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

Optimal. Leaf size=208 \[ -\frac {(a+b) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d \sqrt {e}}+\frac {(a+b) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d \sqrt {e}}-\frac {(a-b) \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d \sqrt {e}}+\frac {(a-b) \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d \sqrt {e}} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 208, normalized size of antiderivative = 1.00, 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} \[ -\frac {(a+b) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d \sqrt {e}}+\frac {(a+b) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d \sqrt {e}}-\frac {(a-b) \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d \sqrt {e}}+\frac {(a-b) \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d \sqrt {e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

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 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 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 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]

Rubi steps

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

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Mathematica [C]  time = 0.08, size = 83, normalized size = 0.40 \[ -\frac {\sqrt [4]{-1} \sqrt {\tan (c+d x)} \left ((a-i b) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+(a+i b) \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )\right )}{d \sqrt {e \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

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fricas [B]  time = 0.56, size = 2884, normalized size = 13.87 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.23, size = 327, normalized size = 1.57 \[ \frac {a \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {e \tan \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \tan \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{4 d e}+\frac {a \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \tan \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{2 d e}-\frac {a \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \tan \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{2 d e}+\frac {b \sqrt {2}\, \ln \left (\frac {e \tan \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \tan \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{4 d \left (e^{2}\right )^{\frac {1}{4}}}+\frac {b \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \tan \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{2 d \left (e^{2}\right )^{\frac {1}{4}}}-\frac {b \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \tan \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{2 d \left (e^{2}\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 168, normalized size = 0.81 \[ \frac {\frac {2 \, \sqrt {2} {\left (a + b\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} + 2 \, \sqrt {e \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}} + \frac {2 \, \sqrt {2} {\left (a + b\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} - 2 \, \sqrt {e \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}} + \frac {\sqrt {2} {\left (a - b\right )} \log \left (e \tan \left (d x + c\right ) + \sqrt {2} \sqrt {e \tan \left (d x + c\right )} \sqrt {e} + e\right )}{\sqrt {e}} - \frac {\sqrt {2} {\left (a - b\right )} \log \left (e \tan \left (d x + c\right ) - \sqrt {2} \sqrt {e \tan \left (d x + c\right )} \sqrt {e} + e\right )}{\sqrt {e}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.78, size = 120, normalized size = 0.58 \[ \frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {e\,\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {e}}\right )}{d\,\sqrt {e}}-\frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {e\,\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {e}}\right )}{d\,\sqrt {e}}-\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {e\,\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {e}}\right )\,1{}\mathrm {i}}{d\,\sqrt {e}}-\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {e\,\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {e}}\right )\,1{}\mathrm {i}}{d\,\sqrt {e}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((-1)^(1/4)*b*atan(((-1)^(1/4)*(e*tan(c + d*x))^(1/2))/e^(1/2)))/(d*e^(1/2)) - ((-1)^(1/4)*a*atanh(((-1)^(1/4)
*(e*tan(c + d*x))^(1/2))/e^(1/2))*1i)/(d*e^(1/2)) - ((-1)^(1/4)*a*atan(((-1)^(1/4)*(e*tan(c + d*x))^(1/2))/e^(
1/2))*1i)/(d*e^(1/2)) - ((-1)^(1/4)*b*atanh(((-1)^(1/4)*(e*tan(c + d*x))^(1/2))/e^(1/2)))/(d*e^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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