3.13.67 \(\int \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)} \, dx\) [1267]

Optimal. Leaf size=218 \[ -\frac {i \sqrt {a-i b} \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f}+\frac {i \sqrt {a+i b} \sqrt {c+i d} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f}+\frac {2 \sqrt {b} \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{f} \]

[Out]

-I*arctanh((c-I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a-I*b)^(1/2)/(c+d*tan(f*x+e))^(1/2))*(a-I*b)^(1/2)*(c-I*d)^(1
/2)/f+I*arctanh((c+I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a+I*b)^(1/2)/(c+d*tan(f*x+e))^(1/2))*(a+I*b)^(1/2)*(c+I*
d)^(1/2)/f+2*arctanh(d^(1/2)*(a+b*tan(f*x+e))^(1/2)/b^(1/2)/(c+d*tan(f*x+e))^(1/2))*b^(1/2)*d^(1/2)/f

________________________________________________________________________________________

Rubi [A]
time = 0.69, antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3656, 920, 65, 223, 212, 6857, 95, 214} \begin {gather*} -\frac {i \sqrt {a-i b} \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f}+\frac {i \sqrt {a+i b} \sqrt {c+i d} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f}+\frac {2 \sqrt {b} \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]],x]

[Out]

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

Rule 65

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 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 214

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

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 920

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[e*(g/c), In
t[(d + e*x)^(m - 1)*(f + g*x)^(n - 1), x], x] + Dist[1/c, Int[Simp[c*d*f - a*e*g + (c*e*f + c*d*g)*x, x]*(d +
e*x)^(m - 1)*((f + g*x)^(n - 1)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0]
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[m, 0] && GtQ[n, 0]

Rule 3656

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Wit
h[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + b*ff*x)^m*((c + d*ff*x)^n/(1 + ff^2*x^2)), x]
, x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] &&
NeQ[c^2 + d^2, 0]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 2.15, size = 261, normalized size = 1.20 \begin {gather*} \frac {i \sqrt {-a+i b} \sqrt {-c+i d} \tanh ^{-1}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )+i \sqrt {a+i b} \sqrt {c+i d} \tanh ^{-1}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )+\frac {2 \sqrt {d} \sqrt {b c-a d} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b c-a d}}\right ) \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}}}{\sqrt {c+d \tan (e+f x)}}}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]],x]

[Out]

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

________________________________________________________________________________________

Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \sqrt {a +b \tan \left (f x +e \right )}\, \sqrt {c +d \tan \left (f x +e \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a + b \tan {\left (e + f x \right )}} \sqrt {c + d \tan {\left (e + f x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choi
ce was done

________________________________________________________________________________________

Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

________________________________________________________________________________________