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

Optimal. Leaf size=162 \[ \frac{(a-b) \tan ^{-1}\left (1-\sqrt{2} \sqrt{-\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{(a-b) \tan ^{-1}\left (\sqrt{2} \sqrt{-\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{(a+b) \log \left (-\tan (c+d x)-\sqrt{2} \sqrt{-\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{(a+b) \log \left (-\tan (c+d x)+\sqrt{2} \sqrt{-\tan (c+d x)}+1\right )}{2 \sqrt{2} d} \]

[Out]

((a - b)*ArcTan[1 - Sqrt[2]*Sqrt[-Tan[c + d*x]]])/(Sqrt[2]*d) - ((a - b)*ArcTan[1 + Sqrt[2]*Sqrt[-Tan[c + d*x]
]])/(Sqrt[2]*d) + ((a + b)*Log[1 - Sqrt[2]*Sqrt[-Tan[c + d*x]] - Tan[c + d*x]])/(2*Sqrt[2]*d) - ((a + b)*Log[1
+ Sqrt[2]*Sqrt[-Tan[c + d*x]] - Tan[c + d*x]])/(2*Sqrt[2]*d)

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Rubi [A]  time = 0.110348, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{(a-b) \tan ^{-1}\left (1-\sqrt{2} \sqrt{-\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{(a-b) \tan ^{-1}\left (\sqrt{2} \sqrt{-\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{(a+b) \log \left (-\tan (c+d x)-\sqrt{2} \sqrt{-\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{(a+b) \log \left (-\tan (c+d x)+\sqrt{2} \sqrt{-\tan (c+d x)}+1\right )}{2 \sqrt{2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a - b)*ArcTan[1 - Sqrt[2]*Sqrt[-Tan[c + d*x]]])/(Sqrt[2]*d) - ((a - b)*ArcTan[1 + Sqrt[2]*Sqrt[-Tan[c + d*x]
]])/(Sqrt[2]*d) + ((a + b)*Log[1 - Sqrt[2]*Sqrt[-Tan[c + d*x]] - Tan[c + d*x]])/(2*Sqrt[2]*d) - ((a + b)*Log[1
+ Sqrt[2]*Sqrt[-Tan[c + d*x]] - Tan[c + d*x]])/(2*Sqrt[2]*d)

Rule 3534

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

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

Mathematica [C]  time = 0.144577, size = 82, normalized size = 0.51 \[ \frac{\sqrt [4]{-1} \tan ^{\frac{3}{2}}(c+d x) \left ((a-i b) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )+(a+i b) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )\right )}{d (-\tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1)^(1/4)*((a - I*b)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + (a + I*b)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]
])*Tan[c + d*x]^(3/2))/(d*(-Tan[c + d*x])^(3/2))

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Maple [A]  time = 0.032, size = 230, normalized size = 1.4 \begin{align*} -{\frac{a\sqrt{2}}{2\,d}\arctan \left ( 1+\sqrt{2}\sqrt{-\tan \left ( dx+c \right ) } \right ) }-{\frac{a\sqrt{2}}{2\,d}\arctan \left ( -1+\sqrt{2}\sqrt{-\tan \left ( dx+c \right ) } \right ) }-{\frac{a\sqrt{2}}{4\,d}\ln \left ({ \left ( 1+\sqrt{2}\sqrt{-\tan \left ( dx+c \right ) }-\tan \left ( dx+c \right ) \right ) \left ( 1-\sqrt{2}\sqrt{-\tan \left ( dx+c \right ) }-\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }+{\frac{b\sqrt{2}}{4\,d}\ln \left ({ \left ( 1-\sqrt{2}\sqrt{-\tan \left ( dx+c \right ) }-\tan \left ( dx+c \right ) \right ) \left ( 1+\sqrt{2}\sqrt{-\tan \left ( dx+c \right ) }-\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }+{\frac{b\sqrt{2}}{2\,d}\arctan \left ( 1+\sqrt{2}\sqrt{-\tan \left ( dx+c \right ) } \right ) }+{\frac{b\sqrt{2}}{2\,d}\arctan \left ( -1+\sqrt{2}\sqrt{-\tan \left ( dx+c \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d*a*2^(1/2)*arctan(1+2^(1/2)*(-tan(d*x+c))^(1/2))-1/2/d*a*2^(1/2)*arctan(-1+2^(1/2)*(-tan(d*x+c))^(1/2))-
1/4/d*a*2^(1/2)*ln((1+2^(1/2)*(-tan(d*x+c))^(1/2)-tan(d*x+c))/(1-2^(1/2)*(-tan(d*x+c))^(1/2)-tan(d*x+c)))+1/4/
d*b*2^(1/2)*ln((1-2^(1/2)*(-tan(d*x+c))^(1/2)-tan(d*x+c))/(1+2^(1/2)*(-tan(d*x+c))^(1/2)-tan(d*x+c)))+1/2/d*b*
2^(1/2)*arctan(1+2^(1/2)*(-tan(d*x+c))^(1/2))+1/2/d*b*2^(1/2)*arctan(-1+2^(1/2)*(-tan(d*x+c))^(1/2))

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Maxima [A]  time = 1.78247, size = 184, normalized size = 1.14 \begin{align*} -\frac{2 \, \sqrt{2}{\left (a - b\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{-\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt{2}{\left (a - b\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{-\tan \left (d x + c\right )}\right )}\right ) + \sqrt{2}{\left (a + b\right )} \log \left (\sqrt{2} \sqrt{-\tan \left (d x + c\right )} - \tan \left (d x + c\right ) + 1\right ) - \sqrt{2}{\left (a + b\right )} \log \left (-\sqrt{2} \sqrt{-\tan \left (d x + c\right )} - \tan \left (d x + c\right ) + 1\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*sqrt(2)*(a - b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(-tan(d*x + c)))) + 2*sqrt(2)*(a - b)*arctan(-1/2*
sqrt(2)*(sqrt(2) - 2*sqrt(-tan(d*x + c)))) + sqrt(2)*(a + b)*log(sqrt(2)*sqrt(-tan(d*x + c)) - tan(d*x + c) +
1) - sqrt(2)*(a + b)*log(-sqrt(2)*sqrt(-tan(d*x + c)) - tan(d*x + c) + 1))/d

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Fricas [B]  time = 2.95735, size = 5760, normalized size = 35.56 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*sqrt(2)*d^4*sqrt(-(2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) - a^4 - 2*a^2*b^2 - b^4)/(a^4 - 2*a^2*b
^2 + b^4))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(3/4)*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4)*arctan(((a^8 + 2*a^6*b^2 - 2*
a^2*b^6 - b^8)*d^4*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4) + sqrt(2)*(a*d^7*sqrt((
a^4 + 2*a^2*b^2 + b^4)/d^4)*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4) + (a^2*b + b^3)*d^5*sqrt((a^4 - 2*a^2*b^2 + b^4)
/d^4))*sqrt(-(2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) - a^4 - 2*a^2*b^2 - b^4)/(a^4 - 2*a^2*b^2 + b^4))*sq
rt(((a^6 - a^4*b^2 - a^2*b^4 + b^6)*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c) + sqrt(2)*((a^4*b - 2*a
^2*b^3 + b^5)*d^3*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c) + (a^7 - a^5*b^2 - a^3*b^4 + a*b^6)*d*cos(d*x
+ c))*sqrt(-(2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) - a^4 - 2*a^2*b^2 - b^4)/(a^4 - 2*a^2*b^2 + b^4))*sq
rt(-sin(d*x + c)/cos(d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(1/4) - (a^8 - 2*a^4*b^4 + b^8)*sin(d*x + c))/cos
(d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(3/4) + sqrt(2)*((a^5 - a*b^4)*d^7*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*
sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4) + (a^6*b + a^4*b^3 - a^2*b^5 - b^7)*d^5*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4))*s
qrt(-(2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) - a^4 - 2*a^2*b^2 - b^4)/(a^4 - 2*a^2*b^2 + b^4))*sqrt(-sin(
d*x + c)/cos(d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(3/4))/(a^12 + 2*a^10*b^2 - a^8*b^4 - 4*a^6*b^6 - a^4*b^8
+ 2*a^2*b^10 + b^12)) + 4*sqrt(2)*d^4*sqrt(-(2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) - a^4 - 2*a^2*b^2 -
b^4)/(a^4 - 2*a^2*b^2 + b^4))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(3/4)*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4)*arctan(-((
a^8 + 2*a^6*b^2 - 2*a^2*b^6 - b^8)*d^4*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4) - s
qrt(2)*(a*d^7*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4) + (a^2*b + b^3)*d^5*sqrt((a^
4 - 2*a^2*b^2 + b^4)/d^4))*sqrt(-(2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) - a^4 - 2*a^2*b^2 - b^4)/(a^4 -
2*a^2*b^2 + b^4))*sqrt(((a^6 - a^4*b^2 - a^2*b^4 + b^6)*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c) - s
qrt(2)*((a^4*b - 2*a^2*b^3 + b^5)*d^3*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c) + (a^7 - a^5*b^2 - a^3*b^
4 + a*b^6)*d*cos(d*x + c))*sqrt(-(2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) - a^4 - 2*a^2*b^2 - b^4)/(a^4 -
2*a^2*b^2 + b^4))*sqrt(-sin(d*x + c)/cos(d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(1/4) - (a^8 - 2*a^4*b^4 + b^
8)*sin(d*x + c))/cos(d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(3/4) - sqrt(2)*((a^5 - a*b^4)*d^7*sqrt((a^4 + 2*
a^2*b^2 + b^4)/d^4)*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4) + (a^6*b + a^4*b^3 - a^2*b^5 - b^7)*d^5*sqrt((a^4 - 2*a^
2*b^2 + b^4)/d^4))*sqrt(-(2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) - a^4 - 2*a^2*b^2 - b^4)/(a^4 - 2*a^2*b^
2 + b^4))*sqrt(-sin(d*x + c)/cos(d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(3/4))/(a^12 + 2*a^10*b^2 - a^8*b^4 -
4*a^6*b^6 - a^4*b^8 + 2*a^2*b^10 + b^12)) + sqrt(2)*(2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) + a^4 + 2*a^
2*b^2 + b^4)*sqrt(-(2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) - a^4 - 2*a^2*b^2 - b^4)/(a^4 - 2*a^2*b^2 + b^
4))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(1/4)*log(((a^6 - a^4*b^2 - a^2*b^4 + b^6)*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/
d^4)*cos(d*x + c) + sqrt(2)*((a^4*b - 2*a^2*b^3 + b^5)*d^3*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c) + (a
^7 - a^5*b^2 - a^3*b^4 + a*b^6)*d*cos(d*x + c))*sqrt(-(2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) - a^4 - 2*a
^2*b^2 - b^4)/(a^4 - 2*a^2*b^2 + b^4))*sqrt(-sin(d*x + c)/cos(d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(1/4) -
(a^8 - 2*a^4*b^4 + b^8)*sin(d*x + c))/cos(d*x + c)) - sqrt(2)*(2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) + a
^4 + 2*a^2*b^2 + b^4)*sqrt(-(2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) - a^4 - 2*a^2*b^2 - b^4)/(a^4 - 2*a^2
*b^2 + b^4))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(1/4)*log(((a^6 - a^4*b^2 - a^2*b^4 + b^6)*d^2*sqrt((a^4 + 2*a^2*b^
2 + b^4)/d^4)*cos(d*x + c) - sqrt(2)*((a^4*b - 2*a^2*b^3 + b^5)*d^3*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x
+ c) + (a^7 - a^5*b^2 - a^3*b^4 + a*b^6)*d*cos(d*x + c))*sqrt(-(2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) -
a^4 - 2*a^2*b^2 - b^4)/(a^4 - 2*a^2*b^2 + b^4))*sqrt(-sin(d*x + c)/cos(d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)
^(1/4) - (a^8 - 2*a^4*b^4 + b^8)*sin(d*x + c))/cos(d*x + c)))/(a^4 + 2*a^2*b^2 + b^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \tan{\left (c + d x \right )}}{\sqrt{- \tan{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.28932, size = 219, normalized size = 1.35 \begin{align*} -\frac{{\left (\sqrt{2} a - \sqrt{2} b\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{-\tan \left (d x + c\right )}\right )}\right )}{2 \, d} - \frac{{\left (\sqrt{2} a - \sqrt{2} b\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{-\tan \left (d x + c\right )}\right )}\right )}{2 \, d} - \frac{{\left (\sqrt{2} a + \sqrt{2} b\right )} \log \left (\sqrt{2} \sqrt{-\tan \left (d x + c\right )} - \tan \left (d x + c\right ) + 1\right )}{4 \, d} + \frac{{\left (\sqrt{2} a + \sqrt{2} b\right )} \log \left (-\sqrt{2} \sqrt{-\tan \left (d x + c\right )} - \tan \left (d x + c\right ) + 1\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(sqrt(2)*a - sqrt(2)*b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(-tan(d*x + c))))/d - 1/2*(sqrt(2)*a - sqrt(2
)*b)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(-tan(d*x + c))))/d - 1/4*(sqrt(2)*a + sqrt(2)*b)*log(sqrt(2)*sqrt(-
tan(d*x + c)) - tan(d*x + c) + 1)/d + 1/4*(sqrt(2)*a + sqrt(2)*b)*log(-sqrt(2)*sqrt(-tan(d*x + c)) - tan(d*x +
c) + 1)/d