3.7.31 \(\int \frac {\sqrt {a+b \sqrt {c+d x}}}{x^3} \, dx\) [631]

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

[Out]

-1/16*b*d^2*arctanh((a+b*(d*x+c)^(1/2))^(1/2)/(a-b*c^(1/2))^(1/2))*(2*a-3*b*c^(1/2))/c^(3/2)/(a-b*c^(1/2))^(3/
2)+1/16*b*d^2*arctanh((a+b*(d*x+c)^(1/2))^(1/2)/(a+b*c^(1/2))^(1/2))*(2*a+3*b*c^(1/2))/c^(3/2)/(a+b*c^(1/2))^(
3/2)-1/2*(a+b*(d*x+c)^(1/2))^(1/2)/x^2+1/8*b*d*(b*c-a*(d*x+c)^(1/2))*(a+b*(d*x+c)^(1/2))^(1/2)/c/(-b^2*c+a^2)/
x

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Rubi [A]
time = 0.31, antiderivative size = 224, normalized size of antiderivative = 1.00, 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 = {378, 1412, 835, 12, 755, 841, 1180, 213} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sqrt[c + d*x]]/x^3,x]

[Out]

-1/2*Sqrt[a + b*Sqrt[c + d*x]]/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 12

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

Rule 213

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

Rule 378

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 755

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 835

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 841

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 1180

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 1412

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]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b \sqrt {c+d x}}}{x^3} \, dx &=d^2 \text {Subst}\left (\int \frac {\sqrt {a+b \sqrt {x}}}{(-c+x)^3} \, dx,x,c+d x\right )\\ &=\left (2 d^2\right ) \text {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 \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 1.81, size = 217, normalized size = 0.97 \begin {gather*} \frac {-\frac {2 \sqrt {c} \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 )}{\left (a^2-b^2 c\right ) x^2}+\frac {b \left (2 a+3 b \sqrt {c}\right ) d^2 \tan ^{-1}\left (\frac {\sqrt {a+b \sqrt {c+d x}}}{\sqrt {-a-b \sqrt {c}}}\right )}{\left (-a-b \sqrt {c}\right )^{3/2}}+\frac {b \left (-2 a+3 b \sqrt {c}\right ) d^2 \tan ^{-1}\left (\frac {\sqrt {a+b \sqrt {c+d x}}}{\sqrt {-a+b \sqrt {c}}}\right )}{\left (-a+b \sqrt {c}\right )^{3/2}}}{16 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(372\) vs. \(2(174)=348\).
time = 0.14, size = 373, normalized size = 1.67

method	result	size
derivativedivides	\(4 d^{2} b^{4} \left (-\frac {\frac {a \left (a +b \sqrt {d x +c}\right )^{\frac {7}{2}}}{32 b^{2} c \left (-b^{2} c +a^{2}\right )}-\frac {\left (b^{2} c +3 a^{2}\right ) \left (a +b \sqrt {d x +c}\right )^{\frac {5}{2}}}{32 b^{2} c \left (-b^{2} c +a^{2}\right )}+\frac {a \left (b^{2} c +3 a^{2}\right ) \left (a +b \sqrt {d x +c}\right )^{\frac {3}{2}}}{32 b^{2} c \left (-b^{2} c +a^{2}\right )}-\frac {\left (-3 b^{2} c +a^{2}\right ) \sqrt {a +b \sqrt {d x +c}}}{32 b^{2} c}}{\left (\left (a +b \sqrt {d x +c}\right )^{2}-2 a \left (a +b \sqrt {d x +c}\right )-b^{2} c +a^{2}\right )^{2}}-\frac {\frac {\left (-3 b^{2} c +a \sqrt {b^{2} c}+2 a^{2}\right ) \arctan \left (\frac {\sqrt {a +b \sqrt {d x +c}}}{\sqrt {-\sqrt {b^{2} c}-a}}\right )}{2 \sqrt {b^{2} c}\, \sqrt {-\sqrt {b^{2} c}-a}}+\frac {\left (3 b^{2} c +a \sqrt {b^{2} c}-2 a^{2}\right ) \arctan \left (\frac {\sqrt {a +b \sqrt {d x +c}}}{\sqrt {\sqrt {b^{2} c}-a}}\right )}{2 \sqrt {b^{2} c}\, \sqrt {\sqrt {b^{2} c}-a}}}{32 b^{2} c \left (-b^{2} c +a^{2}\right )}\right )\)	\(373\)
default	\(4 d^{2} b^{4} \left (-\frac {\frac {a \left (a +b \sqrt {d x +c}\right )^{\frac {7}{2}}}{32 b^{2} c \left (-b^{2} c +a^{2}\right )}-\frac {\left (b^{2} c +3 a^{2}\right ) \left (a +b \sqrt {d x +c}\right )^{\frac {5}{2}}}{32 b^{2} c \left (-b^{2} c +a^{2}\right )}+\frac {a \left (b^{2} c +3 a^{2}\right ) \left (a +b \sqrt {d x +c}\right )^{\frac {3}{2}}}{32 b^{2} c \left (-b^{2} c +a^{2}\right )}-\frac {\left (-3 b^{2} c +a^{2}\right ) \sqrt {a +b \sqrt {d x +c}}}{32 b^{2} c}}{\left (\left (a +b \sqrt {d x +c}\right )^{2}-2 a \left (a +b \sqrt {d x +c}\right )-b^{2} c +a^{2}\right )^{2}}-\frac {\frac {\left (-3 b^{2} c +a \sqrt {b^{2} c}+2 a^{2}\right ) \arctan \left (\frac {\sqrt {a +b \sqrt {d x +c}}}{\sqrt {-\sqrt {b^{2} c}-a}}\right )}{2 \sqrt {b^{2} c}\, \sqrt {-\sqrt {b^{2} c}-a}}+\frac {\left (3 b^{2} c +a \sqrt {b^{2} c}-2 a^{2}\right ) \arctan \left (\frac {\sqrt {a +b \sqrt {d x +c}}}{\sqrt {\sqrt {b^{2} c}-a}}\right )}{2 \sqrt {b^{2} c}\, \sqrt {\sqrt {b^{2} c}-a}}}{32 b^{2} c \left (-b^{2} c +a^{2}\right )}\right )\)	\(373\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*d^2*b^4*(-(1/32*a/b^2/c/(-b^2*c+a^2)*(a+b*(d*x+c)^(1/2))^(7/2)-1/32*(b^2*c+3*a^2)/b^2/c/(-b^2*c+a^2)*(a+b*(d
*x+c)^(1/2))^(5/2)+1/32*a*(b^2*c+3*a^2)/b^2/c/(-b^2*c+a^2)*(a+b*(d*x+c)^(1/2))^(3/2)-1/32*(-3*b^2*c+a^2)/b^2/c
*(a+b*(d*x+c)^(1/2))^(1/2))/((a+b*(d*x+c)^(1/2))^2-2*a*(a+b*(d*x+c)^(1/2))-b^2*c+a^2)^2-1/32/b^2/c/(-b^2*c+a^2
)*(1/2*(-3*b^2*c+a*(b^2*c)^(1/2)+2*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))+1/2*(3*b^2*c+a*(b^2*c)^(1/2)-2*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))))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 2856 vs. \(2 (175) = 350\).
time = 0.62, size = 2856, normalized size = 12.75 \begin {gather*} \text {Too large to display} \end {gather*}

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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + b \sqrt {c + d x}}}{x^{3}}\, dx \end {gather*}

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)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 895 vs. \(2 (175) = 350\).
time = 3.33, size = 895, normalized size = 4.00 \begin {gather*} \frac {\frac {{\left ({\left (b^{3} c^{2} - a^{2} b c\right )}^{2} a b^{3} \sqrt {c} d^{3} - {\left (3 \, b^{7} c^{3} - 4 \, a^{2} b^{5} c^{2} + a^{4} b^{3} c\right )} d^{3} {\left | b^{3} c^{2} - a^{2} b c \right |} + {\left (3 \, a b^{9} c^{\frac {9}{2}} - 8 \, a^{3} b^{7} c^{\frac {7}{2}} + 7 \, a^{5} b^{5} c^{\frac {5}{2}} - 2 \, a^{7} b^{3} c^{\frac {3}{2}}\right )} d^{3}\right )} \arctan \left (\frac {\sqrt {\sqrt {d x + c} b + a}}{\sqrt {-\frac {a b^{2} c^{2} - a^{3} c + \sqrt {{\left (a b^{2} c^{2} - a^{3} c\right )}^{2} + {\left (b^{4} c^{3} - 2 \, a^{2} b^{2} c^{2} + a^{4} c\right )} {\left (b^{2} c^{2} - a^{2} c\right )}}}{b^{2} c^{2} - a^{2} c}}}\right )}{{\left (b^{5} c^{\frac {9}{2}} - a b^{4} c^{4} - 2 \, a^{2} b^{3} c^{\frac {7}{2}} + 2 \, a^{3} b^{2} c^{3} + a^{4} b c^{\frac {5}{2}} - a^{5} c^{2}\right )} \sqrt {-b \sqrt {c} - a} {\left | b^{3} c^{2} - a^{2} b c \right |}} + \frac {{\left ({\left (b^{3} c^{2} - a^{2} b c\right )}^{2} a b^{3} d^{3} + {\left (3 \, b^{7} c^{\frac {5}{2}} - 4 \, a^{2} b^{5} c^{\frac {3}{2}} + a^{4} b^{3} \sqrt {c}\right )} d^{3} {\left | b^{3} c^{2} - a^{2} b c \right |} + {\left (3 \, a b^{9} c^{4} - 8 \, a^{3} b^{7} c^{3} + 7 \, a^{5} b^{5} c^{2} - 2 \, a^{7} b^{3} c\right )} d^{3}\right )} \arctan \left (\frac {\sqrt {\sqrt {d x + c} b + a}}{\sqrt {-\frac {a b^{2} c^{2} - a^{3} c - \sqrt {{\left (a b^{2} c^{2} - a^{3} c\right )}^{2} + {\left (b^{4} c^{3} - 2 \, a^{2} b^{2} c^{2} + a^{4} c\right )} {\left (b^{2} c^{2} - a^{2} c\right )}}}{b^{2} c^{2} - a^{2} c}}}\right )}{{\left (b^{5} c^{4} + a b^{4} c^{\frac {7}{2}} - 2 \, a^{2} b^{3} c^{3} - 2 \, a^{3} b^{2} c^{\frac {5}{2}} + a^{4} b c^{2} + a^{5} c^{\frac {3}{2}}\right )} \sqrt {b \sqrt {c} - a} {\left | b^{3} c^{2} - a^{2} b c \right |}} - \frac {2 \, {\left (3 \, \sqrt {\sqrt {d x + c} b + a} b^{7} c^{2} d^{3} + {\left (\sqrt {d x + c} b + a\right )}^{\frac {5}{2}} b^{5} c d^{3} - {\left (\sqrt {d x + c} b + a\right )}^{\frac {3}{2}} a b^{5} c d^{3} - 4 \, \sqrt {\sqrt {d x + c} b + a} a^{2} b^{5} c d^{3} - {\left (\sqrt {d x + c} b + a\right )}^{\frac {7}{2}} a b^{3} d^{3} + 3 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {5}{2}} a^{2} b^{3} d^{3} - 3 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {3}{2}} a^{3} b^{3} d^{3} + \sqrt {\sqrt {d x + c} b + a} a^{4} b^{3} d^{3}\right )}}{{\left (b^{2} c^{2} - a^{2} c\right )} {\left (b^{2} c - {\left (\sqrt {d x + c} b + a\right )}^{2} + 2 \, {\left (\sqrt {d x + c} b + a\right )} a - a^{2}\right )}^{2}}}{16 \, b d} \end {gather*}

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]

1/16*(((b^3*c^2 - a^2*b*c)^2*a*b^3*sqrt(c)*d^3 - (3*b^7*c^3 - 4*a^2*b^5*c^2 + a^4*b^3*c)*d^3*abs(b^3*c^2 - a^2
*b*c) + (3*a*b^9*c^(9/2) - 8*a^3*b^7*c^(7/2) + 7*a^5*b^5*c^(5/2) - 2*a^7*b^3*c^(3/2))*d^3)*arctan(sqrt(sqrt(d*
x + c)*b + a)/sqrt(-(a*b^2*c^2 - a^3*c + sqrt((a*b^2*c^2 - a^3*c)^2 + (b^4*c^3 - 2*a^2*b^2*c^2 + a^4*c)*(b^2*c
^2 - a^2*c)))/(b^2*c^2 - a^2*c)))/((b^5*c^(9/2) - a*b^4*c^4 - 2*a^2*b^3*c^(7/2) + 2*a^3*b^2*c^3 + a^4*b*c^(5/2
) - a^5*c^2)*sqrt(-b*sqrt(c) - a)*abs(b^3*c^2 - a^2*b*c)) + ((b^3*c^2 - a^2*b*c)^2*a*b^3*d^3 + (3*b^7*c^(5/2)
- 4*a^2*b^5*c^(3/2) + a^4*b^3*sqrt(c))*d^3*abs(b^3*c^2 - a^2*b*c) + (3*a*b^9*c^4 - 8*a^3*b^7*c^3 + 7*a^5*b^5*c
^2 - 2*a^7*b^3*c)*d^3)*arctan(sqrt(sqrt(d*x + c)*b + a)/sqrt(-(a*b^2*c^2 - a^3*c - sqrt((a*b^2*c^2 - a^3*c)^2
+ (b^4*c^3 - 2*a^2*b^2*c^2 + a^4*c)*(b^2*c^2 - a^2*c)))/(b^2*c^2 - a^2*c)))/((b^5*c^4 + a*b^4*c^(7/2) - 2*a^2*
b^3*c^3 - 2*a^3*b^2*c^(5/2) + a^4*b*c^2 + a^5*c^(3/2))*sqrt(b*sqrt(c) - a)*abs(b^3*c^2 - a^2*b*c)) - 2*(3*sqrt
(sqrt(d*x + c)*b + a)*b^7*c^2*d^3 + (sqrt(d*x + c)*b + a)^(5/2)*b^5*c*d^3 - (sqrt(d*x + c)*b + a)^(3/2)*a*b^5*
c*d^3 - 4*sqrt(sqrt(d*x + c)*b + a)*a^2*b^5*c*d^3 - (sqrt(d*x + c)*b + a)^(7/2)*a*b^3*d^3 + 3*(sqrt(d*x + c)*b
+ a)^(5/2)*a^2*b^3*d^3 - 3*(sqrt(d*x + c)*b + a)^(3/2)*a^3*b^3*d^3 + sqrt(sqrt(d*x + c)*b + a)*a^4*b^3*d^3)/(
(b^2*c^2 - a^2*c)*(b^2*c - (sqrt(d*x + c)*b + a)^2 + 2*(sqrt(d*x + c)*b + a)*a - a^2)^2))/(b*d)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {a+b\,\sqrt {c+d\,x}}}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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