3.44 \(\int \frac{1}{(a+b \tan (c+d \sqrt{x}))^2} \, dx\)

Optimal. Leaf size=204 \[ -\frac{2 i a b \text{PolyLog}\left (2,-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{d^2 \left (a^2+b^2\right )^2}+\frac{2 b \left (2 a d \sqrt{x}+b\right ) \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{d^2 \left (a^2+b^2\right )^2}-\frac{2 b \sqrt{x}}{d \left (a^2+b^2\right ) \left (a+b \tan \left (c+d \sqrt{x}\right )\right )}+\frac{\left (2 a d \sqrt{x}+b\right )^2}{2 a d^2 (a+i b) \left (a^2+b^2\right )}-\frac{x}{a^2+b^2} \]

[Out]

(b + 2*a*d*Sqrt[x])^2/(2*a*(a + I*b)*(a^2 + b^2)*d^2) - x/(a^2 + b^2) + (2*b*(b + 2*a*d*Sqrt[x])*Log[1 + ((a^2
+ b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2])/((a^2 + b^2)^2*d^2) - ((2*I)*a*b*PolyLog[2, -(((a^2 + b^2)*E^
((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/((a^2 + b^2)^2*d^2) - (2*b*Sqrt[x])/((a^2 + b^2)*d*(a + b*Tan[c + d*Sq
rt[x]]))

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Rubi [A]  time = 0.255395, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {3739, 3733, 3732, 2190, 2279, 2391} \[ -\frac{2 i a b \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{d^2 \left (a^2+b^2\right )^2}+\frac{2 b \left (2 a d \sqrt{x}+b\right ) \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{d^2 \left (a^2+b^2\right )^2}-\frac{2 b \sqrt{x}}{d \left (a^2+b^2\right ) \left (a+b \tan \left (c+d \sqrt{x}\right )\right )}+\frac{\left (2 a d \sqrt{x}+b\right )^2}{2 a d^2 (a+i b) \left (a^2+b^2\right )}-\frac{x}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*Sqrt[x]])^(-2),x]

[Out]

(b + 2*a*d*Sqrt[x])^2/(2*a*(a + I*b)*(a^2 + b^2)*d^2) - x/(a^2 + b^2) + (2*b*(b + 2*a*d*Sqrt[x])*Log[1 + ((a^2
+ b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2])/((a^2 + b^2)^2*d^2) - ((2*I)*a*b*PolyLog[2, -(((a^2 + b^2)*E^
((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/((a^2 + b^2)^2*d^2) - (2*b*Sqrt[x])/((a^2 + b^2)*d*(a + b*Tan[c + d*Sq
rt[x]]))

Rule 3739

Int[((a_.) + (b_.)*Tan[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*Ta
n[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[1/n, 0] && IntegerQ[p]

Rule 3733

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

Rule 3732

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

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b \tan \left (c+d \sqrt{x}\right )\right )^2} \, dx &=2 \operatorname{Subst}\left (\int \frac{x}{(a+b \tan (c+d x))^2} \, dx,x,\sqrt{x}\right )\\ &=-\frac{x}{a^2+b^2}-\frac{2 b \sqrt{x}}{\left (a^2+b^2\right ) d \left (a+b \tan \left (c+d \sqrt{x}\right )\right )}+\frac{2 \operatorname{Subst}\left (\int \frac{b+2 a d x}{a+b \tan (c+d x)} \, dx,x,\sqrt{x}\right )}{\left (a^2+b^2\right ) d}\\ &=\frac{\left (b+2 a d \sqrt{x}\right )^2}{2 a (a+i b) \left (a^2+b^2\right ) d^2}-\frac{x}{a^2+b^2}-\frac{2 b \sqrt{x}}{\left (a^2+b^2\right ) d \left (a+b \tan \left (c+d \sqrt{x}\right )\right )}+\frac{(4 i b) \operatorname{Subst}\left (\int \frac{e^{2 i (c+d x)} (b+2 a d x)}{(a+i b)^2+\left (a^2+b^2\right ) e^{2 i (c+d x)}} \, dx,x,\sqrt{x}\right )}{\left (a^2+b^2\right ) d}\\ &=\frac{\left (b+2 a d \sqrt{x}\right )^2}{2 a (a+i b) \left (a^2+b^2\right ) d^2}-\frac{x}{a^2+b^2}+\frac{2 b \left (b+2 a d \sqrt{x}\right ) \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right )^2 d^2}-\frac{2 b \sqrt{x}}{\left (a^2+b^2\right ) d \left (a+b \tan \left (c+d \sqrt{x}\right )\right )}-\frac{(4 a b) \operatorname{Subst}\left (\int \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt{x}\right )}{\left (a^2+b^2\right )^2 d}\\ &=\frac{\left (b+2 a d \sqrt{x}\right )^2}{2 a (a+i b) \left (a^2+b^2\right ) d^2}-\frac{x}{a^2+b^2}+\frac{2 b \left (b+2 a d \sqrt{x}\right ) \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right )^2 d^2}-\frac{2 b \sqrt{x}}{\left (a^2+b^2\right ) d \left (a+b \tan \left (c+d \sqrt{x}\right )\right )}+\frac{(2 i a b) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{\left (a^2+b^2\right ) x}{(a+i b)^2}\right )}{x} \, dx,x,e^{2 i \left (c+d \sqrt{x}\right )}\right )}{\left (a^2+b^2\right )^2 d^2}\\ &=\frac{\left (b+2 a d \sqrt{x}\right )^2}{2 a (a+i b) \left (a^2+b^2\right ) d^2}-\frac{x}{a^2+b^2}+\frac{2 b \left (b+2 a d \sqrt{x}\right ) \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right )^2 d^2}-\frac{2 i a b \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right )^2 d^2}-\frac{2 b \sqrt{x}}{\left (a^2+b^2\right ) d \left (a+b \tan \left (c+d \sqrt{x}\right )\right )}\\ \end{align*}

Mathematica [B]  time = 6.51845, size = 517, normalized size = 2.53 \[ \frac{\sec ^2\left (c+d \sqrt{x}\right ) \left (a \cos \left (c+d \sqrt{x}\right )+b \sin \left (c+d \sqrt{x}\right )\right ) \left (-2 a b \left (a \cos \left (c+d \sqrt{x}\right )+b \sin \left (c+d \sqrt{x}\right )\right ) \left (a \left (i \text{PolyLog}\left (2,e^{2 i \left (\tan ^{-1}\left (\frac{a}{b}\right )+c+d \sqrt{x}\right )}\right )-i \left (\pi -2 \tan ^{-1}\left (\frac{a}{b}\right )\right ) \left (c+d \sqrt{x}\right )-2 \left (\tan ^{-1}\left (\frac{a}{b}\right )+c+d \sqrt{x}\right ) \log \left (1-e^{2 i \left (\tan ^{-1}\left (\frac{a}{b}\right )+c+d \sqrt{x}\right )}\right )+2 \tan ^{-1}\left (\frac{a}{b}\right ) \log \left (\sin \left (\tan ^{-1}\left (\frac{a}{b}\right )+c+d \sqrt{x}\right )\right )-\pi \log \left (1+e^{-2 i \left (c+d \sqrt{x}\right )}\right )+\pi \log \left (\cos \left (c+d \sqrt{x}\right )\right )\right )+b \sqrt{\frac{a^2}{b^2}+1} e^{i \tan ^{-1}\left (\frac{a}{b}\right )} \left (c+d \sqrt{x}\right )^2\right )+2 b^2 d \sqrt{x} \left (a^2+b^2\right ) \sin \left (c+d \sqrt{x}\right )-a \left (a^2+b^2\right ) \left (c-d \sqrt{x}\right ) \left (c+d \sqrt{x}\right ) \left (a \cos \left (c+d \sqrt{x}\right )+b \sin \left (c+d \sqrt{x}\right )\right )-2 b^2 \left (a \cos \left (c+d \sqrt{x}\right )+b \sin \left (c+d \sqrt{x}\right )\right ) \left (b \left (c+d \sqrt{x}\right )-a \log \left (a \cos \left (c+d \sqrt{x}\right )+b \sin \left (c+d \sqrt{x}\right )\right )\right )+4 a b c \left (a \cos \left (c+d \sqrt{x}\right )+b \sin \left (c+d \sqrt{x}\right )\right ) \left (b \left (c+d \sqrt{x}\right )-a \log \left (a \cos \left (c+d \sqrt{x}\right )+b \sin \left (c+d \sqrt{x}\right )\right )\right )\right )}{a d^2 \left (a^2+b^2\right )^2 \left (a+b \tan \left (c+d \sqrt{x}\right )\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*Tan[c + d*Sqrt[x]])^(-2),x]

[Out]

(Sec[c + d*Sqrt[x]]^2*(a*Cos[c + d*Sqrt[x]] + b*Sin[c + d*Sqrt[x]])*(2*b^2*(a^2 + b^2)*d*Sqrt[x]*Sin[c + d*Sqr
t[x]] - a*(a^2 + b^2)*(c - d*Sqrt[x])*(c + d*Sqrt[x])*(a*Cos[c + d*Sqrt[x]] + b*Sin[c + d*Sqrt[x]]) - 2*b^2*(b
*(c + d*Sqrt[x]) - a*Log[a*Cos[c + d*Sqrt[x]] + b*Sin[c + d*Sqrt[x]]])*(a*Cos[c + d*Sqrt[x]] + b*Sin[c + d*Sqr
t[x]]) + 4*a*b*c*(b*(c + d*Sqrt[x]) - a*Log[a*Cos[c + d*Sqrt[x]] + b*Sin[c + d*Sqrt[x]]])*(a*Cos[c + d*Sqrt[x]
] + b*Sin[c + d*Sqrt[x]]) - 2*a*b*(Sqrt[1 + a^2/b^2]*b*E^(I*ArcTan[a/b])*(c + d*Sqrt[x])^2 + a*((-I)*(c + d*Sq
rt[x])*(Pi - 2*ArcTan[a/b]) - Pi*Log[1 + E^((-2*I)*(c + d*Sqrt[x]))] - 2*(c + d*Sqrt[x] + ArcTan[a/b])*Log[1 -
E^((2*I)*(c + d*Sqrt[x] + ArcTan[a/b]))] + Pi*Log[Cos[c + d*Sqrt[x]]] + 2*ArcTan[a/b]*Log[Sin[c + d*Sqrt[x] +
ArcTan[a/b]]] + I*PolyLog[2, E^((2*I)*(c + d*Sqrt[x] + ArcTan[a/b]))]))*(a*Cos[c + d*Sqrt[x]] + b*Sin[c + d*S
qrt[x]])))/(a*(a^2 + b^2)^2*d^2*(a + b*Tan[c + d*Sqrt[x]])^2)

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Maple [F]  time = 0.263, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\tan \left ( c+d\sqrt{x} \right ) \right ) ^{-2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 2.8726, size = 1355, normalized size = 6.64 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*((a^3 - I*a^2*b + a*b^2 - I*b^3)*d^2*x + (2*I*a*b^2 - 2*b^3 - 2*(-I*a*b^2 - b^3)*cos(2*d*sqrt(x) + 2*c) - (2
*a*b^2 - 2*I*b^3)*sin(2*d*sqrt(x) + 2*c))*arctan2(-b*cos(2*d*sqrt(x) + 2*c) + a*sin(2*d*sqrt(x) + 2*c) + b, a*
cos(2*d*sqrt(x) + 2*c) + b*sin(2*d*sqrt(x) + 2*c) + a) + ((-4*I*a^2*b - 4*a*b^2)*d*sqrt(x)*cos(2*d*sqrt(x) + 2
*c) + 4*(a^2*b - I*a*b^2)*d*sqrt(x)*sin(2*d*sqrt(x) + 2*c) + (-4*I*a^2*b + 4*a*b^2)*d*sqrt(x))*arctan2((2*a*b*
cos(2*d*sqrt(x) + 2*c) - (a^2 - b^2)*sin(2*d*sqrt(x) + 2*c))/(a^2 + b^2), (2*a*b*sin(2*d*sqrt(x) + 2*c) + a^2
+ b^2 + (a^2 - b^2)*cos(2*d*sqrt(x) + 2*c))/(a^2 + b^2)) + ((a^3 - 3*I*a^2*b - 3*a*b^2 + I*b^3)*d^2*x - 4*(I*a
*b^2 + b^3)*d*sqrt(x))*cos(2*d*sqrt(x) + 2*c) + (-2*I*a^2*b + 2*a*b^2 + (-2*I*a^2*b - 2*a*b^2)*cos(2*d*sqrt(x)
+ 2*c) + 2*(a^2*b - I*a*b^2)*sin(2*d*sqrt(x) + 2*c))*dilog((I*a + b)*e^(2*I*d*sqrt(x) + 2*I*c)/(-I*a + b)) +
(a*b^2 + I*b^3 + (a*b^2 - I*b^3)*cos(2*d*sqrt(x) + 2*c) + (I*a*b^2 + b^3)*sin(2*d*sqrt(x) + 2*c))*log((a^2 + b
^2)*cos(2*d*sqrt(x) + 2*c)^2 + 4*a*b*sin(2*d*sqrt(x) + 2*c) + (a^2 + b^2)*sin(2*d*sqrt(x) + 2*c)^2 + a^2 + b^2
+ 2*(a^2 - b^2)*cos(2*d*sqrt(x) + 2*c)) + (2*(a^2*b - I*a*b^2)*d*sqrt(x)*cos(2*d*sqrt(x) + 2*c) + (2*I*a^2*b
+ 2*a*b^2)*d*sqrt(x)*sin(2*d*sqrt(x) + 2*c) + 2*(a^2*b + I*a*b^2)*d*sqrt(x))*log(((a^2 + b^2)*cos(2*d*sqrt(x)
+ 2*c)^2 + 4*a*b*sin(2*d*sqrt(x) + 2*c) + (a^2 + b^2)*sin(2*d*sqrt(x) + 2*c)^2 + a^2 + b^2 + 2*(a^2 - b^2)*cos
(2*d*sqrt(x) + 2*c))/(a^2 + b^2)) + ((I*a^3 + 3*a^2*b - 3*I*a*b^2 - b^3)*d^2*x + (4*a*b^2 - 4*I*b^3)*d*sqrt(x)
)*sin(2*d*sqrt(x) + 2*c))/((2*a^5 - 2*I*a^4*b + 4*a^3*b^2 - 4*I*a^2*b^3 + 2*a*b^4 - 2*I*b^5)*d^2*cos(2*d*sqrt(
x) + 2*c) + (2*I*a^5 + 2*a^4*b + 4*I*a^3*b^2 + 4*a^2*b^3 + 2*I*a*b^4 + 2*b^5)*d^2*sin(2*d*sqrt(x) + 2*c) + (2*
a^5 + 2*I*a^4*b + 4*a^3*b^2 + 4*I*a^2*b^3 + 2*a*b^4 + 2*I*b^5)*d^2)

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Fricas [B]  time = 1.68268, size = 2016, normalized size = 9.88 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*b^3*d*sqrt(x) - (a^3 - a*b^2)*d^2*x + (a^3 - a*b^2)*d^2 - (-I*a*b^2*tan(d*sqrt(x) + c) - I*a^2*b)*dilog(-(
(2*I*a*b + 2*b^2)*tan(d*sqrt(x) + c)^2 + 2*a^2 - 2*I*a*b + (2*I*a^2 + 4*a*b - 2*I*b^2)*tan(d*sqrt(x) + c))/((a
^2 + b^2)*tan(d*sqrt(x) + c)^2 + a^2 + b^2) + 1) - (I*a*b^2*tan(d*sqrt(x) + c) + I*a^2*b)*dilog(-((-2*I*a*b +
2*b^2)*tan(d*sqrt(x) + c)^2 + 2*a^2 + 2*I*a*b + (-2*I*a^2 + 4*a*b + 2*I*b^2)*tan(d*sqrt(x) + c))/((a^2 + b^2)*
tan(d*sqrt(x) + c)^2 + a^2 + b^2) + 1) - 2*(a^2*b*d*sqrt(x) + a^2*b*c + (a*b^2*d*sqrt(x) + a*b^2*c)*tan(d*sqrt
(x) + c))*log(((2*I*a*b + 2*b^2)*tan(d*sqrt(x) + c)^2 + 2*a^2 - 2*I*a*b + (2*I*a^2 + 4*a*b - 2*I*b^2)*tan(d*sq
rt(x) + c))/((a^2 + b^2)*tan(d*sqrt(x) + c)^2 + a^2 + b^2)) - 2*(a^2*b*d*sqrt(x) + a^2*b*c + (a*b^2*d*sqrt(x)
+ a*b^2*c)*tan(d*sqrt(x) + c))*log(((-2*I*a*b + 2*b^2)*tan(d*sqrt(x) + c)^2 + 2*a^2 + 2*I*a*b + (-2*I*a^2 + 4*
a*b + 2*I*b^2)*tan(d*sqrt(x) + c))/((a^2 + b^2)*tan(d*sqrt(x) + c)^2 + a^2 + b^2)) + (2*a^2*b*c - a*b^2 + (2*a
*b^2*c - b^3)*tan(d*sqrt(x) + c))*log(((I*a*b + b^2)*tan(d*sqrt(x) + c)^2 - a^2 + I*a*b + (I*a^2 + I*b^2)*tan(
d*sqrt(x) + c))/(tan(d*sqrt(x) + c)^2 + 1)) + (2*a^2*b*c - a*b^2 + (2*a*b^2*c - b^3)*tan(d*sqrt(x) + c))*log((
(I*a*b - b^2)*tan(d*sqrt(x) + c)^2 + a^2 + I*a*b + (I*a^2 + I*b^2)*tan(d*sqrt(x) + c))/(tan(d*sqrt(x) + c)^2 +
1)) - (2*a*b^2*d*sqrt(x) + (a^2*b - b^3)*d^2*x - (a^2*b - b^3)*d^2)*tan(d*sqrt(x) + c))/((a^4*b + 2*a^2*b^3 +
b^5)*d^2*tan(d*sqrt(x) + c) + (a^5 + 2*a^3*b^2 + a*b^4)*d^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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