*b*c - 5*a^2*d - b^2*d)*ArcTanh[(Sqr
t[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/((a^2 + b^2)^2*(b*c - a*d)^(3/2)*f) - (b^2*Sqrt[c + d*Tan[e +
f*x]])/((a^2 + b^2)*(b*c - a*d)*f*(a + b*Tan[e + f*x]))

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Rubi [A]  time = 0.8548, antiderivative size = 244, 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 = {3569, 3653, 3539, 3537, 63, 208, 3634} \[ -\frac{b^2 \sqrt{c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))}-\frac{b^{3/2} \left (-5 a^2 d+4 a b c-b^2 d\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{f \left (a^2+b^2\right )^2 (b c-a d)^{3/2}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (a-i b)^2 \sqrt{c-i d}}+\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (a+i b)^2 \sqrt{c+i d}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*Tan[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]]),x]

[Out]

((-I)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((a - I*b)^2*Sqrt[c - I*d]*f) + (I*ArcTanh[Sqrt[c + d*T
an[e + f*x]]/Sqrt[c + I*d]])/((a + I*b)^2*Sqrt[c + I*d]*f) - (b^(3/2)*(4*a*b*c - 5*a^2*d - b^2*d)*ArcTanh[(Sqr
t[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/((a^2 + b^2)^2*(b*c - a*d)^(3/2)*f) - (b^2*Sqrt[c + d*Tan[e +
f*x]])/((a^2 + b^2)*(b*c - a*d)*f*(a + b*Tan[e + f*x]))

Rule 3569

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 + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d)), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*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] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))

Rule 3653

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*
x])^n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 +
b^2), Int[((c + d*Tan[e + f*x])^n*(1 + Tan[e + f*x]^2))/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e,
f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0] &&  !LeQ[n, -
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]

Rule 3634

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

Rubi steps

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

Mathematica [A]  time = 2.25134, size = 258, normalized size = 1.06 \[ \frac{\frac{b^{3/2} \left (5 a^2 d-4 a b c+b^2 d\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{\left (a^2+b^2\right ) \sqrt{b c-a d}}-\frac{i \left (\frac{(a-i b)^2 (a d-b c) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{\sqrt{c+i d}}+\frac{(a+i b)^2 (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{\sqrt{c-i d}}\right )}{a^2+b^2}-\frac{b^2 \sqrt{c+d \tan (e+f x)}}{a+b \tan (e+f x)}}{f \left (a^2+b^2\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*Tan[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]]),x]

[Out]

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

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Maple [B]  time = 0.102, size = 6548, normalized size = 26.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c+d*tan(f*x+e))^(1/2)/(a+b*tan(f*x+e))^2,x)

[Out]

result too large to display

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+d*tan(f*x+e))^(1/2)/(a+b*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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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(1/(c+d*tan(f*x+e))^(1/2)/(a+b*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

Timed out

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

[Out]

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

[Out]

integrate(1/((b*tan(f*x + e) + a)^2*sqrt(d*tan(f*x + e) + c)), x)