3.8.30 \(\int \frac {1}{x^3 \sqrt {a+b x} (c+d x)^{5/2}} \, dx\)
Optimal. Leaf size=277 \[ \frac {d \sqrt {a+b x} \left (-35 a^2 d^2+18 a b c d+9 b^2 c^2\right )}{12 a^2 c^3 (c+d x)^{3/2} (b c-a d)}+\frac {\sqrt {a+b x} (7 a d+3 b c)}{4 a^2 c^2 x (c+d x)^{3/2}}-\frac {\left (35 a^2 d^2+10 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{5/2} c^{9/2}}+\frac {d \sqrt {a+b x} \left (105 a^3 d^3-145 a^2 b c d^2+15 a b^2 c^2 d+9 b^3 c^3\right )}{12 a^2 c^4 \sqrt {c+d x} (b c-a d)^2}-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}} \]
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Rubi [A] time = 0.25, antiderivative size = 277, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used =
{103, 151, 152, 12, 93, 208} \begin {gather*} \frac {d \sqrt {a+b x} \left (-145 a^2 b c d^2+105 a^3 d^3+15 a b^2 c^2 d+9 b^3 c^3\right )}{12 a^2 c^4 \sqrt {c+d x} (b c-a d)^2}+\frac {d \sqrt {a+b x} \left (-35 a^2 d^2+18 a b c d+9 b^2 c^2\right )}{12 a^2 c^3 (c+d x)^{3/2} (b c-a d)}-\frac {\left (35 a^2 d^2+10 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{5/2} c^{9/2}}+\frac {\sqrt {a+b x} (7 a d+3 b c)}{4 a^2 c^2 x (c+d x)^{3/2}}-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[1/(x^3*Sqrt[a + b*x]*(c + d*x)^(5/2)),x]
[Out]
(d*(9*b^2*c^2 + 18*a*b*c*d - 35*a^2*d^2)*Sqrt[a + b*x])/(12*a^2*c^3*(b*c - a*d)*(c + d*x)^(3/2)) - Sqrt[a + b*
x]/(2*a*c*x^2*(c + d*x)^(3/2)) + ((3*b*c + 7*a*d)*Sqrt[a + b*x])/(4*a^2*c^2*x*(c + d*x)^(3/2)) + (d*(9*b^3*c^3
+ 15*a*b^2*c^2*d - 145*a^2*b*c*d^2 + 105*a^3*d^3)*Sqrt[a + b*x])/(12*a^2*c^4*(b*c - a*d)^2*Sqrt[c + d*x]) - (
(3*b^2*c^2 + 10*a*b*c*d + 35*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(5/2)*c^(
9/2))
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 93
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]
Rule 103
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])
Rule 151
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]
Rule 152
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]
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]
Rubi steps
\begin {align*} \int \frac {1}{x^3 \sqrt {a+b x} (c+d x)^{5/2}} \, dx &=-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}-\frac {\int \frac {\frac {1}{2} (3 b c+7 a d)+3 b d x}{x^2 \sqrt {a+b x} (c+d x)^{5/2}} \, dx}{2 a c}\\ &=-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}+\frac {(3 b c+7 a d) \sqrt {a+b x}}{4 a^2 c^2 x (c+d x)^{3/2}}+\frac {\int \frac {\frac {1}{4} \left (3 b^2 c^2+10 a b c d+35 a^2 d^2\right )+b d (3 b c+7 a d) x}{x \sqrt {a+b x} (c+d x)^{5/2}} \, dx}{2 a^2 c^2}\\ &=\frac {d \left (9 b^2 c^2+18 a b c d-35 a^2 d^2\right ) \sqrt {a+b x}}{12 a^2 c^3 (b c-a d) (c+d x)^{3/2}}-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}+\frac {(3 b c+7 a d) \sqrt {a+b x}}{4 a^2 c^2 x (c+d x)^{3/2}}-\frac {\int \frac {-\frac {3}{8} (b c-a d) \left (3 b^2 c^2+10 a b c d+35 a^2 d^2\right )-\frac {1}{4} b d \left (9 b^2 c^2+18 a b c d-35 a^2 d^2\right ) x}{x \sqrt {a+b x} (c+d x)^{3/2}} \, dx}{3 a^2 c^3 (b c-a d)}\\ &=\frac {d \left (9 b^2 c^2+18 a b c d-35 a^2 d^2\right ) \sqrt {a+b x}}{12 a^2 c^3 (b c-a d) (c+d x)^{3/2}}-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}+\frac {(3 b c+7 a d) \sqrt {a+b x}}{4 a^2 c^2 x (c+d x)^{3/2}}+\frac {d \left (9 b^3 c^3+15 a b^2 c^2 d-145 a^2 b c d^2+105 a^3 d^3\right ) \sqrt {a+b x}}{12 a^2 c^4 (b c-a d)^2 \sqrt {c+d x}}+\frac {2 \int \frac {3 (b c-a d)^2 \left (3 b^2 c^2+10 a b c d+35 a^2 d^2\right )}{16 x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{3 a^2 c^4 (b c-a d)^2}\\ &=\frac {d \left (9 b^2 c^2+18 a b c d-35 a^2 d^2\right ) \sqrt {a+b x}}{12 a^2 c^3 (b c-a d) (c+d x)^{3/2}}-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}+\frac {(3 b c+7 a d) \sqrt {a+b x}}{4 a^2 c^2 x (c+d x)^{3/2}}+\frac {d \left (9 b^3 c^3+15 a b^2 c^2 d-145 a^2 b c d^2+105 a^3 d^3\right ) \sqrt {a+b x}}{12 a^2 c^4 (b c-a d)^2 \sqrt {c+d x}}+\frac {\left (3 b^2 c^2+10 a b c d+35 a^2 d^2\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{8 a^2 c^4}\\ &=\frac {d \left (9 b^2 c^2+18 a b c d-35 a^2 d^2\right ) \sqrt {a+b x}}{12 a^2 c^3 (b c-a d) (c+d x)^{3/2}}-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}+\frac {(3 b c+7 a d) \sqrt {a+b x}}{4 a^2 c^2 x (c+d x)^{3/2}}+\frac {d \left (9 b^3 c^3+15 a b^2 c^2 d-145 a^2 b c d^2+105 a^3 d^3\right ) \sqrt {a+b x}}{12 a^2 c^4 (b c-a d)^2 \sqrt {c+d x}}+\frac {\left (3 b^2 c^2+10 a b c d+35 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{4 a^2 c^4}\\ &=\frac {d \left (9 b^2 c^2+18 a b c d-35 a^2 d^2\right ) \sqrt {a+b x}}{12 a^2 c^3 (b c-a d) (c+d x)^{3/2}}-\frac {\sqrt {a+b x}}{2 a c x^2 (c+d x)^{3/2}}+\frac {(3 b c+7 a d) \sqrt {a+b x}}{4 a^2 c^2 x (c+d x)^{3/2}}+\frac {d \left (9 b^3 c^3+15 a b^2 c^2 d-145 a^2 b c d^2+105 a^3 d^3\right ) \sqrt {a+b x}}{12 a^2 c^4 (b c-a d)^2 \sqrt {c+d x}}-\frac {\left (3 b^2 c^2+10 a b c d+35 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{5/2} c^{9/2}}\\ \end {align*}
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Mathematica [A] time = 0.54, size = 274, normalized size = 0.99 \begin {gather*} \frac {\frac {d \sqrt {a+b x} \left (35 a^2 d^2-18 a b c d-9 b^2 c^2\right )}{a c^2 (a d-b c)}-\frac {(c+d x) \left (3 \sqrt {c+d x} (b c-a d)^2 \left (35 a^2 d^2+10 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )-\sqrt {a} \sqrt {c} d \sqrt {a+b x} \left (105 a^3 d^3-145 a^2 b c d^2+15 a b^2 c^2 d+9 b^3 c^3\right )\right )}{a^{3/2} c^{7/2} (b c-a d)^2}+\frac {3 \sqrt {a+b x} (7 a d+3 b c)}{a c x}-\frac {6 \sqrt {a+b x}}{x^2}}{12 a c (c+d x)^{3/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[1/(x^3*Sqrt[a + b*x]*(c + d*x)^(5/2)),x]
[Out]
((d*(-9*b^2*c^2 - 18*a*b*c*d + 35*a^2*d^2)*Sqrt[a + b*x])/(a*c^2*(-(b*c) + a*d)) - (6*Sqrt[a + b*x])/x^2 + (3*
(3*b*c + 7*a*d)*Sqrt[a + b*x])/(a*c*x) - ((c + d*x)*(-(Sqrt[a]*Sqrt[c]*d*(9*b^3*c^3 + 15*a*b^2*c^2*d - 145*a^2
*b*c*d^2 + 105*a^3*d^3)*Sqrt[a + b*x]) + 3*(b*c - a*d)^2*(3*b^2*c^2 + 10*a*b*c*d + 35*a^2*d^2)*Sqrt[c + d*x]*A
rcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])]))/(a^(3/2)*c^(7/2)*(b*c - a*d)^2))/(12*a*c*(c + d*x)^(
3/2))
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IntegrateAlgebraic [A] time = 0.44, size = 368, normalized size = 1.33 \begin {gather*} \frac {\left (-35 a^2 d^2-10 a b c d-3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{5/2} c^{9/2}}+\frac {\sqrt {a+b x} \left (105 a^5 d^4-\frac {175 a^4 c d^4 (a+b x)}{c+d x}-180 a^4 b c d^3+54 a^3 b^2 c^2 d^2+\frac {56 a^3 c^2 d^4 (a+b x)^2}{(c+d x)^2}+\frac {300 a^3 b c^2 d^3 (a+b x)}{c+d x}+12 a^2 b^3 c^3 d-\frac {90 a^2 b^2 c^3 d^2 (a+b x)}{c+d x}+\frac {8 a^2 c^3 d^4 (a+b x)^3}{(c+d x)^3}-\frac {96 a^2 b c^3 d^3 (a+b x)^2}{(c+d x)^2}+\frac {9 b^4 c^5 (a+b x)}{c+d x}-15 a b^4 c^4+\frac {12 a b^3 c^4 d (a+b x)}{c+d x}\right )}{12 a^2 c^4 \sqrt {c+d x} (a d-b c)^2 \left (a-\frac {c (a+b x)}{c+d x}\right )^2} \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[1/(x^3*Sqrt[a + b*x]*(c + d*x)^(5/2)),x]
[Out]
(Sqrt[a + b*x]*(-15*a*b^4*c^4 + 12*a^2*b^3*c^3*d + 54*a^3*b^2*c^2*d^2 - 180*a^4*b*c*d^3 + 105*a^5*d^4 + (8*a^2
*c^3*d^4*(a + b*x)^3)/(c + d*x)^3 - (96*a^2*b*c^3*d^3*(a + b*x)^2)/(c + d*x)^2 + (56*a^3*c^2*d^4*(a + b*x)^2)/
(c + d*x)^2 + (9*b^4*c^5*(a + b*x))/(c + d*x) + (12*a*b^3*c^4*d*(a + b*x))/(c + d*x) - (90*a^2*b^2*c^3*d^2*(a
+ b*x))/(c + d*x) + (300*a^3*b*c^2*d^3*(a + b*x))/(c + d*x) - (175*a^4*c*d^4*(a + b*x))/(c + d*x)))/(12*a^2*c^
4*(-(b*c) + a*d)^2*Sqrt[c + d*x]*(a - (c*(a + b*x))/(c + d*x))^2) + ((-3*b^2*c^2 - 10*a*b*c*d - 35*a^2*d^2)*Ar
cTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(5/2)*c^(9/2))
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fricas [B] time = 9.62, size = 1186, normalized size = 4.28
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x^3/(d*x+c)^(5/2)/(b*x+a)^(1/2),x, algorithm="fricas")
[Out]
[1/48*(3*((3*b^4*c^4*d^2 + 4*a*b^3*c^3*d^3 + 18*a^2*b^2*c^2*d^4 - 60*a^3*b*c*d^5 + 35*a^4*d^6)*x^4 + 2*(3*b^4*
c^5*d + 4*a*b^3*c^4*d^2 + 18*a^2*b^2*c^3*d^3 - 60*a^3*b*c^2*d^4 + 35*a^4*c*d^5)*x^3 + (3*b^4*c^6 + 4*a*b^3*c^5
*d + 18*a^2*b^2*c^4*d^2 - 60*a^3*b*c^3*d^3 + 35*a^4*c^2*d^4)*x^2)*sqrt(a*c)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*
c*d + a^2*d^2)*x^2 - 4*(2*a*c + (b*c + a*d)*x)*sqrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x
)/x^2) - 4*(6*a^2*b^2*c^6 - 12*a^3*b*c^5*d + 6*a^4*c^4*d^2 - (9*a*b^3*c^4*d^2 + 15*a^2*b^2*c^3*d^3 - 145*a^3*b
*c^2*d^4 + 105*a^4*c*d^5)*x^3 - 2*(9*a*b^3*c^5*d + 12*a^2*b^2*c^4*d^2 - 99*a^3*b*c^3*d^3 + 70*a^4*c^2*d^4)*x^2
- 3*(3*a*b^3*c^6 + a^2*b^2*c^5*d - 11*a^3*b*c^4*d^2 + 7*a^4*c^3*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/((a^3*b^
2*c^7*d^2 - 2*a^4*b*c^6*d^3 + a^5*c^5*d^4)*x^4 + 2*(a^3*b^2*c^8*d - 2*a^4*b*c^7*d^2 + a^5*c^6*d^3)*x^3 + (a^3*
b^2*c^9 - 2*a^4*b*c^8*d + a^5*c^7*d^2)*x^2), 1/24*(3*((3*b^4*c^4*d^2 + 4*a*b^3*c^3*d^3 + 18*a^2*b^2*c^2*d^4 -
60*a^3*b*c*d^5 + 35*a^4*d^6)*x^4 + 2*(3*b^4*c^5*d + 4*a*b^3*c^4*d^2 + 18*a^2*b^2*c^3*d^3 - 60*a^3*b*c^2*d^4 +
35*a^4*c*d^5)*x^3 + (3*b^4*c^6 + 4*a*b^3*c^5*d + 18*a^2*b^2*c^4*d^2 - 60*a^3*b*c^3*d^3 + 35*a^4*c^2*d^4)*x^2)*
sqrt(-a*c)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x + c)/(a*b*c*d*x^2 + a^2*c^2 +
(a*b*c^2 + a^2*c*d)*x)) - 2*(6*a^2*b^2*c^6 - 12*a^3*b*c^5*d + 6*a^4*c^4*d^2 - (9*a*b^3*c^4*d^2 + 15*a^2*b^2*c^
3*d^3 - 145*a^3*b*c^2*d^4 + 105*a^4*c*d^5)*x^3 - 2*(9*a*b^3*c^5*d + 12*a^2*b^2*c^4*d^2 - 99*a^3*b*c^3*d^3 + 70
*a^4*c^2*d^4)*x^2 - 3*(3*a*b^3*c^6 + a^2*b^2*c^5*d - 11*a^3*b*c^4*d^2 + 7*a^4*c^3*d^3)*x)*sqrt(b*x + a)*sqrt(d
*x + c))/((a^3*b^2*c^7*d^2 - 2*a^4*b*c^6*d^3 + a^5*c^5*d^4)*x^4 + 2*(a^3*b^2*c^8*d - 2*a^4*b*c^7*d^2 + a^5*c^6
*d^3)*x^3 + (a^3*b^2*c^9 - 2*a^4*b*c^8*d + a^5*c^7*d^2)*x^2)]
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giac [B] time = 7.18, size = 1234, normalized size = 4.45
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x^3/(d*x+c)^(5/2)/(b*x+a)^(1/2),x, algorithm="giac")
[Out]
-2/3*sqrt(b*x + a)*((11*b^4*c^5*d^5*abs(b) - 9*a*b^3*c^4*d^6*abs(b))*(b*x + a)/(b^4*c^10*d - 2*a*b^3*c^9*d^2 +
a^2*b^2*c^8*d^3) + 3*(4*b^5*c^6*d^4*abs(b) - 7*a*b^4*c^5*d^5*abs(b) + 3*a^2*b^3*c^4*d^6*abs(b))/(b^4*c^10*d -
2*a*b^3*c^9*d^2 + a^2*b^2*c^8*d^3))/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) - 1/4*(3*sqrt(b*d)*b^4*c^2 + 10*sqr
t(b*d)*a*b^3*c*d + 35*sqrt(b*d)*a^2*b^2*d^2)*arctan(-1/2*(b^2*c + a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*
c + (b*x + a)*b*d - a*b*d))^2)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*a^2*b*c^4*abs(b)) + 1/2*(3*sqrt(b*d)*b^10*c
^5 - sqrt(b*d)*a*b^9*c^4*d - 26*sqrt(b*d)*a^2*b^8*c^3*d^2 + 54*sqrt(b*d)*a^3*b^7*c^2*d^3 - 41*sqrt(b*d)*a^4*b^
6*c*d^4 + 11*sqrt(b*d)*a^5*b^5*d^5 - 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d
))^2*b^8*c^4 - 28*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^7*c^3*d + 50
*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^6*c^2*d^2 + 20*sqrt(b*d)*(s
qrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^3*b^5*c*d^3 - 33*sqrt(b*d)*(sqrt(b*d)*sqrt(b
*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^4*d^4 + 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2
*c + (b*x + a)*b*d - a*b*d))^4*b^6*c^3 + 39*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d -
a*b*d))^4*a*b^5*c^2*d + 31*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^2*b^4
*c*d^2 + 33*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^3*b^3*d^3 - 3*sqrt(b
*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^4*c^2 - 10*sqrt(b*d)*(sqrt(b*d)*sqrt(b
*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a*b^3*c*d - 11*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*
c + (b*x + a)*b*d - a*b*d))^6*a^2*b^2*d^2)/((b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 - 2*(sqrt(b*d)*sqrt(b*x + a)
- sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^2*c - 2*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b
*d))^2*a*b*d + (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4)^2*a^2*c^4*abs(b))
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maple [B] time = 0.04, size = 1288, normalized size = 4.65
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/x^3/(d*x+c)^(5/2)/(b*x+a)^(1/2),x)
[Out]
-1/24*(b*x+a)^(1/2)/a^2/c^4*(-280*x^2*a^3*c*d^4*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-36*x^2*b^3*c^4*d*(a*c)^(1/
2)*((b*x+a)*(d*x+c))^(1/2)-42*x*a^3*c^2*d^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-24*a^2*b*c^4*d*(a*c)^(1/2)*((b
*x+a)*(d*x+c))^(1/2)-18*x^3*b^3*c^3*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-210*x^3*a^3*d^5*(a*c)^(1/2)*((b*x+
a)*(d*x+c))^(1/2)+9*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^2*b^4*c^6+105*ln((a*d*x+
b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^4*a^4*d^6-180*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*
x+a)*(d*x+c))^(1/2))/x)*x^4*a^3*b*c*d^5+54*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^4
*a^2*b^2*c^2*d^4+12*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^4*a*b^3*c^3*d^3-360*ln((
a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^3*a^3*b*c^2*d^4+108*ln((a*d*x+b*c*x+2*a*c+2*(a*c
)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^3*a^2*b^2*c^3*d^3+24*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c
))^(1/2))/x)*x^3*a*b^3*c^4*d^2-180*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^2*a^3*b*c
^3*d^3+54*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^2*a^2*b^2*c^4*d^2+12*ln((a*d*x+b*c
*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^2*a*b^3*c^5*d-18*x*b^3*c^5*(a*c)^(1/2)*((b*x+a)*(d*x+c))^
(1/2)+12*a^3*c^3*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+12*a*b^2*c^5*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+9*ln
((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^4*b^4*c^4*d^2+210*ln((a*d*x+b*c*x+2*a*c+2*(a*c
)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^3*a^4*c*d^5+18*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/
2))/x)*x^3*b^4*c^5*d+105*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^2*a^4*c^2*d^4+290*x
^3*a^2*b*c*d^4*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-30*x^3*a*b^2*c^2*d^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+39
6*x^2*a^2*b*c^2*d^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-48*x^2*a*b^2*c^3*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/
2)+66*x*a^2*b*c^3*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-6*x*a*b^2*c^4*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))
/(a*d-b*c)^2/(a*c)^(1/2)/x^2/((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(3/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {b x + a} {\left (d x + c\right )}^{\frac {5}{2}} x^{3}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x^3/(d*x+c)^(5/2)/(b*x+a)^(1/2),x, algorithm="maxima")
[Out]
integrate(1/(sqrt(b*x + a)*(d*x + c)^(5/2)*x^3), x)
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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\,x}\,{\left (c+d\,x\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(x^3*(a + b*x)^(1/2)*(c + d*x)^(5/2)),x)
[Out]
int(1/(x^3*(a + b*x)^(1/2)*(c + d*x)^(5/2)), x)
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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 x} \left (c + d x\right )^{\frac {5}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x**3/(d*x+c)**(5/2)/(b*x+a)**(1/2),x)
[Out]
Integral(1/(x**3*sqrt(a + b*x)*(c + d*x)**(5/2)), x)
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