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

Optimal. Leaf size=87 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3539, 3537, 63, 208} \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 3537

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

Rule 3539

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

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 107, normalized size = 1.23 \[ \frac {\sqrt {a-i b} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d (-a+i b)}+\frac {\sqrt {a+i b} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d (-a-i b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.00, size = 2122, normalized size = 24.39 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(2)*(a^2 + b^2)*d^4*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*sqrt(b^2/((a^4 +
2*a^2*b^2 + b^4)*d^4))*(1/((a^2 + b^2)*d^4))^(3/4)*arctan(-((a^4 + 2*a^2*b^2 + b^4)*d^4*sqrt(b^2/((a^4 + 2*a^
2*b^2 + b^4)*d^4))*sqrt(1/((a^2 + b^2)*d^4)) + (a^3 + a*b^2)*d^2*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - sqr
t(2)*((a^5 + 2*a^3*b^2 + a*b^4)*d^7*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*sqrt(1/((a^2 + b^2)*d^4)) + (a^4 +
2*a^2*b^2 + b^4)*d^5*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)))*sqrt(((a^2 + b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4))
*cos(d*x + c) + sqrt(2)*((a^2 + b^2)*d^3*sqrt(1/((a^2 + b^2)*d^4))*cos(d*x + c) + a*d*cos(d*x + c))*sqrt((a*co
s(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2
)*(1/((a^2 + b^2)*d^4))^(1/4) + a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1
/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*(1/((a^2 + b^2)*d^4))^(3/4) + sqrt(2)*((a^5 + 2*a^3*b^2 + a*b^4)*d^7*sqr
t(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*sqrt(1/((a^2 + b^2)*d^4)) + (a^4 + 2*a^2*b^2 + b^4)*d^5*sqrt(b^2/((a^4 +
2*a^2*b^2 + b^4)*d^4)))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/(
(a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*(1/((a^2 + b^2)*d^4))^(3/4))/b^2) - sqrt(2)*(a^2 + b^2)*d^4*sqrt(-((a^3 +
a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*(1/((a^2 + b^2)
*d^4))^(3/4)*arctan(((a^4 + 2*a^2*b^2 + b^4)*d^4*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*sqrt(1/((a^2 + b^2)*d
^4)) + (a^3 + a*b^2)*d^2*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + sqrt(2)*((a^5 + 2*a^3*b^2 + a*b^4)*d^7*sqrt
(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*sqrt(1/((a^2 + b^2)*d^4)) + (a^4 + 2*a^2*b^2 + b^4)*d^5*sqrt(b^2/((a^4 + 2
*a^2*b^2 + b^4)*d^4)))*sqrt(((a^2 + b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4))*cos(d*x + c) - sqrt(2)*((a^2 + b^2)*d^3
*sqrt(1/((a^2 + b^2)*d^4))*cos(d*x + c) + a*d*cos(d*x + c))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c
))*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*(1/((a^2 + b^2)*d^4))^(1/4) + a*cos(d*
x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*(1
/((a^2 + b^2)*d^4))^(3/4) - sqrt(2)*((a^5 + 2*a^3*b^2 + a*b^4)*d^7*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*sqr
t(1/((a^2 + b^2)*d^4)) + (a^4 + 2*a^2*b^2 + b^4)*d^5*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)))*sqrt((a*cos(d*x
+ c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*(1/(
(a^2 + b^2)*d^4))^(3/4))/b^2) - 1/4*sqrt(2)*(a*d^2*sqrt(1/((a^2 + b^2)*d^4)) + 1)*sqrt(-((a^3 + a*b^2)*d^2*sqr
t(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*(1/((a^2 + b^2)*d^4))^(1/4)*log(((a^2 + b^2)*d^2*sqrt(1/((a^2 + b^2)*
d^4))*cos(d*x + c) + sqrt(2)*((a^2 + b^2)*d^3*sqrt(1/((a^2 + b^2)*d^4))*cos(d*x + c) + a*d*cos(d*x + c))*sqrt(
(a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2
)/b^2)*(1/((a^2 + b^2)*d^4))^(1/4) + a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c)) + 1/4*sqrt(2)*(a*d^2*sqrt(
1/((a^2 + b^2)*d^4)) + 1)*sqrt(-((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*(1/((a^2 + b^2)
*d^4))^(1/4)*log(((a^2 + b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4))*cos(d*x + c) - sqrt(2)*((a^2 + b^2)*d^3*sqrt(1/((a
^2 + b^2)*d^4))*cos(d*x + c) + a*d*cos(d*x + c))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-((
a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) - a^2 - b^2)/b^2)*(1/((a^2 + b^2)*d^4))^(1/4) + a*cos(d*x + c) + b*
sin(d*x + c))/cos(d*x + c))

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

[Out]

Timed out

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maple [B]  time = 0.29, size = 479, normalized size = 5.51 \[ -\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4 d \sqrt {a^{2}+b^{2}}}-\frac {\arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) a}{d \sqrt {a^{2}+b^{2}}\, \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{4 d \sqrt {a^{2}+b^{2}}}+\frac {\arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) a}{d \sqrt {a^{2}+b^{2}}\, \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {\arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(tan(d*x+c)/(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 = 4.75, size = 781, normalized size = 8.98 \[ -2\,\mathrm {atanh}\left (\frac {32\,b^2\,\sqrt {\frac {a}{4\,a^2\,d^2+4\,b^2\,d^2}-\frac {b\,1{}\mathrm {i}}{4\,a^2\,d^2+4\,b^2\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{\frac {16\,b^2}{d}-\frac {64\,a^2\,b^2\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {a\,b^3\,d^2\,64{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}}-\frac {128\,a^2\,b^2\,\sqrt {\frac {a}{4\,a^2\,d^2+4\,b^2\,d^2}-\frac {b\,1{}\mathrm {i}}{4\,a^2\,d^2+4\,b^2\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{\frac {64\,b^4}{d}+\frac {64\,a^2\,b^2}{d}-\frac {256\,a^2\,b^4\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {a^3\,b^3\,d^2\,256{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}-\frac {256\,a^4\,b^2\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {a\,b^5\,d^2\,256{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}}+\frac {a\,b^3\,\sqrt {\frac {a}{4\,a^2\,d^2+4\,b^2\,d^2}-\frac {b\,1{}\mathrm {i}}{4\,a^2\,d^2+4\,b^2\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,128{}\mathrm {i}}{\frac {64\,b^4}{d}+\frac {64\,a^2\,b^2}{d}-\frac {256\,a^2\,b^4\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {a^3\,b^3\,d^2\,256{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}-\frac {256\,a^4\,b^2\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {a\,b^5\,d^2\,256{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}}\right )\,\sqrt {\frac {a-b\,1{}\mathrm {i}}{4\,a^2\,d^2+4\,b^2\,d^2}}-\mathrm {atanh}\left (\frac {16\,b^2\,\sqrt {\frac {1}{a\,d^2-b\,d^2\,1{}\mathrm {i}}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{\frac {16\,b^2}{d}-\frac {16\,a\,b^2\,d^2}{a\,d^3-b\,d^3\,1{}\mathrm {i}}}+\frac {16\,a\,b^2\,\sqrt {\frac {1}{a\,d^2-b\,d^2\,1{}\mathrm {i}}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{\frac {b^3\,16{}\mathrm {i}}{d}-\frac {16\,a\,b^2}{d}-\frac {a\,b^3\,d^2\,16{}\mathrm {i}}{a\,d^3-b\,d^3\,1{}\mathrm {i}}+\frac {16\,a^2\,b^2\,d^2}{a\,d^3-b\,d^3\,1{}\mathrm {i}}}\right )\,\sqrt {\frac {1}{a\,d^2-b\,d^2\,1{}\mathrm {i}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- 2*atanh((32*b^2*(a/(4*a^2*d^2 + 4*b^2*d^2) - (b*1i)/(4*a^2*d^2 + 4*b^2*d^2))^(1/2)*(a + b*tan(c + d*x))^(1/2
))/((16*b^2)/d - (64*a^2*b^2*d^2)/(4*a^2*d^3 + 4*b^2*d^3) + (a*b^3*d^2*64i)/(4*a^2*d^3 + 4*b^2*d^3)) - (128*a^
2*b^2*(a/(4*a^2*d^2 + 4*b^2*d^2) - (b*1i)/(4*a^2*d^2 + 4*b^2*d^2))^(1/2)*(a + b*tan(c + d*x))^(1/2))/((64*b^4)
/d + (64*a^2*b^2)/d - (256*a^2*b^4*d^2)/(4*a^2*d^3 + 4*b^2*d^3) + (a^3*b^3*d^2*256i)/(4*a^2*d^3 + 4*b^2*d^3) -
(256*a^4*b^2*d^2)/(4*a^2*d^3 + 4*b^2*d^3) + (a*b^5*d^2*256i)/(4*a^2*d^3 + 4*b^2*d^3)) + (a*b^3*(a/(4*a^2*d^2
+ 4*b^2*d^2) - (b*1i)/(4*a^2*d^2 + 4*b^2*d^2))^(1/2)*(a + b*tan(c + d*x))^(1/2)*128i)/((64*b^4)/d + (64*a^2*b^
2)/d - (256*a^2*b^4*d^2)/(4*a^2*d^3 + 4*b^2*d^3) + (a^3*b^3*d^2*256i)/(4*a^2*d^3 + 4*b^2*d^3) - (256*a^4*b^2*d
^2)/(4*a^2*d^3 + 4*b^2*d^3) + (a*b^5*d^2*256i)/(4*a^2*d^3 + 4*b^2*d^3)))*((a - b*1i)/(4*a^2*d^2 + 4*b^2*d^2))^
(1/2) - atanh((16*b^2*(1/(a*d^2 - b*d^2*1i))^(1/2)*(a + b*tan(c + d*x))^(1/2))/((16*b^2)/d - (16*a*b^2*d^2)/(a
*d^3 - b*d^3*1i)) + (16*a*b^2*(1/(a*d^2 - b*d^2*1i))^(1/2)*(a + b*tan(c + d*x))^(1/2))/((b^3*16i)/d - (16*a*b^
2)/d - (a*b^3*d^2*16i)/(a*d^3 - b*d^3*1i) + (16*a^2*b^2*d^2)/(a*d^3 - b*d^3*1i)))*(1/(a*d^2 - b*d^2*1i))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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