3.7.52 \(\int \frac {1}{x^3 \sqrt {a+b \sqrt {c+d x}}} \, dx\) [652]

Optimal. Leaf size=261 \[ -\frac {\left (a-b \sqrt {c+d x}\right ) \sqrt {a+b \sqrt {c+d x}}}{2 \left (a^2-b^2 c\right ) x^2}-\frac {b d \sqrt {a+b \sqrt {c+d x}} \left (6 a b c-\left (a^2+5 b^2 c\right ) \sqrt {c+d x}\right )}{8 c \left (a^2-b^2 c\right )^2 x}+\frac {b \left (2 a-5 b \sqrt {c}\right ) d^2 \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {c+d x}}}{\sqrt {a-b \sqrt {c}}}\right )}{16 \left (a-b \sqrt {c}\right )^{5/2} c^{3/2}}-\frac {b \left (2 a+5 b \sqrt {c}\right ) d^2 \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {c+d x}}}{\sqrt {a+b \sqrt {c}}}\right )}{16 \left (a+b \sqrt {c}\right )^{5/2} c^{3/2}} \]

[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-5*b*c^(1/2))/c^(3/2)/(a-b*c^(1/2))^(5/2
)-1/16*b*d^2*arctanh((a+b*(d*x+c)^(1/2))^(1/2)/(a+b*c^(1/2))^(1/2))*(2*a+5*b*c^(1/2))/c^(3/2)/(a+b*c^(1/2))^(5
/2)-1/2*(a-b*(d*x+c)^(1/2))*(a+b*(d*x+c)^(1/2))^(1/2)/(-b^2*c+a^2)/x^2-1/8*b*d*(6*a*b*c-(5*b^2*c+a^2)*(d*x+c)^
(1/2))*(a+b*(d*x+c)^(1/2))^(1/2)/c/(-b^2*c+a^2)^2/x

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Rubi [A]
time = 0.35, antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {378, 1412, 837, 841, 1180, 213} \begin {gather*} -\frac {\left (a-b \sqrt {c+d x}\right ) \sqrt {a+b \sqrt {c+d x}}}{2 x^2 \left (a^2-b^2 c\right )}-\frac {b d \sqrt {a+b \sqrt {c+d x}} \left (6 a b c-\left (a^2+5 b^2 c\right ) \sqrt {c+d x}\right )}{8 c x \left (a^2-b^2 c\right )^2}+\frac {b d^2 \left (2 a-5 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 )^{5/2}}-\frac {b d^2 \left (2 a+5 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 )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*Sqrt[a + b*Sqrt[c + d*x]]),x]

[Out]

-1/2*((a - b*Sqrt[c + d*x])*Sqrt[a + b*Sqrt[c + d*x]])/((a^2 - b^2*c)*x^2) - (b*d*Sqrt[a + b*Sqrt[c + d*x]]*(6
*a*b*c - (a^2 + 5*b^2*c)*Sqrt[c + d*x]))/(8*c*(a^2 - b^2*c)^2*x) + (b*(2*a - 5*b*Sqrt[c])*d^2*ArcTanh[Sqrt[a +
b*Sqrt[c + d*x]]/Sqrt[a - b*Sqrt[c]]])/(16*(a - b*Sqrt[c])^(5/2)*c^(3/2)) - (b*(2*a + 5*b*Sqrt[c])*d^2*ArcTan
h[Sqrt[a + b*Sqrt[c + d*x]]/Sqrt[a + b*Sqrt[c]]])/(16*(a + b*Sqrt[c])^(5/2)*c^(3/2))

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 837

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(
m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] +
Dist[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^
2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g},
x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (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 {1}{x^3 \sqrt {a+b \sqrt {c+d x}}} \, dx &=d^2 \text {Subst}\left (\int \frac {1}{\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 {\left (a-b \sqrt {c+d x}\right ) \sqrt {a+b \sqrt {c+d x}}}{2 \left (a^2-b^2 c\right ) x^2}+\frac {d^2 \text {Subst}\left (\int \frac {-\frac {1}{2} a b c+\frac {5}{2} b^2 c x}{\sqrt {a+b x} \left (-c+x^2\right )^2} \, dx,x,\sqrt {c+d x}\right )}{2 c \left (a^2-b^2 c\right )}\\ &=-\frac {\left (a-b \sqrt {c+d x}\right ) \sqrt {a+b \sqrt {c+d x}}}{2 \left (a^2-b^2 c\right ) x^2}-\frac {b d \sqrt {a+b \sqrt {c+d x}} \left (6 a b c-\left (a^2+5 b^2 c\right ) \sqrt {c+d x}\right )}{8 c \left (a^2-b^2 c\right )^2 x}+\frac {d^2 \text {Subst}\left (\int \frac {\frac {1}{2} a b c \left (a^2-4 b^2 c\right )+\frac {1}{4} b^2 c \left (a^2+5 b^2 c\right ) x}{\sqrt {a+b x} \left (-c+x^2\right )} \, dx,x,\sqrt {c+d x}\right )}{4 c^2 \left (a^2-b^2 c\right )^2}\\ &=-\frac {\left (a-b \sqrt {c+d x}\right ) \sqrt {a+b \sqrt {c+d x}}}{2 \left (a^2-b^2 c\right ) x^2}-\frac {b d \sqrt {a+b \sqrt {c+d x}} \left (6 a b c-\left (a^2+5 b^2 c\right ) \sqrt {c+d x}\right )}{8 c \left (a^2-b^2 c\right )^2 x}+\frac {d^2 \text {Subst}\left (\int \frac {\frac {1}{2} a b^2 c \left (a^2-4 b^2 c\right )-\frac {1}{4} a b^2 c \left (a^2+5 b^2 c\right )+\frac {1}{4} b^2 c \left (a^2+5 b^2 c\right ) x^2}{a^2-b^2 c-2 a x^2+x^4} \, dx,x,\sqrt {a+b \sqrt {c+d x}}\right )}{2 c^2 \left (a^2-b^2 c\right )^2}\\ &=-\frac {\left (a-b \sqrt {c+d x}\right ) \sqrt {a+b \sqrt {c+d x}}}{2 \left (a^2-b^2 c\right ) x^2}-\frac {b d \sqrt {a+b \sqrt {c+d x}} \left (6 a b c-\left (a^2+5 b^2 c\right ) \sqrt {c+d x}\right )}{8 c \left (a^2-b^2 c\right )^2 x}-\frac {\left (b \left (2 a-5 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 )^2 c^{3/2}}+\frac {\left (b \left (2 a+5 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 )^2 c^{3/2}}\\ &=-\frac {\left (a-b \sqrt {c+d x}\right ) \sqrt {a+b \sqrt {c+d x}}}{2 \left (a^2-b^2 c\right ) x^2}-\frac {b d \sqrt {a+b \sqrt {c+d x}} \left (6 a b c-\left (a^2+5 b^2 c\right ) \sqrt {c+d x}\right )}{8 c \left (a^2-b^2 c\right )^2 x}+\frac {b \left (2 a-5 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 )^{5/2} c^{3/2}}-\frac {b \left (2 a+5 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 )^{5/2} c^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 1.34, size = 249, normalized size = 0.95 \begin {gather*} \frac {-\frac {2 \sqrt {c} \sqrt {a+b \sqrt {c+d x}} \left (4 a^3 c+b^3 c (4 c-5 d x) \sqrt {c+d x}-a^2 b \sqrt {c+d x} (4 c+d x)+2 a b^2 c (-2 c+3 d x)\right )}{\left (a^2-b^2 c\right )^2 x^2}+\frac {b \left (2 a+5 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 )^{5/2}}+\frac {b \left (-2 a+5 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 )^{5/2}}}{16 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*Sqrt[a + b*Sqrt[c + d*x]]),x]

[Out]

((-2*Sqrt[c]*Sqrt[a + b*Sqrt[c + d*x]]*(4*a^3*c + b^3*c*(4*c - 5*d*x)*Sqrt[c + d*x] - a^2*b*Sqrt[c + d*x]*(4*c
+ d*x) + 2*a*b^2*c*(-2*c + 3*d*x)))/((a^2 - b^2*c)^2*x^2) + (b*(2*a + 5*b*Sqrt[c])*d^2*ArcTan[Sqrt[a + b*Sqrt
[c + d*x]]/Sqrt[-a - b*Sqrt[c]]])/(-a - b*Sqrt[c])^(5/2) + (b*(-2*a + 5*b*Sqrt[c])*d^2*ArcTan[Sqrt[a + b*Sqrt[
c + d*x]]/Sqrt[-a + b*Sqrt[c]]])/(-a + b*Sqrt[c])^(5/2))/(16*c^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(426\) vs. \(2(209)=418\).
time = 0.21, size = 427, normalized size = 1.64

method	result	size
derivativedivides	\(4 d^{2} b^{4} \left (-\frac {\frac {-\frac {\left (5 \sqrt {b^{2} c}+2 a \right ) \left (a +b \sqrt {d x +c}\right )^{\frac {3}{2}}}{4 \left (b^{2} c +2 a \sqrt {b^{2} c}+a^{2}\right )}+\frac {\left (7 \sqrt {b^{2} c}+2 a \right ) \sqrt {a +b \sqrt {d x +c}}}{4 \sqrt {b^{2} c}+4 a}}{\left (-b \sqrt {d x +c}+\sqrt {b^{2} c}\right )^{2}}-\frac {\left (5 \sqrt {b^{2} c}+2 a \right ) \arctan \left (\frac {\sqrt {a +b \sqrt {d x +c}}}{\sqrt {-\sqrt {b^{2} c}-a}}\right )}{4 \left (b^{2} c +2 a \sqrt {b^{2} c}+a^{2}\right ) \sqrt {-\sqrt {b^{2} c}-a}}}{16 b^{2} c \sqrt {b^{2} c}}+\frac {\frac {-\frac {\left (-5 \sqrt {b^{2} c}+2 a \right ) \left (a +b \sqrt {d x +c}\right )^{\frac {3}{2}}}{4 \left (b^{2} c -2 a \sqrt {b^{2} c}+a^{2}\right )}+\frac {\left (-7 \sqrt {b^{2} c}+2 a \right ) \sqrt {a +b \sqrt {d x +c}}}{-4 \sqrt {b^{2} c}+4 a}}{\left (-b \sqrt {d x +c}-\sqrt {b^{2} c}\right )^{2}}-\frac {\left (5 \sqrt {b^{2} c}-2 a \right ) \arctan \left (\frac {\sqrt {a +b \sqrt {d x +c}}}{\sqrt {\sqrt {b^{2} c}-a}}\right )}{4 \left (-b^{2} c +2 a \sqrt {b^{2} c}-a^{2}\right ) \sqrt {\sqrt {b^{2} c}-a}}}{16 b^{2} c \sqrt {b^{2} c}}\right )\)	\(427\)
default	\(4 d^{2} b^{4} \left (-\frac {\frac {-\frac {\left (5 \sqrt {b^{2} c}+2 a \right ) \left (a +b \sqrt {d x +c}\right )^{\frac {3}{2}}}{4 \left (b^{2} c +2 a \sqrt {b^{2} c}+a^{2}\right )}+\frac {\left (7 \sqrt {b^{2} c}+2 a \right ) \sqrt {a +b \sqrt {d x +c}}}{4 \sqrt {b^{2} c}+4 a}}{\left (-b \sqrt {d x +c}+\sqrt {b^{2} c}\right )^{2}}-\frac {\left (5 \sqrt {b^{2} c}+2 a \right ) \arctan \left (\frac {\sqrt {a +b \sqrt {d x +c}}}{\sqrt {-\sqrt {b^{2} c}-a}}\right )}{4 \left (b^{2} c +2 a \sqrt {b^{2} c}+a^{2}\right ) \sqrt {-\sqrt {b^{2} c}-a}}}{16 b^{2} c \sqrt {b^{2} c}}+\frac {\frac {-\frac {\left (-5 \sqrt {b^{2} c}+2 a \right ) \left (a +b \sqrt {d x +c}\right )^{\frac {3}{2}}}{4 \left (b^{2} c -2 a \sqrt {b^{2} c}+a^{2}\right )}+\frac {\left (-7 \sqrt {b^{2} c}+2 a \right ) \sqrt {a +b \sqrt {d x +c}}}{-4 \sqrt {b^{2} c}+4 a}}{\left (-b \sqrt {d x +c}-\sqrt {b^{2} c}\right )^{2}}-\frac {\left (5 \sqrt {b^{2} c}-2 a \right ) \arctan \left (\frac {\sqrt {a +b \sqrt {d x +c}}}{\sqrt {\sqrt {b^{2} c}-a}}\right )}{4 \left (-b^{2} c +2 a \sqrt {b^{2} c}-a^{2}\right ) \sqrt {\sqrt {b^{2} c}-a}}}{16 b^{2} c \sqrt {b^{2} c}}\right )\)	\(427\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate(1/(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. 4390 vs. \(2 (212) = 424\).
time = 1.28, size = 4390, normalized size = 16.82 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*((b^4*c^3 - 2*a^2*b^2*c^2 + a^4*c)*x^2*sqrt(-((105*a*b^8*c^3 + 70*a^3*b^6*c^2 - 35*a^5*b^4*c + 4*a^7*b^2
)*d^4 + (b^10*c^8 - 5*a^2*b^8*c^7 + 10*a^4*b^6*c^6 - 10*a^6*b^4*c^5 + 5*a^8*b^2*c^4 - a^10*c^3)*sqrt((625*b^18
*c^4 + 7700*a^2*b^16*c^3 + 21966*a^4*b^14*c^2 - 10780*a^6*b^12*c + 1225*a^8*b^10)*d^8/(b^20*c^13 - 10*a^2*b^18
*c^12 + 45*a^4*b^16*c^11 - 120*a^6*b^14*c^10 + 210*a^8*b^12*c^9 - 252*a^10*b^10*c^8 + 210*a^12*b^8*c^7 - 120*a
^14*b^6*c^6 + 45*a^16*b^4*c^5 - 10*a^18*b^2*c^4 + a^20*c^3)))/(b^10*c^8 - 5*a^2*b^8*c^7 + 10*a^4*b^6*c^6 - 10*
a^6*b^4*c^5 + 5*a^8*b^2*c^4 - a^10*c^3))*log((625*b^12*c^3 + 3750*a^2*b^10*c^2 - 1491*a^4*b^8*c + 140*a^6*b^6)
*sqrt(sqrt(d*x + c)*b + a)*d^6 + ((325*a*b^12*c^5 + 1977*a^3*b^10*c^4 - 609*a^5*b^8*c^3 + 35*a^7*b^6*c^2)*d^4
- (5*b^14*c^10 - 16*a^2*b^12*c^9 + 3*a^4*b^10*c^8 + 50*a^6*b^8*c^7 - 85*a^8*b^6*c^6 + 60*a^10*b^4*c^5 - 19*a^1
2*b^2*c^4 + 2*a^14*c^3)*sqrt((625*b^18*c^4 + 7700*a^2*b^16*c^3 + 21966*a^4*b^14*c^2 - 10780*a^6*b^12*c + 1225*
a^8*b^10)*d^8/(b^20*c^13 - 10*a^2*b^18*c^12 + 45*a^4*b^16*c^11 - 120*a^6*b^14*c^10 + 210*a^8*b^12*c^9 - 252*a^
10*b^10*c^8 + 210*a^12*b^8*c^7 - 120*a^14*b^6*c^6 + 45*a^16*b^4*c^5 - 10*a^18*b^2*c^4 + a^20*c^3)))*sqrt(-((10
5*a*b^8*c^3 + 70*a^3*b^6*c^2 - 35*a^5*b^4*c + 4*a^7*b^2)*d^4 + (b^10*c^8 - 5*a^2*b^8*c^7 + 10*a^4*b^6*c^6 - 10
*a^6*b^4*c^5 + 5*a^8*b^2*c^4 - a^10*c^3)*sqrt((625*b^18*c^4 + 7700*a^2*b^16*c^3 + 21966*a^4*b^14*c^2 - 10780*a
^6*b^12*c + 1225*a^8*b^10)*d^8/(b^20*c^13 - 10*a^2*b^18*c^12 + 45*a^4*b^16*c^11 - 120*a^6*b^14*c^10 + 210*a^8*
b^12*c^9 - 252*a^10*b^10*c^8 + 210*a^12*b^8*c^7 - 120*a^14*b^6*c^6 + 45*a^16*b^4*c^5 - 10*a^18*b^2*c^4 + a^20*
c^3)))/(b^10*c^8 - 5*a^2*b^8*c^7 + 10*a^4*b^6*c^6 - 10*a^6*b^4*c^5 + 5*a^8*b^2*c^4 - a^10*c^3))) - (b^4*c^3 -
2*a^2*b^2*c^2 + a^4*c)*x^2*sqrt(-((105*a*b^8*c^3 + 70*a^3*b^6*c^2 - 35*a^5*b^4*c + 4*a^7*b^2)*d^4 + (b^10*c^8
- 5*a^2*b^8*c^7 + 10*a^4*b^6*c^6 - 10*a^6*b^4*c^5 + 5*a^8*b^2*c^4 - a^10*c^3)*sqrt((625*b^18*c^4 + 7700*a^2*b^
16*c^3 + 21966*a^4*b^14*c^2 - 10780*a^6*b^12*c + 1225*a^8*b^10)*d^8/(b^20*c^13 - 10*a^2*b^18*c^12 + 45*a^4*b^1
6*c^11 - 120*a^6*b^14*c^10 + 210*a^8*b^12*c^9 - 252*a^10*b^10*c^8 + 210*a^12*b^8*c^7 - 120*a^14*b^6*c^6 + 45*a
^16*b^4*c^5 - 10*a^18*b^2*c^4 + a^20*c^3)))/(b^10*c^8 - 5*a^2*b^8*c^7 + 10*a^4*b^6*c^6 - 10*a^6*b^4*c^5 + 5*a^
8*b^2*c^4 - a^10*c^3))*log((625*b^12*c^3 + 3750*a^2*b^10*c^2 - 1491*a^4*b^8*c + 140*a^6*b^6)*sqrt(sqrt(d*x + c
)*b + a)*d^6 - ((325*a*b^12*c^5 + 1977*a^3*b^10*c^4 - 609*a^5*b^8*c^3 + 35*a^7*b^6*c^2)*d^4 - (5*b^14*c^10 - 1
6*a^2*b^12*c^9 + 3*a^4*b^10*c^8 + 50*a^6*b^8*c^7 - 85*a^8*b^6*c^6 + 60*a^10*b^4*c^5 - 19*a^12*b^2*c^4 + 2*a^14
*c^3)*sqrt((625*b^18*c^4 + 7700*a^2*b^16*c^3 + 21966*a^4*b^14*c^2 - 10780*a^6*b^12*c + 1225*a^8*b^10)*d^8/(b^2
0*c^13 - 10*a^2*b^18*c^12 + 45*a^4*b^16*c^11 - 120*a^6*b^14*c^10 + 210*a^8*b^12*c^9 - 252*a^10*b^10*c^8 + 210*
a^12*b^8*c^7 - 120*a^14*b^6*c^6 + 45*a^16*b^4*c^5 - 10*a^18*b^2*c^4 + a^20*c^3)))*sqrt(-((105*a*b^8*c^3 + 70*a
^3*b^6*c^2 - 35*a^5*b^4*c + 4*a^7*b^2)*d^4 + (b^10*c^8 - 5*a^2*b^8*c^7 + 10*a^4*b^6*c^6 - 10*a^6*b^4*c^5 + 5*a
^8*b^2*c^4 - a^10*c^3)*sqrt((625*b^18*c^4 + 7700*a^2*b^16*c^3 + 21966*a^4*b^14*c^2 - 10780*a^6*b^12*c + 1225*a
^8*b^10)*d^8/(b^20*c^13 - 10*a^2*b^18*c^12 + 45*a^4*b^16*c^11 - 120*a^6*b^14*c^10 + 210*a^8*b^12*c^9 - 252*a^1
0*b^10*c^8 + 210*a^12*b^8*c^7 - 120*a^14*b^6*c^6 + 45*a^16*b^4*c^5 - 10*a^18*b^2*c^4 + a^20*c^3)))/(b^10*c^8 -
5*a^2*b^8*c^7 + 10*a^4*b^6*c^6 - 10*a^6*b^4*c^5 + 5*a^8*b^2*c^4 - a^10*c^3))) + (b^4*c^3 - 2*a^2*b^2*c^2 + a^
4*c)*x^2*sqrt(-((105*a*b^8*c^3 + 70*a^3*b^6*c^2 - 35*a^5*b^4*c + 4*a^7*b^2)*d^4 - (b^10*c^8 - 5*a^2*b^8*c^7 +
10*a^4*b^6*c^6 - 10*a^6*b^4*c^5 + 5*a^8*b^2*c^4 - a^10*c^3)*sqrt((625*b^18*c^4 + 7700*a^2*b^16*c^3 + 21966*a^4
*b^14*c^2 - 10780*a^6*b^12*c + 1225*a^8*b^10)*d^8/(b^20*c^13 - 10*a^2*b^18*c^12 + 45*a^4*b^16*c^11 - 120*a^6*b
^14*c^10 + 210*a^8*b^12*c^9 - 252*a^10*b^10*c^8 + 210*a^12*b^8*c^7 - 120*a^14*b^6*c^6 + 45*a^16*b^4*c^5 - 10*a
^18*b^2*c^4 + a^20*c^3)))/(b^10*c^8 - 5*a^2*b^8*c^7 + 10*a^4*b^6*c^6 - 10*a^6*b^4*c^5 + 5*a^8*b^2*c^4 - a^10*c
^3))*log((625*b^12*c^3 + 3750*a^2*b^10*c^2 - 1491*a^4*b^8*c + 140*a^6*b^6)*sqrt(sqrt(d*x + c)*b + a)*d^6 + ((3
25*a*b^12*c^5 + 1977*a^3*b^10*c^4 - 609*a^5*b^8*c^3 + 35*a^7*b^6*c^2)*d^4 + (5*b^14*c^10 - 16*a^2*b^12*c^9 + 3
*a^4*b^10*c^8 + 50*a^6*b^8*c^7 - 85*a^8*b^6*c^6 + 60*a^10*b^4*c^5 - 19*a^12*b^2*c^4 + 2*a^14*c^3)*sqrt((625*b^
18*c^4 + 7700*a^2*b^16*c^3 + 21966*a^4*b^14*c^2 - 10780*a^6*b^12*c + 1225*a^8*b^10)*d^8/(b^20*c^13 - 10*a^2*b^
18*c^12 + 45*a^4*b^16*c^11 - 120*a^6*b^14*c^10 + 210*a^8*b^12*c^9 - 252*a^10*b^10*c^8 + 210*a^12*b^8*c^7 - 120
*a^14*b^6*c^6 + 45*a^16*b^4*c^5 - 10*a^18*b^2*c^4 + a^20*c^3)))*sqrt(-((105*a*b^8*c^3 + 70*a^3*b^6*c^2 - 35*a^
5*b^4*c + 4*a^7*b^2)*d^4 - (b^10*c^8 - 5*a^2*b^8*c^7 + 10*a^4*b^6*c^6 - 10*a^6*b^4*c^5 + 5*a^8*b^2*c^4 - a^10*
c^3)*sqrt((625*b^18*c^4 + 7700*a^2*b^16*c^3 + 21966*a^4*b^14*c^2 - 10780*a^6*b^12*c + 1225*a^8*b^10)*d^8/(b^20
*c^13 - 10*a^2*b^18*c^12 + 45*a^4*b^16*c^11 - 1...

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{3} \sqrt {a + b \sqrt {c + d x}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**3*sqrt(a + b*sqrt(c + d*x))), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 1303 vs. \(2 (212) = 424\).
time = 3.97, size = 1303, normalized size = 4.99 \begin {gather*} \frac {\frac {{\left ({\left (b^{5} c^{3} - 2 \, a^{2} b^{3} c^{2} + a^{4} b c\right )}^{2} {\left (5 \, b^{6} c + a^{2} b^{4}\right )} d^{3} - {\left (13 \, a b^{10} c^{\frac {7}{2}} - 27 \, a^{3} b^{8} c^{\frac {5}{2}} + 15 \, a^{5} b^{6} c^{\frac {3}{2}} - a^{7} b^{4} \sqrt {c}\right )} d^{3} {\left | b^{5} c^{3} - 2 \, a^{2} b^{3} c^{2} + a^{4} b c \right |} + 2 \, {\left (4 \, a^{2} b^{14} c^{6} - 17 \, a^{4} b^{12} c^{5} + 28 \, a^{6} b^{10} c^{4} - 22 \, a^{8} b^{8} c^{3} + 8 \, a^{10} b^{6} c^{2} - a^{12} b^{4} c\right )} d^{3}\right )} \arctan \left (\frac {\sqrt {\sqrt {d x + c} b + a}}{\sqrt {-\frac {a b^{4} c^{3} - 2 \, a^{3} b^{2} c^{2} + a^{5} c + \sqrt {{\left (a b^{4} c^{3} - 2 \, a^{3} b^{2} c^{2} + a^{5} c\right )}^{2} + {\left (b^{6} c^{4} - 3 \, a^{2} b^{4} c^{3} + 3 \, a^{4} b^{2} c^{2} - a^{6} c\right )} {\left (b^{4} c^{3} - 2 \, a^{2} b^{2} c^{2} + a^{4} c\right )}}}{b^{4} c^{3} - 2 \, a^{2} b^{2} c^{2} + a^{4} c}}}\right )}{{\left (b^{9} c^{6} - a b^{8} c^{\frac {11}{2}} - 4 \, a^{2} b^{7} c^{5} + 4 \, a^{3} b^{6} c^{\frac {9}{2}} + 6 \, a^{4} b^{5} c^{4} - 6 \, a^{5} b^{4} c^{\frac {7}{2}} - 4 \, a^{6} b^{3} c^{3} + 4 \, a^{7} b^{2} c^{\frac {5}{2}} + a^{8} b c^{2} - a^{9} c^{\frac {3}{2}}\right )} \sqrt {-b \sqrt {c} - a} {\left | b^{5} c^{3} - 2 \, a^{2} b^{3} c^{2} + a^{4} b c \right |}} + \frac {{\left ({\left (b^{5} c^{3} - 2 \, a^{2} b^{3} c^{2} + a^{4} b c\right )}^{2} {\left (5 \, b^{6} c + a^{2} b^{4}\right )} d^{3} + {\left (13 \, a b^{10} c^{\frac {7}{2}} - 27 \, a^{3} b^{8} c^{\frac {5}{2}} + 15 \, a^{5} b^{6} c^{\frac {3}{2}} - a^{7} b^{4} \sqrt {c}\right )} d^{3} {\left | b^{5} c^{3} - 2 \, a^{2} b^{3} c^{2} + a^{4} b c \right |} + 2 \, {\left (4 \, a^{2} b^{14} c^{6} - 17 \, a^{4} b^{12} c^{5} + 28 \, a^{6} b^{10} c^{4} - 22 \, a^{8} b^{8} c^{3} + 8 \, a^{10} b^{6} c^{2} - a^{12} b^{4} c\right )} d^{3}\right )} \arctan \left (\frac {\sqrt {\sqrt {d x + c} b + a}}{\sqrt {-\frac {a b^{4} c^{3} - 2 \, a^{3} b^{2} c^{2} + a^{5} c - \sqrt {{\left (a b^{4} c^{3} - 2 \, a^{3} b^{2} c^{2} + a^{5} c\right )}^{2} + {\left (b^{6} c^{4} - 3 \, a^{2} b^{4} c^{3} + 3 \, a^{4} b^{2} c^{2} - a^{6} c\right )} {\left (b^{4} c^{3} - 2 \, a^{2} b^{2} c^{2} + a^{4} c\right )}}}{b^{4} c^{3} - 2 \, a^{2} b^{2} c^{2} + a^{4} c}}}\right )}{{\left (b^{9} c^{6} + a b^{8} c^{\frac {11}{2}} - 4 \, a^{2} b^{7} c^{5} - 4 \, a^{3} b^{6} c^{\frac {9}{2}} + 6 \, a^{4} b^{5} c^{4} + 6 \, a^{5} b^{4} c^{\frac {7}{2}} - 4 \, a^{6} b^{3} c^{3} - 4 \, a^{7} b^{2} c^{\frac {5}{2}} + a^{8} b c^{2} + a^{9} c^{\frac {3}{2}}\right )} \sqrt {b \sqrt {c} - a} {\left | b^{5} c^{3} - 2 \, a^{2} b^{3} c^{2} + a^{4} b c \right |}} - \frac {2 \, {\left (9 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {3}{2}} b^{8} c^{2} d^{3} - 19 \, \sqrt {\sqrt {d x + c} b + a} a b^{8} c^{2} d^{3} - 5 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {7}{2}} b^{6} c d^{3} + 21 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {5}{2}} a b^{6} c d^{3} - 30 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {3}{2}} a^{2} b^{6} c d^{3} + 18 \, \sqrt {\sqrt {d x + c} b + a} a^{3} b^{6} c d^{3} - {\left (\sqrt {d x + c} b + a\right )}^{\frac {7}{2}} a^{2} b^{4} d^{3} + 3 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {5}{2}} a^{3} b^{4} d^{3} - 3 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {3}{2}} a^{4} b^{4} d^{3} + \sqrt {\sqrt {d x + c} b + a} a^{5} b^{4} d^{3}\right )}}{{\left (b^{4} c^{3} - 2 \, a^{2} b^{2} c^{2} + a^{4} 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^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(((b^5*c^3 - 2*a^2*b^3*c^2 + a^4*b*c)^2*(5*b^6*c + a^2*b^4)*d^3 - (13*a*b^10*c^(7/2) - 27*a^3*b^8*c^(5/2)
+ 15*a^5*b^6*c^(3/2) - a^7*b^4*sqrt(c))*d^3*abs(b^5*c^3 - 2*a^2*b^3*c^2 + a^4*b*c) + 2*(4*a^2*b^14*c^6 - 17*a
^4*b^12*c^5 + 28*a^6*b^10*c^4 - 22*a^8*b^8*c^3 + 8*a^10*b^6*c^2 - a^12*b^4*c)*d^3)*arctan(sqrt(sqrt(d*x + c)*b
+ a)/sqrt(-(a*b^4*c^3 - 2*a^3*b^2*c^2 + a^5*c + sqrt((a*b^4*c^3 - 2*a^3*b^2*c^2 + a^5*c)^2 + (b^6*c^4 - 3*a^2
*b^4*c^3 + 3*a^4*b^2*c^2 - a^6*c)*(b^4*c^3 - 2*a^2*b^2*c^2 + a^4*c)))/(b^4*c^3 - 2*a^2*b^2*c^2 + a^4*c)))/((b^
9*c^6 - a*b^8*c^(11/2) - 4*a^2*b^7*c^5 + 4*a^3*b^6*c^(9/2) + 6*a^4*b^5*c^4 - 6*a^5*b^4*c^(7/2) - 4*a^6*b^3*c^3
+ 4*a^7*b^2*c^(5/2) + a^8*b*c^2 - a^9*c^(3/2))*sqrt(-b*sqrt(c) - a)*abs(b^5*c^3 - 2*a^2*b^3*c^2 + a^4*b*c)) +
((b^5*c^3 - 2*a^2*b^3*c^2 + a^4*b*c)^2*(5*b^6*c + a^2*b^4)*d^3 + (13*a*b^10*c^(7/2) - 27*a^3*b^8*c^(5/2) + 15
*a^5*b^6*c^(3/2) - a^7*b^4*sqrt(c))*d^3*abs(b^5*c^3 - 2*a^2*b^3*c^2 + a^4*b*c) + 2*(4*a^2*b^14*c^6 - 17*a^4*b^
12*c^5 + 28*a^6*b^10*c^4 - 22*a^8*b^8*c^3 + 8*a^10*b^6*c^2 - a^12*b^4*c)*d^3)*arctan(sqrt(sqrt(d*x + c)*b + a)
/sqrt(-(a*b^4*c^3 - 2*a^3*b^2*c^2 + a^5*c - sqrt((a*b^4*c^3 - 2*a^3*b^2*c^2 + a^5*c)^2 + (b^6*c^4 - 3*a^2*b^4*
c^3 + 3*a^4*b^2*c^2 - a^6*c)*(b^4*c^3 - 2*a^2*b^2*c^2 + a^4*c)))/(b^4*c^3 - 2*a^2*b^2*c^2 + a^4*c)))/((b^9*c^6
+ a*b^8*c^(11/2) - 4*a^2*b^7*c^5 - 4*a^3*b^6*c^(9/2) + 6*a^4*b^5*c^4 + 6*a^5*b^4*c^(7/2) - 4*a^6*b^3*c^3 - 4*
a^7*b^2*c^(5/2) + a^8*b*c^2 + a^9*c^(3/2))*sqrt(b*sqrt(c) - a)*abs(b^5*c^3 - 2*a^2*b^3*c^2 + a^4*b*c)) - 2*(9*
(sqrt(d*x + c)*b + a)^(3/2)*b^8*c^2*d^3 - 19*sqrt(sqrt(d*x + c)*b + a)*a*b^8*c^2*d^3 - 5*(sqrt(d*x + c)*b + a)
^(7/2)*b^6*c*d^3 + 21*(sqrt(d*x + c)*b + a)^(5/2)*a*b^6*c*d^3 - 30*(sqrt(d*x + c)*b + a)^(3/2)*a^2*b^6*c*d^3 +
18*sqrt(sqrt(d*x + c)*b + a)*a^3*b^6*c*d^3 - (sqrt(d*x + c)*b + a)^(7/2)*a^2*b^4*d^3 + 3*(sqrt(d*x + c)*b + a
)^(5/2)*a^3*b^4*d^3 - 3*(sqrt(d*x + c)*b + a)^(3/2)*a^4*b^4*d^3 + sqrt(sqrt(d*x + c)*b + a)*a^5*b^4*d^3)/((b^4
*c^3 - 2*a^2*b^2*c^2 + a^4*c)*(b^2*c - (sqrt(d*x + c)*b + a)^2 + 2*(sqrt(d*x + c)*b + a)*a - a^2)^2))/(b^2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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