3.505 \(\int \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=382 \[ -\frac {b \log \left (-\sqrt {2} \sqrt {\sqrt {a^2+b^2}+a} \sqrt {a+b \tan (c+d x)}+\sqrt {a^2+b^2}+a+b \tan (c+d x)\right )}{2 \sqrt {2} d \sqrt {\sqrt {a^2+b^2}+a}}+\frac {b \log \left (\sqrt {2} \sqrt {\sqrt {a^2+b^2}+a} \sqrt {a+b \tan (c+d x)}+\sqrt {a^2+b^2}+a+b \tan (c+d x)\right )}{2 \sqrt {2} d \sqrt {\sqrt {a^2+b^2}+a}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a^2+b^2}+a}-\sqrt {2} \sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {a^2+b^2}}}\right )}{\sqrt {2} d \sqrt {a-\sqrt {a^2+b^2}}}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a^2+b^2}+a}+\sqrt {2} \sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {a^2+b^2}}}\right )}{\sqrt {2} d \sqrt {a-\sqrt {a^2+b^2}}}+\frac {2 (a+b \tan (c+d x))^{3/2}}{3 b d} \]

[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)+2/3*(a+b*
tan(d*x+c))^(3/2)/b/d

________________________________________________________________________________________

Rubi [A]  time = 0.31, antiderivative size = 382, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3543, 3485, 700, 1129, 634, 618, 206, 628} \[ -\frac {b \log \left (-\sqrt {2} \sqrt {\sqrt {a^2+b^2}+a} \sqrt {a+b \tan (c+d x)}+\sqrt {a^2+b^2}+a+b \tan (c+d x)\right )}{2 \sqrt {2} d \sqrt {\sqrt {a^2+b^2}+a}}+\frac {b \log \left (\sqrt {2} \sqrt {\sqrt {a^2+b^2}+a} \sqrt {a+b \tan (c+d x)}+\sqrt {a^2+b^2}+a+b \tan (c+d x)\right )}{2 \sqrt {2} d \sqrt {\sqrt {a^2+b^2}+a}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a^2+b^2}+a}-\sqrt {2} \sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {a^2+b^2}}}\right )}{\sqrt {2} d \sqrt {a-\sqrt {a^2+b^2}}}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a^2+b^2}+a}+\sqrt {2} \sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {a^2+b^2}}}\right )}{\sqrt {2} d \sqrt {a-\sqrt {a^2+b^2}}}+\frac {2 (a+b \tan (c+d x))^{3/2}}{3 b d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^2*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]])/S
qrt[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*L
og[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) + (2*(a + b*Tan[c + d*x])^(3/2))/(3*b*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]

Rule 3543

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
(d^2*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e
+ f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[m, -1] &&  !(EqQ[m, 2] && EqQ
[a, 0])

Rubi steps

\begin {align*} \int \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)} \, dx &=\frac {2 (a+b \tan (c+d x))^{3/2}}{3 b d}-\int \sqrt {a+b \tan (c+d x)} \, dx\\ &=\frac {2 (a+b \tan (c+d x))^{3/2}}{3 b d}-\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 (a+b \tan (c+d x))^{3/2}}{3 b 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 {2 (a+b \tan (c+d x))^{3/2}}{3 b 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 {2 (a+b \tan (c+d x))^{3/2}}{3 b 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 {2 (a+b \tan (c+d x))^{3/2}}{3 b 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}+\frac {2 (a+b \tan (c+d x))^{3/2}}{3 b d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.25, size = 115, normalized size = 0.30 \[ \frac {2 (a+b \tan (c+d x))^{3/2}}{3 b d}+\frac {i \sqrt {a-i b} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {i \sqrt {a+i b} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

________________________________________________________________________________________

fricas [B]  time = 0.57, size = 1586, normalized size = 4.15 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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)^2,x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

maple [A]  time = 0.27, size = 499, normalized size = 1.31 \[ \frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3 b d}-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4 d b}+\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4 d b}-\frac {b \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{4 b d}-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{4 b d}+\frac {b \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(a+b*tan(d*x+c))^(3/2)/b/d-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))+1/4/d/b*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*ln(b*ta
n(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/(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))-1/4/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))+b/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))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*tan(d*x + c) + a)*tan(d*x + c)^2, x)

________________________________________________________________________________________

mupad [B]  time = 5.47, size = 231, normalized size = 0.60 \[ \mathrm {atanh}\left (\frac {d^3\,\sqrt {-\frac {a-b\,1{}\mathrm {i}}{d^2}}\,\left (\frac {16\,\left (b^4-a^2\,b^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^2}+\frac {16\,a\,b^2\,\left (a-b\,1{}\mathrm {i}\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^2}\right )}{16\,\left (a^2\,b^3+b^5\right )}\right )\,\sqrt {-\frac {a-b\,1{}\mathrm {i}}{d^2}}+\mathrm {atanh}\left (\frac {d^3\,\sqrt {-\frac {a+b\,1{}\mathrm {i}}{d^2}}\,\left (\frac {16\,\left (b^4-a^2\,b^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^2}+\frac {16\,a\,b^2\,\left (a+b\,1{}\mathrm {i}\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^2}\right )}{16\,\left (a^2\,b^3+b^5\right )}\right )\,\sqrt {-\frac {a+b\,1{}\mathrm {i}}{d^2}}+\frac {2\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{3\,b\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^2*(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*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) + 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) + (2*(a + b*tan(c + d*x))^(3/2))/(3*b*d)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \tan {\left (c + d x \right )}} \tan ^{2}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*tan(c + d*x))*tan(c + d*x)**2, x)

________________________________________________________________________________________