3.611 \(\int \frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\)

Optimal. Leaf size=115 \[ -\frac {i \sqrt {-b+i a} \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {i \sqrt {b+i a} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3575, 910, 93, 205, 208} \[ -\frac {i \sqrt {-b+i a} \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {i \sqrt {b+i a} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Sqrt[I*a - b]*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - (I*Sqrt[I*a + b]*
ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d

Rule 93

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 205

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

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 910

Int[((d_.) + (e_.)*(x_))^(m_)/(Sqrt[(f_.) + (g_.)*(x_)]*((a_.) + (c_.)*(x_)^2)), x_Symbol] :> Int[ExpandIntegr
and[1/(Sqrt[d + e*x]*Sqrt[f + g*x]), (d + e*x)^(m + 1/2)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] &
& NeQ[c*d^2 + a*e^2, 0] && IGtQ[m + 1/2, 0]

Rule 3575

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.10, size = 123, normalized size = 1.07 \[ \frac {(-1)^{3/4} \left (\sqrt {-a+i b} \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\sqrt {a+i b} \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1)^(3/4)*(Sqrt[-a + I*b]*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] -
Sqrt[a + I*b]*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]))/d

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

maple [B]  time = 0.73, size = 1084638, normalized size = 9431.63 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {b \tan \left (d x + c\right ) + a}}{\sqrt {\tan \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 6.08, size = 224, normalized size = 1.95 \[ -\mathrm {atan}\left (\frac {\sqrt {a}\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {b+a\,1{}\mathrm {i}}{d^2}}-d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {b+a\,1{}\mathrm {i}}{d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{a+b\,\mathrm {tan}\left (c+d\,x\right )-\sqrt {a}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}\right )\,\sqrt {-\frac {b+a\,1{}\mathrm {i}}{d^2}}\,1{}\mathrm {i}-\mathrm {atan}\left (\frac {d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {-b+a\,1{}\mathrm {i}}{d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}-\sqrt {a}\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {-b+a\,1{}\mathrm {i}}{d^2}}}{a+b\,\mathrm {tan}\left (c+d\,x\right )-\sqrt {a}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}\right )\,\sqrt {\frac {-b+a\,1{}\mathrm {i}}{d^2}}\,1{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- atan((a^(1/2)*d*tan(c + d*x)^(1/2)*(-(a*1i + b)/d^2)^(1/2) - d*tan(c + d*x)^(1/2)*(-(a*1i + b)/d^2)^(1/2)*(a
+ b*tan(c + d*x))^(1/2))/(a + b*tan(c + d*x) - a^(1/2)*(a + b*tan(c + d*x))^(1/2)))*(-(a*1i + b)/d^2)^(1/2)*1
i - atan((d*tan(c + d*x)^(1/2)*((a*1i - b)/d^2)^(1/2)*(a + b*tan(c + d*x))^(1/2) - a^(1/2)*d*tan(c + d*x)^(1/2
)*((a*1i - b)/d^2)^(1/2))/(a + b*tan(c + d*x) - a^(1/2)*(a + b*tan(c + d*x))^(1/2)))*((a*1i - b)/d^2)^(1/2)*1i

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________