3.532 \(\int \frac{\tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx\)
Optimal. Leaf size=87 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d \sqrt{a-i b}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d \sqrt{a+i b}} \]
[Out]
-(ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]]/(Sqrt[a - I*b]*d)) - ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a
+ I*b]]/(Sqrt[a + I*b]*d)
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Rubi [A] time = 0.137044, antiderivative size = 87, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used =
{3539, 3537, 63, 208} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d \sqrt{a-i b}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d \sqrt{a+i b}} \]
Antiderivative was successfully verified.
[In]
Int[Tan[c + d*x]/Sqrt[a + b*Tan[c + d*x]],x]
[Out]
-(ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]]/(Sqrt[a - I*b]*d)) - ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a
+ I*b]]/(Sqrt[a + I*b]*d)
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{\tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx &=\frac{1}{2} i \int \frac{1-i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx-\frac{1}{2} i \int \frac{1+i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a-i b x}} \, dx,x,i \tan (c+d x)\right )}{2 d}+\frac{\operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a+i b x}} \, dx,x,-i \tan (c+d x)\right )}{2 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \frac{1}{-1+\frac{i a}{b}-\frac{i x^2}{b}} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{b d}+\frac{i \operatorname{Subst}\left (\int \frac{1}{-1-\frac{i a}{b}+\frac{i x^2}{b}} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{b d}\\ &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{\sqrt{a-i b} d}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{\sqrt{a+i b} d}\\ \end{align*}
Mathematica [A] time = 0.0630739, size = 107, normalized size = 1.23 \[ \frac{\sqrt{a-i b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )}{d (-a+i b)}+\frac{\sqrt{a+i b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d (-a-i b)} \]
Antiderivative was successfully verified.
[In]
Integrate[Tan[c + d*x]/Sqrt[a + b*Tan[c + d*x]],x]
[Out]
(Sqrt[a - I*b]*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/((-a + I*b)*d) + (Sqrt[a + I*b]*ArcTanh[Sqrt[a
+ b*Tan[c + d*x]]/Sqrt[a + I*b]])/((-a - I*b)*d)
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Maple [B] time = 0.052, size = 479, normalized size = 5.5 \begin{align*} -{\frac{1}{4\,d}\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}\ln \left ( b\tan \left ( dx+c \right ) +a+\sqrt{a+b\tan \left ( dx+c \right ) }\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}+\sqrt{{a}^{2}+{b}^{2}} \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}}-{\frac{a}{d}\arctan \left ({ \left ( 2\,\sqrt{a+b\tan \left ( dx+c \right ) }+\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a} \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}} \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}{\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}}+{\frac{1}{d}\arctan \left ({ \left ( 2\,\sqrt{a+b\tan \left ( dx+c \right ) }+\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a} \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}} \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}}+{\frac{1}{4\,d}\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}\ln \left ( \sqrt{a+b\tan \left ( dx+c \right ) }\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}-b\tan \left ( dx+c \right ) -a-\sqrt{{a}^{2}+{b}^{2}} \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}}+{\frac{a}{d}\arctan \left ({ \left ( \sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}-2\,\sqrt{a+b\tan \left ( dx+c \right ) } \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}} \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}{\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}}-{\frac{1}{d}\arctan \left ({ \left ( \sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}-2\,\sqrt{a+b\tan \left ( dx+c \right ) } \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}} \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(d*x+c)/(a+b*tan(d*x+c))^(1/2),x)
[Out]
-1/4/d/(a^2+b^2)^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/
2)+2*a)^(1/2)+(a^2+b^2)^(1/2))-1/d/(a^2+b^2)^(1/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1
/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a+1/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((
2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))+1/4/d/(a^2+b^2)^(1/2)*(
2*(a^2+b^2)^(1/2)+2*a)^(1/2)*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^
(1/2))+1/d/(a^2+b^2)^(1/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+
c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a-1/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^
(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))
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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(tan(d*x+c)/(a+b*tan(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 2.92242, size = 4791, normalized size = 55.07 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)/(a+b*tan(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
-sqrt(2)*(a^2 + b^2)*d^4*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*sqrt(b^2/((a^4 +
2*a^2*b^2 + b^4)*d^4))*(1/((a^2 + b^2)*d^4))^(3/4)*arctan(-((a^4 + 2*a^2*b^2 + b^4)*d^4*sqrt(b^2/((a^4 + 2*a^
2*b^2 + b^4)*d^4))*sqrt(1/((a^2 + b^2)*d^4)) + (a^3 + a*b^2)*d^2*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - sqr
t(2)*((a^5 + 2*a^3*b^2 + a*b^4)*d^7*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*sqrt(1/((a^2 + b^2)*d^4)) + (a^4 +
2*a^2*b^2 + b^4)*d^5*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)))*sqrt(((a^2 + b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4))
*cos(d*x + c) + sqrt(2)*((a^2 + b^2)*d^3*sqrt(1/((a^2 + b^2)*d^4))*cos(d*x + c) + a*d*cos(d*x + c))*sqrt((a*co
s(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2
)*(1/((a^2 + b^2)*d^4))^(1/4) + a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1
/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*(1/((a^2 + b^2)*d^4))^(3/4) + sqrt(2)*((a^5 + 2*a^3*b^2 + a*b^4)*d^7*sqr
t(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*sqrt(1/((a^2 + b^2)*d^4)) + (a^4 + 2*a^2*b^2 + b^4)*d^5*sqrt(b^2/((a^4 +
2*a^2*b^2 + b^4)*d^4)))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/(
(a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*(1/((a^2 + b^2)*d^4))^(3/4))/b^2) - sqrt(2)*(a^2 + b^2)*d^4*sqrt(-((a^3 +
a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*(1/((a^2 + b^2)
*d^4))^(3/4)*arctan(((a^4 + 2*a^2*b^2 + b^4)*d^4*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*sqrt(1/((a^2 + b^2)*d
^4)) + (a^3 + a*b^2)*d^2*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + sqrt(2)*((a^5 + 2*a^3*b^2 + a*b^4)*d^7*sqrt
(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*sqrt(1/((a^2 + b^2)*d^4)) + (a^4 + 2*a^2*b^2 + b^4)*d^5*sqrt(b^2/((a^4 + 2
*a^2*b^2 + b^4)*d^4)))*sqrt(((a^2 + b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4))*cos(d*x + c) - sqrt(2)*((a^2 + b^2)*d^3
*sqrt(1/((a^2 + b^2)*d^4))*cos(d*x + c) + a*d*cos(d*x + c))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c
))*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*(1/((a^2 + b^2)*d^4))^(1/4) + a*cos(d*
x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*(1
/((a^2 + b^2)*d^4))^(3/4) - sqrt(2)*((a^5 + 2*a^3*b^2 + a*b^4)*d^7*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*sqr
t(1/((a^2 + b^2)*d^4)) + (a^4 + 2*a^2*b^2 + b^4)*d^5*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)))*sqrt((a*cos(d*x
+ c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*(1/(
(a^2 + b^2)*d^4))^(3/4))/b^2) - 1/4*sqrt(2)*(a*d^2*sqrt(1/((a^2 + b^2)*d^4)) + 1)*sqrt(-((a^3 + a*b^2)*d^2*sqr
t(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*(1/((a^2 + b^2)*d^4))^(1/4)*log(((a^2 + b^2)*d^2*sqrt(1/((a^2 + b^2)*
d^4))*cos(d*x + c) + sqrt(2)*((a^2 + b^2)*d^3*sqrt(1/((a^2 + b^2)*d^4))*cos(d*x + c) + a*d*cos(d*x + c))*sqrt(
(a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2
)/b^2)*(1/((a^2 + b^2)*d^4))^(1/4) + a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c)) + 1/4*sqrt(2)*(a*d^2*sqrt(
1/((a^2 + b^2)*d^4)) + 1)*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*(1/((a^2 + b^2)
*d^4))^(1/4)*log(((a^2 + b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4))*cos(d*x + c) - sqrt(2)*((a^2 + b^2)*d^3*sqrt(1/((a
^2 + b^2)*d^4))*cos(d*x + c) + a*d*cos(d*x + c))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-((
a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*(1/((a^2 + b^2)*d^4))^(1/4) + a*cos(d*x + c) + b*
sin(d*x + c))/cos(d*x + c))
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan{\left (c + d x \right )}}{\sqrt{a + b \tan{\left (c + d x \right )}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)/(a+b*tan(d*x+c))**(1/2),x)
[Out]
Integral(tan(c + d*x)/sqrt(a + b*tan(c + d*x)), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (d x + c\right )}{\sqrt{b \tan \left (d x + c\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)/(a+b*tan(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate(tan(d*x + c)/sqrt(b*tan(d*x + c) + a), x)