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

Optimal. Leaf size=358 \[ \frac {b \log \left (-\sqrt {2} \sqrt {\sqrt {a^2+b^2}+a} \sqrt {a+b \tan (c+d x)}+\sqrt {a^2+b^2}+a+b \tan (c+d x)\right )}{2 \sqrt {2} d \sqrt {\sqrt {a^2+b^2}+a}}-\frac {b \log \left (\sqrt {2} \sqrt {\sqrt {a^2+b^2}+a} \sqrt {a+b \tan (c+d x)}+\sqrt {a^2+b^2}+a+b \tan (c+d x)\right )}{2 \sqrt {2} d \sqrt {\sqrt {a^2+b^2}+a}}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a^2+b^2}+a}-\sqrt {2} \sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {a^2+b^2}}}\right )}{\sqrt {2} d \sqrt {a-\sqrt {a^2+b^2}}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a^2+b^2}+a}+\sqrt {2} \sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {a^2+b^2}}}\right )}{\sqrt {2} d \sqrt {a-\sqrt {a^2+b^2}}} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 358, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3485, 700, 1129, 634, 618, 206, 628} \[ \frac {b \log \left (-\sqrt {2} \sqrt {\sqrt {a^2+b^2}+a} \sqrt {a+b \tan (c+d x)}+\sqrt {a^2+b^2}+a+b \tan (c+d x)\right )}{2 \sqrt {2} d \sqrt {\sqrt {a^2+b^2}+a}}-\frac {b \log \left (\sqrt {2} \sqrt {\sqrt {a^2+b^2}+a} \sqrt {a+b \tan (c+d x)}+\sqrt {a^2+b^2}+a+b \tan (c+d x)\right )}{2 \sqrt {2} d \sqrt {\sqrt {a^2+b^2}+a}}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a^2+b^2}+a}-\sqrt {2} \sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {a^2+b^2}}}\right )}{\sqrt {2} d \sqrt {a-\sqrt {a^2+b^2}}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a^2+b^2}+a}+\sqrt {2} \sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {a^2+b^2}}}\right )}{\sqrt {2} d \sqrt {a-\sqrt {a^2+b^2}}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Tan[c + d*x]],x]

[Out]

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

Rule 206

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

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; 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 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 700

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

Rule 1129

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r = Rt[2*q - b/
c, 2]}, Dist[1/(2*c*r), Int[x^(m - 1)/(q - r*x + x^2), x], x] - Dist[1/(2*c*r), Int[x^(m - 1)/(q + r*x + x^2),
x], x]]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && GeQ[m, 1] && LtQ[m, 3] && NegQ[b^2 - 4*a*c]

Rule 3485

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[(a + x)^n/(b^2 + x^2), x], x
, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 + b^2, 0]

Rubi steps

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

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Mathematica [C]  time = 0.03, size = 87, normalized size = 0.24 \[ -\frac {i \left (\sqrt {a-i b} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )-\sqrt {a+i b} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Tan[c + d*x]],x]

[Out]

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

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fricas [B]  time = 0.49, size = 1465, normalized size = 4.09 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.29, size = 654, normalized size = 1.83 \[ \frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4 d b}-\frac {a^{2} \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d b \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4 d b}+\frac {\left (a^{2}+b^{2}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d b \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{4 b d}+\frac {a^{2} \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d b \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{4 b d}-\frac {\left (a^{2}+b^{2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d b \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/d/b*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a*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/b*a^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))-1/4/d/b*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(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/b*(a^2+b^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))-1/4/
b/d*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a*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/b*a^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))+1/4/b/d*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(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/b*(a^2+b^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))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is b-a positive, negative or zero?

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mupad [B]  time = 4.18, size = 213, normalized size = 0.59 \[ -\mathrm {atanh}\left (\frac {d^3\,\sqrt {-\frac {a-b\,1{}\mathrm {i}}{d^2}}\,\left (\frac {16\,\left (b^4-a^2\,b^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^2}+\frac {16\,a\,b^2\,\left (a-b\,1{}\mathrm {i}\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^2}\right )}{16\,\left (a^2\,b^3+b^5\right )}\right )\,\sqrt {-\frac {a-b\,1{}\mathrm {i}}{d^2}}-\mathrm {atanh}\left (\frac {d^3\,\sqrt {-\frac {a+b\,1{}\mathrm {i}}{d^2}}\,\left (\frac {16\,\left (b^4-a^2\,b^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^2}+\frac {16\,a\,b^2\,\left (a+b\,1{}\mathrm {i}\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^2}\right )}{16\,\left (a^2\,b^3+b^5\right )}\right )\,\sqrt {-\frac {a+b\,1{}\mathrm {i}}{d^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x))^(1/2),x)

[Out]

- atanh((d^3*(-(a - b*1i)/d^2)^(1/2)*((16*(b^4 - a^2*b^2)*(a + b*tan(c + d*x))^(1/2))/d^2 + (16*a*b^2*(a - b*1
i)*(a + b*tan(c + d*x))^(1/2))/d^2))/(16*(b^5 + a^2*b^3)))*(-(a - b*1i)/d^2)^(1/2) - atanh((d^3*(-(a + b*1i)/d
^2)^(1/2)*((16*(b^4 - a^2*b^2)*(a + b*tan(c + d*x))^(1/2))/d^2 + (16*a*b^2*(a + b*1i)*(a + b*tan(c + d*x))^(1/
2))/d^2))/(16*(b^5 + a^2*b^3)))*(-(a + b*1i)/d^2)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*tan(c + d*x)), x)

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