3.631 \(\int \frac{\sqrt{a+b \sqrt{c+d x}}}{x^3} \, dx\)

Optimal. Leaf size=224 \[ \frac{b d \left (b c-a \sqrt{c+d x}\right ) \sqrt{a+b \sqrt{c+d x}}}{8 c x \left (a^2-b^2 c\right )}-\frac{b d^2 \left (2 a-3 b \sqrt{c}\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c+d x}}}{\sqrt{a-b \sqrt{c}}}\right )}{16 c^{3/2} \left (a-b \sqrt{c}\right )^{3/2}}+\frac{b d^2 \left (2 a+3 b \sqrt{c}\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c+d x}}}{\sqrt{a+b \sqrt{c}}}\right )}{16 c^{3/2} \left (a+b \sqrt{c}\right )^{3/2}}-\frac{\sqrt{a+b \sqrt{c+d x}}}{2 x^2} \]

[Out]

-Sqrt[a + b*Sqrt[c + d*x]]/(2*x^2) + (b*d*(b*c - a*Sqrt[c + d*x])*Sqrt[a + b*Sqrt[c + d*x]])/(8*c*(a^2 - b^2*c
)*x) - (b*(2*a - 3*b*Sqrt[c])*d^2*ArcTanh[Sqrt[a + b*Sqrt[c + d*x]]/Sqrt[a - b*Sqrt[c]]])/(16*(a - b*Sqrt[c])^
(3/2)*c^(3/2)) + (b*(2*a + 3*b*Sqrt[c])*d^2*ArcTanh[Sqrt[a + b*Sqrt[c + d*x]]/Sqrt[a + b*Sqrt[c]]])/(16*(a + b
*Sqrt[c])^(3/2)*c^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.428875, antiderivative size = 224, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {371, 1398, 821, 12, 741, 827, 1166, 207} \[ \frac{b d \left (b c-a \sqrt{c+d x}\right ) \sqrt{a+b \sqrt{c+d x}}}{8 c x \left (a^2-b^2 c\right )}-\frac{b d^2 \left (2 a-3 b \sqrt{c}\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c+d x}}}{\sqrt{a-b \sqrt{c}}}\right )}{16 c^{3/2} \left (a-b \sqrt{c}\right )^{3/2}}+\frac{b d^2 \left (2 a+3 b \sqrt{c}\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c+d x}}}{\sqrt{a+b \sqrt{c}}}\right )}{16 c^{3/2} \left (a+b \sqrt{c}\right )^{3/2}}-\frac{\sqrt{a+b \sqrt{c+d x}}}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sqrt[c + d*x]]/x^3,x]

[Out]

-Sqrt[a + b*Sqrt[c + d*x]]/(2*x^2) + (b*d*(b*c - a*Sqrt[c + d*x])*Sqrt[a + b*Sqrt[c + d*x]])/(8*c*(a^2 - b^2*c
)*x) - (b*(2*a - 3*b*Sqrt[c])*d^2*ArcTanh[Sqrt[a + b*Sqrt[c + d*x]]/Sqrt[a - b*Sqrt[c]]])/(16*(a - b*Sqrt[c])^
(3/2)*c^(3/2)) + (b*(2*a + 3*b*Sqrt[c])*d^2*ArcTanh[Sqrt[a + b*Sqrt[c + d*x]]/Sqrt[a + b*Sqrt[c]]])/(16*(a + b
*Sqrt[c])^(3/2)*c^(3/2))

Rule 371

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]
] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]

Rule 1398

Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{g = Denominator[n]}, D
ist[g, Subst[Int[x^(g - 1)*(d + e*x^(g*n))^q*(a + c*x^(2*g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, c, d, e, p
, q}, x] && EqQ[n2, 2*n] && FractionQ[n]

Rule 821

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^m*
(a + c*x^2)^(p + 1)*(a*g - c*f*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*
x^2)^(p + 1)*Simp[a*e*g*m - c*d*f*(2*p + 3) - c*e*f*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x
] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 827

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

Rule 1166

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

Rule 207

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

Rubi steps

\begin{align*} \int \frac{\sqrt{a+b \sqrt{c+d x}}}{x^3} \, dx &=d^2 \operatorname{Subst}\left (\int \frac{\sqrt{a+b \sqrt{x}}}{(-c+x)^3} \, dx,x,c+d x\right )\\ &=\left (2 d^2\right ) \operatorname{Subst}\left (\int \frac{x \sqrt{a+b x}}{\left (-c+x^2\right )^3} \, dx,x,\sqrt{c+d x}\right )\\ &=-\frac{\sqrt{a+b \sqrt{c+d x}}}{2 x^2}-\frac{d^2 \operatorname{Subst}\left (\int -\frac{b c}{2 \sqrt{a+b x} \left (-c+x^2\right )^2} \, dx,x,\sqrt{c+d x}\right )}{2 c}\\ &=-\frac{\sqrt{a+b \sqrt{c+d x}}}{2 x^2}+\frac{1}{4} \left (b d^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x} \left (-c+x^2\right )^2} \, dx,x,\sqrt{c+d x}\right )\\ &=-\frac{\sqrt{a+b \sqrt{c+d x}}}{2 x^2}+\frac{b d \left (b c-a \sqrt{c+d x}\right ) \sqrt{a+b \sqrt{c+d x}}}{8 c \left (a^2-b^2 c\right ) x}+\frac{\left (b d^2\right ) \operatorname{Subst}\left (\int \frac{\frac{1}{2} \left (-2 a^2+3 b^2 c\right )-\frac{a b x}{2}}{\sqrt{a+b x} \left (-c+x^2\right )} \, dx,x,\sqrt{c+d x}\right )}{8 c \left (a^2-b^2 c\right )}\\ &=-\frac{\sqrt{a+b \sqrt{c+d x}}}{2 x^2}+\frac{b d \left (b c-a \sqrt{c+d x}\right ) \sqrt{a+b \sqrt{c+d x}}}{8 c \left (a^2-b^2 c\right ) x}+\frac{\left (b d^2\right ) \operatorname{Subst}\left (\int \frac{\frac{a^2 b}{2}+\frac{1}{2} b \left (-2 a^2+3 b^2 c\right )-\frac{1}{2} a b x^2}{a^2-b^2 c-2 a x^2+x^4} \, dx,x,\sqrt{a+b \sqrt{c+d x}}\right )}{4 c \left (a^2-b^2 c\right )}\\ &=-\frac{\sqrt{a+b \sqrt{c+d x}}}{2 x^2}+\frac{b d \left (b c-a \sqrt{c+d x}\right ) \sqrt{a+b \sqrt{c+d x}}}{8 c \left (a^2-b^2 c\right ) x}+\frac{\left (b \left (2 a-3 b \sqrt{c}\right ) d^2\right ) \operatorname{Subst}\left (\int \frac{1}{-a+b \sqrt{c}+x^2} \, dx,x,\sqrt{a+b \sqrt{c+d x}}\right )}{16 \left (a-b \sqrt{c}\right ) c^{3/2}}-\frac{\left (b \left (2 a+3 b \sqrt{c}\right ) d^2\right ) \operatorname{Subst}\left (\int \frac{1}{-a-b \sqrt{c}+x^2} \, dx,x,\sqrt{a+b \sqrt{c+d x}}\right )}{16 \left (a+b \sqrt{c}\right ) c^{3/2}}\\ &=-\frac{\sqrt{a+b \sqrt{c+d x}}}{2 x^2}+\frac{b d \left (b c-a \sqrt{c+d x}\right ) \sqrt{a+b \sqrt{c+d x}}}{8 c \left (a^2-b^2 c\right ) x}-\frac{b \left (2 a-3 b \sqrt{c}\right ) d^2 \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c+d x}}}{\sqrt{a-b \sqrt{c}}}\right )}{16 \left (a-b \sqrt{c}\right )^{3/2} c^{3/2}}+\frac{b \left (2 a+3 b \sqrt{c}\right ) d^2 \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c+d x}}}{\sqrt{a+b \sqrt{c}}}\right )}{16 \left (a+b \sqrt{c}\right )^{3/2} c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.437401, size = 258, normalized size = 1.15 \[ \frac{-\left (a-b \sqrt{c}\right ) \left (2 \sqrt{c} \left (a+b \sqrt{c}\right ) \sqrt{a+b \sqrt{c+d x}} \left (4 a^2 c+a b d x \sqrt{c+d x}-b^2 c (4 c+d x)\right )-b d^2 x^2 \sqrt{a+b \sqrt{c}} \left (2 a^2+a b \sqrt{c}-3 b^2 c\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c+d x}}}{\sqrt{a+b \sqrt{c}}}\right )\right )-b d^2 x^2 \left (2 a-3 b \sqrt{c}\right ) \sqrt{a-b \sqrt{c}} \left (a+b \sqrt{c}\right )^2 \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c+d x}}}{\sqrt{a-b \sqrt{c}}}\right )}{16 c^{3/2} x^2 \left (a^2-b^2 c\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Sqrt[c + d*x]]/x^3,x]

[Out]

(-(b*(2*a - 3*b*Sqrt[c])*Sqrt[a - b*Sqrt[c]]*(a + b*Sqrt[c])^2*d^2*x^2*ArcTanh[Sqrt[a + b*Sqrt[c + d*x]]/Sqrt[
a - b*Sqrt[c]]]) - (a - b*Sqrt[c])*(2*(a + b*Sqrt[c])*Sqrt[c]*Sqrt[a + b*Sqrt[c + d*x]]*(4*a^2*c + a*b*d*x*Sqr
t[c + d*x] - b^2*c*(4*c + d*x)) - b*Sqrt[a + b*Sqrt[c]]*(2*a^2 + a*b*Sqrt[c] - 3*b^2*c)*d^2*x^2*ArcTanh[Sqrt[a
+ b*Sqrt[c + d*x]]/Sqrt[a + b*Sqrt[c]]]))/(16*c^(3/2)*(a^2 - b^2*c)^2*x^2)

________________________________________________________________________________________

Maple [B]  time = 0.029, size = 784, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*(d*x+c)^(1/2))^(1/2)/x^3,x)

[Out]

-1/8*b^2*d^2/(b^2*(d*x+c)-b^2*c)^2*a/c/(-b^2*c+a^2)*(a+b*(d*x+c)^(1/2))^(7/2)+1/8*b^4*d^2/(b^2*(d*x+c)-b^2*c)^
2/(-b^2*c+a^2)*(a+b*(d*x+c)^(1/2))^(5/2)+3/8*b^2*d^2/(b^2*(d*x+c)-b^2*c)^2/c/(-b^2*c+a^2)*(a+b*(d*x+c)^(1/2))^
(5/2)*a^2-1/8*b^4*d^2/(b^2*(d*x+c)-b^2*c)^2*a/(-b^2*c+a^2)*(a+b*(d*x+c)^(1/2))^(3/2)-3/8*b^2*d^2/(b^2*(d*x+c)-
b^2*c)^2*a^3/c/(-b^2*c+a^2)*(a+b*(d*x+c)^(1/2))^(3/2)-3/8*b^4*d^2/(b^2*(d*x+c)-b^2*c)^2*(a+b*(d*x+c)^(1/2))^(1
/2)+1/8*b^2*d^2/(b^2*(d*x+c)-b^2*c)^2/c*(a+b*(d*x+c)^(1/2))^(1/2)*a^2+3/16*b^4*d^2/(-b^2*c+a^2)/(b^2*c)^(1/2)/
(-(b^2*c)^(1/2)-a)^(1/2)*arctan((a+b*(d*x+c)^(1/2))^(1/2)/(-(b^2*c)^(1/2)-a)^(1/2))-1/16*b^2*d^2/c/(-b^2*c+a^2
)/(-(b^2*c)^(1/2)-a)^(1/2)*arctan((a+b*(d*x+c)^(1/2))^(1/2)/(-(b^2*c)^(1/2)-a)^(1/2))*a-1/8*b^2*d^2/c/(-b^2*c+
a^2)/(b^2*c)^(1/2)/(-(b^2*c)^(1/2)-a)^(1/2)*arctan((a+b*(d*x+c)^(1/2))^(1/2)/(-(b^2*c)^(1/2)-a)^(1/2))*a^2-3/1
6*b^4*d^2/(-b^2*c+a^2)/(b^2*c)^(1/2)/((b^2*c)^(1/2)-a)^(1/2)*arctan((a+b*(d*x+c)^(1/2))^(1/2)/((b^2*c)^(1/2)-a
)^(1/2))-1/16*b^2*d^2/c/(-b^2*c+a^2)/((b^2*c)^(1/2)-a)^(1/2)*arctan((a+b*(d*x+c)^(1/2))^(1/2)/((b^2*c)^(1/2)-a
)^(1/2))*a+1/8*b^2*d^2/c/(-b^2*c+a^2)/(b^2*c)^(1/2)/((b^2*c)^(1/2)-a)^(1/2)*arctan((a+b*(d*x+c)^(1/2))^(1/2)/(
(b^2*c)^(1/2)-a)^(1/2))*a^2

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\sqrt{d x + c} b + a}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(d*x+c)^(1/2))^(1/2)/x^3,x, algorithm="maxima")

[Out]

integrate(sqrt(sqrt(d*x + c)*b + a)/x^3, x)

________________________________________________________________________________________

Fricas [B]  time = 3.44089, size = 5759, normalized size = 25.71 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(d*x+c)^(1/2))^(1/2)/x^3,x, algorithm="fricas")

[Out]

1/32*((b^2*c^2 - a^2*c)*x^2*sqrt(-((15*a*b^6*c^2 - 15*a^3*b^4*c + 4*a^5*b^2)*d^4 + (b^6*c^6 - 3*a^2*b^4*c^5 +
3*a^4*b^2*c^4 - a^6*c^3)*sqrt((81*b^14*c^2 - 90*a^2*b^12*c + 25*a^4*b^10)*d^8/(b^12*c^9 - 6*a^2*b^10*c^8 + 15*
a^4*b^8*c^7 - 20*a^6*b^6*c^6 + 15*a^8*b^4*c^5 - 6*a^10*b^2*c^4 + a^12*c^3)))/(b^6*c^6 - 3*a^2*b^4*c^5 + 3*a^4*
b^2*c^4 - a^6*c^3))*log((81*b^10*c^2 - 81*a^2*b^8*c + 20*a^4*b^6)*sqrt(sqrt(d*x + c)*b + a)*d^6 + ((27*b^10*c^
4 - 24*a^2*b^8*c^3 + 5*a^4*b^6*c^2)*d^4 - 2*(2*a*b^8*c^7 - 7*a^3*b^6*c^6 + 9*a^5*b^4*c^5 - 5*a^7*b^2*c^4 + a^9
*c^3)*sqrt((81*b^14*c^2 - 90*a^2*b^12*c + 25*a^4*b^10)*d^8/(b^12*c^9 - 6*a^2*b^10*c^8 + 15*a^4*b^8*c^7 - 20*a^
6*b^6*c^6 + 15*a^8*b^4*c^5 - 6*a^10*b^2*c^4 + a^12*c^3)))*sqrt(-((15*a*b^6*c^2 - 15*a^3*b^4*c + 4*a^5*b^2)*d^4
+ (b^6*c^6 - 3*a^2*b^4*c^5 + 3*a^4*b^2*c^4 - a^6*c^3)*sqrt((81*b^14*c^2 - 90*a^2*b^12*c + 25*a^4*b^10)*d^8/(b
^12*c^9 - 6*a^2*b^10*c^8 + 15*a^4*b^8*c^7 - 20*a^6*b^6*c^6 + 15*a^8*b^4*c^5 - 6*a^10*b^2*c^4 + a^12*c^3)))/(b^
6*c^6 - 3*a^2*b^4*c^5 + 3*a^4*b^2*c^4 - a^6*c^3))) - (b^2*c^2 - a^2*c)*x^2*sqrt(-((15*a*b^6*c^2 - 15*a^3*b^4*c
+ 4*a^5*b^2)*d^4 + (b^6*c^6 - 3*a^2*b^4*c^5 + 3*a^4*b^2*c^4 - a^6*c^3)*sqrt((81*b^14*c^2 - 90*a^2*b^12*c + 25
*a^4*b^10)*d^8/(b^12*c^9 - 6*a^2*b^10*c^8 + 15*a^4*b^8*c^7 - 20*a^6*b^6*c^6 + 15*a^8*b^4*c^5 - 6*a^10*b^2*c^4
+ a^12*c^3)))/(b^6*c^6 - 3*a^2*b^4*c^5 + 3*a^4*b^2*c^4 - a^6*c^3))*log((81*b^10*c^2 - 81*a^2*b^8*c + 20*a^4*b^
6)*sqrt(sqrt(d*x + c)*b + a)*d^6 - ((27*b^10*c^4 - 24*a^2*b^8*c^3 + 5*a^4*b^6*c^2)*d^4 - 2*(2*a*b^8*c^7 - 7*a^
3*b^6*c^6 + 9*a^5*b^4*c^5 - 5*a^7*b^2*c^4 + a^9*c^3)*sqrt((81*b^14*c^2 - 90*a^2*b^12*c + 25*a^4*b^10)*d^8/(b^1
2*c^9 - 6*a^2*b^10*c^8 + 15*a^4*b^8*c^7 - 20*a^6*b^6*c^6 + 15*a^8*b^4*c^5 - 6*a^10*b^2*c^4 + a^12*c^3)))*sqrt(
-((15*a*b^6*c^2 - 15*a^3*b^4*c + 4*a^5*b^2)*d^4 + (b^6*c^6 - 3*a^2*b^4*c^5 + 3*a^4*b^2*c^4 - a^6*c^3)*sqrt((81
*b^14*c^2 - 90*a^2*b^12*c + 25*a^4*b^10)*d^8/(b^12*c^9 - 6*a^2*b^10*c^8 + 15*a^4*b^8*c^7 - 20*a^6*b^6*c^6 + 15
*a^8*b^4*c^5 - 6*a^10*b^2*c^4 + a^12*c^3)))/(b^6*c^6 - 3*a^2*b^4*c^5 + 3*a^4*b^2*c^4 - a^6*c^3))) + (b^2*c^2 -
a^2*c)*x^2*sqrt(-((15*a*b^6*c^2 - 15*a^3*b^4*c + 4*a^5*b^2)*d^4 - (b^6*c^6 - 3*a^2*b^4*c^5 + 3*a^4*b^2*c^4 -
a^6*c^3)*sqrt((81*b^14*c^2 - 90*a^2*b^12*c + 25*a^4*b^10)*d^8/(b^12*c^9 - 6*a^2*b^10*c^8 + 15*a^4*b^8*c^7 - 20
*a^6*b^6*c^6 + 15*a^8*b^4*c^5 - 6*a^10*b^2*c^4 + a^12*c^3)))/(b^6*c^6 - 3*a^2*b^4*c^5 + 3*a^4*b^2*c^4 - a^6*c^
3))*log((81*b^10*c^2 - 81*a^2*b^8*c + 20*a^4*b^6)*sqrt(sqrt(d*x + c)*b + a)*d^6 + ((27*b^10*c^4 - 24*a^2*b^8*c
^3 + 5*a^4*b^6*c^2)*d^4 + 2*(2*a*b^8*c^7 - 7*a^3*b^6*c^6 + 9*a^5*b^4*c^5 - 5*a^7*b^2*c^4 + a^9*c^3)*sqrt((81*b
^14*c^2 - 90*a^2*b^12*c + 25*a^4*b^10)*d^8/(b^12*c^9 - 6*a^2*b^10*c^8 + 15*a^4*b^8*c^7 - 20*a^6*b^6*c^6 + 15*a
^8*b^4*c^5 - 6*a^10*b^2*c^4 + a^12*c^3)))*sqrt(-((15*a*b^6*c^2 - 15*a^3*b^4*c + 4*a^5*b^2)*d^4 - (b^6*c^6 - 3*
a^2*b^4*c^5 + 3*a^4*b^2*c^4 - a^6*c^3)*sqrt((81*b^14*c^2 - 90*a^2*b^12*c + 25*a^4*b^10)*d^8/(b^12*c^9 - 6*a^2*
b^10*c^8 + 15*a^4*b^8*c^7 - 20*a^6*b^6*c^6 + 15*a^8*b^4*c^5 - 6*a^10*b^2*c^4 + a^12*c^3)))/(b^6*c^6 - 3*a^2*b^
4*c^5 + 3*a^4*b^2*c^4 - a^6*c^3))) - (b^2*c^2 - a^2*c)*x^2*sqrt(-((15*a*b^6*c^2 - 15*a^3*b^4*c + 4*a^5*b^2)*d^
4 - (b^6*c^6 - 3*a^2*b^4*c^5 + 3*a^4*b^2*c^4 - a^6*c^3)*sqrt((81*b^14*c^2 - 90*a^2*b^12*c + 25*a^4*b^10)*d^8/(
b^12*c^9 - 6*a^2*b^10*c^8 + 15*a^4*b^8*c^7 - 20*a^6*b^6*c^6 + 15*a^8*b^4*c^5 - 6*a^10*b^2*c^4 + a^12*c^3)))/(b
^6*c^6 - 3*a^2*b^4*c^5 + 3*a^4*b^2*c^4 - a^6*c^3))*log((81*b^10*c^2 - 81*a^2*b^8*c + 20*a^4*b^6)*sqrt(sqrt(d*x
+ c)*b + a)*d^6 - ((27*b^10*c^4 - 24*a^2*b^8*c^3 + 5*a^4*b^6*c^2)*d^4 + 2*(2*a*b^8*c^7 - 7*a^3*b^6*c^6 + 9*a^
5*b^4*c^5 - 5*a^7*b^2*c^4 + a^9*c^3)*sqrt((81*b^14*c^2 - 90*a^2*b^12*c + 25*a^4*b^10)*d^8/(b^12*c^9 - 6*a^2*b^
10*c^8 + 15*a^4*b^8*c^7 - 20*a^6*b^6*c^6 + 15*a^8*b^4*c^5 - 6*a^10*b^2*c^4 + a^12*c^3)))*sqrt(-((15*a*b^6*c^2
- 15*a^3*b^4*c + 4*a^5*b^2)*d^4 - (b^6*c^6 - 3*a^2*b^4*c^5 + 3*a^4*b^2*c^4 - a^6*c^3)*sqrt((81*b^14*c^2 - 90*a
^2*b^12*c + 25*a^4*b^10)*d^8/(b^12*c^9 - 6*a^2*b^10*c^8 + 15*a^4*b^8*c^7 - 20*a^6*b^6*c^6 + 15*a^8*b^4*c^5 - 6
*a^10*b^2*c^4 + a^12*c^3)))/(b^6*c^6 - 3*a^2*b^4*c^5 + 3*a^4*b^2*c^4 - a^6*c^3))) - 4*(b^2*c*d*x - sqrt(d*x +
c)*a*b*d*x + 4*b^2*c^2 - 4*a^2*c)*sqrt(sqrt(d*x + c)*b + a))/((b^2*c^2 - a^2*c)*x^2)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a + b \sqrt{c + d x}}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(d*x+c)**(1/2))**(1/2)/x**3,x)

[Out]

Integral(sqrt(a + b*sqrt(c + d*x))/x**3, x)

________________________________________________________________________________________

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((a+b*(d*x+c)^(1/2))^(1/2)/x^3,x, algorithm="giac")

[Out]

Timed out