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

Optimal. Leaf size=166 \[ -\frac{(a-b) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{(a-b) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{(a+b) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{(a+b) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{2 b \sqrt{\tan (c+d x)}}{d} \]

[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) + (2*b*Sqrt[Tan[c + d*x]])/d

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Rubi [A]  time = 0.113096, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3528, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{(a-b) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{(a-b) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{(a+b) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{(a+b) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{2 b \sqrt{\tan (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Tan[c + d*x]]*(a + b*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) + (2*b*Sqrt[Tan[c + d*x]])/d

Rule 3528

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

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

Mathematica [C]  time = 0.0761828, size = 79, normalized size = 0.48 \[ \frac{\sqrt [4]{-1} (b+i a) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )-(-1)^{3/4} (a+i b) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )+2 b \sqrt{\tan (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.009, size = 220, normalized size = 1.3 \begin{align*} 2\,{\frac{b\sqrt{\tan \left ( dx+c \right ) }}{d}}-{\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 ) }-{\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{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{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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.62445, size = 182, normalized size = 1.1 \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 ) + 8 \, b \sqrt{\tan \left (d x + c\right )}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c)),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*sq
rt(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) + 8*b*sqrt(tan(d*x + c)))/d

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Fricas [B]  time = 2.42376, size = 5851, normalized size = 35.25 \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)^(1/2)*(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/4*(4*sqrt(2)*d^5*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)*(b*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^3 + a*b^2)*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))*sqr
t(((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^5 - 2*a^3*
b^2 + a*b^4)*d^3*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c) + (a^6*b - a^4*b^3 - a^2*b^5 + b^7)*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))*sqr
t(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^4*b - b^5)*d^7*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*sq
rt((a^4 - 2*a^2*b^2 + b^4)/d^4) + (a^7 + a^5*b^2 - a^3*b^4 - a*b^6)*d^5*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4))*sqr
t(-(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^5*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)*(b*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^3 + a*b^2)*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) - sqrt
(2)*((a^5 - 2*a^3*b^2 + a*b^4)*d^3*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c) + (a^6*b - a^4*b^3 - a^2*b^5
+ b^7)*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)*s
in(d*x + c))/cos(d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(3/4) - sqrt(2)*((a^4*b - b^5)*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^7 + a^5*b^2 - a^3*b^4 - a*b^6)*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^3*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) + (a^4 + 2*a^2*b^
2 + b^4)*d)*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^5 - 2*a^3*b^2 + a*b^4)*d^3*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c) + (a^
6*b - a^4*b^3 - a^2*b^5 + b^7)*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^3*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) + (a^
4 + 2*a^2*b^2 + b^4)*d)*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^5 - 2*a^3*b^2 + a*b^4)*d^3*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*
x + c) + (a^6*b - a^4*b^3 - a^2*b^5 + b^7)*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)) + 8*(a^4*b + 2*a^2*b^3 + b^5)*sqrt(sin(d*x + c)/
cos(d*x + c)))/((a^4 + 2*a^2*b^2 + b^4)*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan{\left (c + d x \right )}\right ) \sqrt{\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)**(1/2)*(a+b*tan(d*x+c)),x)

[Out]

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

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out