3.7.28 \(\int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x^4} \, dx\)

Optimal. Leaf size=228 \[ -\frac {\left (a^3 d^3-5 a^2 b c d^2+15 a b^2 c^2 d+5 b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 \sqrt {a} c^{5/2}}+2 b^{5/2} \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )-\frac {\sqrt {a+b x} \sqrt {c+d x} (5 b c-a d) (a d+b c)}{8 c^2 x}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 x^3}-\frac {(a+b x)^{3/2} \sqrt {c+d x} (a d+5 b c)}{12 c x^2} \]

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Rubi [A]  time = 0.19, antiderivative size = 228, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {97, 149, 157, 63, 217, 206, 93, 208} \begin {gather*} -\frac {\left (-5 a^2 b c d^2+a^3 d^3+15 a b^2 c^2 d+5 b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 \sqrt {a} c^{5/2}}+2 b^{5/2} \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )-\frac {\sqrt {a+b x} \sqrt {c+d x} (5 b c-a d) (a d+b c)}{8 c^2 x}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 x^3}-\frac {(a+b x)^{3/2} \sqrt {c+d x} (a d+5 b c)}{12 c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(5/2)*Sqrt[c + d*x])/x^4,x]

[Out]

-((5*b*c - a*d)*(b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*c^2*x) - ((5*b*c + a*d)*(a + b*x)^(3/2)*Sqrt[c + d
*x])/(12*c*x^2) - ((a + b*x)^(5/2)*Sqrt[c + d*x])/(3*x^3) - ((5*b^3*c^3 + 15*a*b^2*c^2*d - 5*a^2*b*c*d^2 + a^3
*d^3)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(8*Sqrt[a]*c^(5/2)) + 2*b^(5/2)*Sqrt[d]*ArcTan
h[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, 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 97

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

Rule 149

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*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegerQ[m]

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 206

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

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]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

\begin {align*} \int \frac {(a+b x)^{5/2} \sqrt {c+d x}}{x^4} \, dx &=-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 x^3}+\frac {1}{3} \int \frac {(a+b x)^{3/2} \left (\frac {1}{2} (5 b c+a d)+3 b d x\right )}{x^3 \sqrt {c+d x}} \, dx\\ &=-\frac {(5 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 c x^2}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 x^3}+\frac {\int \frac {\sqrt {a+b x} \left (\frac {3}{4} (5 b c-a d) (b c+a d)+6 b^2 c d x\right )}{x^2 \sqrt {c+d x}} \, dx}{6 c}\\ &=-\frac {(5 b c-a d) (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 c^2 x}-\frac {(5 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 c x^2}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 x^3}+\frac {\int \frac {\frac {3}{8} \left (5 b^3 c^3+15 a b^2 c^2 d-5 a^2 b c d^2+a^3 d^3\right )+6 b^3 c^2 d x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{6 c^2}\\ &=-\frac {(5 b c-a d) (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 c^2 x}-\frac {(5 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 c x^2}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 x^3}+\left (b^3 d\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx+\frac {\left (5 b^3 c^3+15 a b^2 c^2 d-5 a^2 b c d^2+a^3 d^3\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{16 c^2}\\ &=-\frac {(5 b c-a d) (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 c^2 x}-\frac {(5 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 c x^2}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 x^3}+\left (2 b^2 d\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x}\right )+\frac {\left (5 b^3 c^3+15 a b^2 c^2 d-5 a^2 b c d^2+a^3 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{8 c^2}\\ &=-\frac {(5 b c-a d) (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 c^2 x}-\frac {(5 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 c x^2}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 x^3}-\frac {\left (5 b^3 c^3+15 a b^2 c^2 d-5 a^2 b c d^2+a^3 d^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 \sqrt {a} c^{5/2}}+\left (2 b^2 d\right ) \operatorname {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )\\ &=-\frac {(5 b c-a d) (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{8 c^2 x}-\frac {(5 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 c x^2}-\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 x^3}-\frac {\left (5 b^3 c^3+15 a b^2 c^2 d-5 a^2 b c d^2+a^3 d^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 \sqrt {a} c^{5/2}}+2 b^{5/2} \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )\\ \end {align*}

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Mathematica [A]  time = 2.11, size = 235, normalized size = 1.03 \begin {gather*} -\frac {\sqrt {a+b x} \sqrt {c+d x} \left (a^2 \left (8 c^2+2 c d x-3 d^2 x^2\right )+2 a b c x (13 c+7 d x)+33 b^2 c^2 x^2\right )}{24 c^2 x^3}-\frac {\left (a^3 d^3-5 a^2 b c d^2+15 a b^2 c^2 d+5 b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{8 \sqrt {a} c^{5/2}}+\frac {2 \sqrt {d} (b c-a d)^{5/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/2} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{(c+d x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(5/2)*Sqrt[c + d*x])/x^4,x]

[Out]

-1/24*(Sqrt[a + b*x]*Sqrt[c + d*x]*(33*b^2*c^2*x^2 + 2*a*b*c*x*(13*c + 7*d*x) + a^2*(8*c^2 + 2*c*d*x - 3*d^2*x
^2)))/(c^2*x^3) + (2*Sqrt[d]*(b*c - a*d)^(5/2)*((b*(c + d*x))/(b*c - a*d))^(5/2)*ArcSinh[(Sqrt[d]*Sqrt[a + b*x
])/Sqrt[b*c - a*d]])/(c + d*x)^(5/2) - ((5*b^3*c^3 + 15*a*b^2*c^2*d - 5*a^2*b*c*d^2 + a^3*d^3)*ArcTanh[(Sqrt[c
]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(8*Sqrt[a]*c^(5/2))

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IntegrateAlgebraic [B]  time = 0.51, size = 491, normalized size = 2.15 \begin {gather*} \frac {\left (-a^3 d^3+5 a^2 b c d^2-15 a b^2 c^2 d-5 b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{8 \sqrt {a} c^{5/2}}+\frac {-\frac {3 a^5 d^3 (c+d x)^{5/2}}{(a+b x)^{5/2}}+\frac {8 a^4 c d^3 (c+d x)^{3/2}}{(a+b x)^{3/2}}+\frac {15 a^4 b c d^2 (c+d x)^{5/2}}{(a+b x)^{5/2}}+\frac {3 a^3 b^2 c^2 d (c+d x)^{5/2}}{(a+b x)^{5/2}}+\frac {3 a^3 c^2 d^3 \sqrt {c+d x}}{\sqrt {a+b x}}-\frac {24 a^3 b c^2 d^2 (c+d x)^{3/2}}{(a+b x)^{3/2}}-\frac {15 a^2 b^3 c^3 (c+d x)^{5/2}}{(a+b x)^{5/2}}-\frac {24 a^2 b^2 c^3 d (c+d x)^{3/2}}{(a+b x)^{3/2}}-\frac {15 a^2 b c^3 d^2 \sqrt {c+d x}}{\sqrt {a+b x}}-\frac {33 b^3 c^5 \sqrt {c+d x}}{\sqrt {a+b x}}+\frac {40 a b^3 c^4 (c+d x)^{3/2}}{(a+b x)^{3/2}}+\frac {45 a b^2 c^4 d \sqrt {c+d x}}{\sqrt {a+b x}}}{24 c^2 \left (c-\frac {a (c+d x)}{a+b x}\right )^3}+2 b^{5/2} \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x)^(5/2)*Sqrt[c + d*x])/x^4,x]

[Out]

((-33*b^3*c^5*Sqrt[c + d*x])/Sqrt[a + b*x] + (45*a*b^2*c^4*d*Sqrt[c + d*x])/Sqrt[a + b*x] - (15*a^2*b*c^3*d^2*
Sqrt[c + d*x])/Sqrt[a + b*x] + (3*a^3*c^2*d^3*Sqrt[c + d*x])/Sqrt[a + b*x] + (40*a*b^3*c^4*(c + d*x)^(3/2))/(a
+ b*x)^(3/2) - (24*a^2*b^2*c^3*d*(c + d*x)^(3/2))/(a + b*x)^(3/2) - (24*a^3*b*c^2*d^2*(c + d*x)^(3/2))/(a + b
*x)^(3/2) + (8*a^4*c*d^3*(c + d*x)^(3/2))/(a + b*x)^(3/2) - (15*a^2*b^3*c^3*(c + d*x)^(5/2))/(a + b*x)^(5/2) +
(3*a^3*b^2*c^2*d*(c + d*x)^(5/2))/(a + b*x)^(5/2) + (15*a^4*b*c*d^2*(c + d*x)^(5/2))/(a + b*x)^(5/2) - (3*a^5
*d^3*(c + d*x)^(5/2))/(a + b*x)^(5/2))/(24*c^2*(c - (a*(c + d*x))/(a + b*x))^3) + ((-5*b^3*c^3 - 15*a*b^2*c^2*
d + 5*a^2*b*c*d^2 - a^3*d^3)*ArcTanh[(Sqrt[a]*Sqrt[c + d*x])/(Sqrt[c]*Sqrt[a + b*x])])/(8*Sqrt[a]*c^(5/2)) + 2
*b^(5/2)*Sqrt[d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])]

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fricas [A]  time = 8.53, size = 1241, normalized size = 5.44

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(d*x+c)^(1/2)/x^4,x, algorithm="fricas")

[Out]

[1/96*(48*sqrt(b*d)*a*b^2*c^3*x^3*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d*x + b*c + a*d)*
sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) + 3*(5*b^3*c^3 + 15*a*b^2*c^2*d - 5*a^2*b*c*d
^2 + a^3*d^3)*sqrt(a*c)*x^3*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)*s
qrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(8*a^3*c^3 + (33*a*b^2*c^3 + 14*a^2*b
*c^2*d - 3*a^3*c*d^2)*x^2 + 2*(13*a^2*b*c^3 + a^3*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*c^3*x^3), -1/96*(9
6*sqrt(-b*d)*a*b^2*c^3*x^3*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^
2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) - 3*(5*b^3*c^3 + 15*a*b^2*c^2*d - 5*a^2*b*c*d^2 + a^3*d^3)*sqrt(a*c)*x^3
*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*(8*a^3*c^3 + (33*a*b^2*c^3 + 14*a^2*b*c^2*d - 3*a^3*c*d^2)*x^2 +
2*(13*a^2*b*c^3 + a^3*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*c^3*x^3), 1/48*(24*sqrt(b*d)*a*b^2*c^3*x^3*lo
g(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c
) + 8*(b^2*c*d + a*b*d^2)*x) + 3*(5*b^3*c^3 + 15*a*b^2*c^2*d - 5*a^2*b*c*d^2 + a^3*d^3)*sqrt(-a*c)*x^3*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*(8*a^3*c^3 + (33*a*b^2*c^3 + 14*a^2*b*c^2*d - 3*a^3*c*d^2)*x^2 + 2*(13*a^2*b*c^3 + a^3*c^2*d)*x)*sqr
t(b*x + a)*sqrt(d*x + c))/(a*c^3*x^3), -1/48*(48*sqrt(-b*d)*a*b^2*c^3*x^3*arctan(1/2*(2*b*d*x + b*c + a*d)*sqr
t(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) - 3*(5*b^3*c^3 + 15*a*b^2
*c^2*d - 5*a^2*b*c*d^2 + a^3*d^3)*sqrt(-a*c)*x^3*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*s
qrt(d*x + c)/(a*b*c*d*x^2 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x)) + 2*(8*a^3*c^3 + (33*a*b^2*c^3 + 14*a^2*b*c^2*d
- 3*a^3*c*d^2)*x^2 + 2*(13*a^2*b*c^3 + a^3*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*c^3*x^3)]

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giac [B]  time = 98.96, size = 2265, normalized size = 9.93

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(d*x+c)^(1/2)/x^4,x, algorithm="giac")

[Out]

-1/24*(24*sqrt(b*d)*b^2*abs(b)*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2) + 3*(5*s
qrt(b*d)*b^4*c^3*abs(b) + 15*sqrt(b*d)*a*b^3*c^2*d*abs(b) - 5*sqrt(b*d)*a^2*b^2*c*d^2*abs(b) + sqrt(b*d)*a^3*b
*d^3*abs(b))*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)*b*c^2) + 2*(33*sqrt(b*d)*b^14*c^8*abs(b) - 184*sqrt(b*d)*a*b^13*c^7*d*abs(b
) + 408*sqrt(b*d)*a^2*b^12*c^6*d^2*abs(b) - 432*sqrt(b*d)*a^3*b^11*c^5*d^3*abs(b) + 170*sqrt(b*d)*a^4*b^10*c^4
*d^4*abs(b) + 72*sqrt(b*d)*a^5*b^9*c^3*d^5*abs(b) - 96*sqrt(b*d)*a^6*b^8*c^2*d^6*abs(b) + 32*sqrt(b*d)*a^7*b^7
*c*d^7*abs(b) - 3*sqrt(b*d)*a^8*b^6*d^8*abs(b) - 165*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x +
a)*b*d - a*b*d))^2*b^12*c^7*abs(b) + 477*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^11*c^6*d*abs(b) - 309*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^10*c^5*d^2*abs(b) - 219*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^3*b
^9*c^4*d^3*abs(b) + 201*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^8*c^
3*d^4*abs(b) + 111*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^5*b^7*c^2*d^5
*abs(b) - 111*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^6*b^6*c*d^6*abs(b)
+ 15*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^7*b^5*d^7*abs(b) + 330*sqr
t(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^10*c^6*abs(b) - 228*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^9*c^5*d*abs(b) - 114*sqrt(b*d)*(sqrt(b*d)*sqr
t(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^2*b^8*c^4*d^2*abs(b) - 312*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^7*c^3*d^3*abs(b) + 198*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a)
- sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^4*b^6*c^2*d^4*abs(b) + 156*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sq
rt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^5*b^5*c*d^5*abs(b) - 30*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c
+ (b*x + a)*b*d - a*b*d))^4*a^6*b^4*d^6*abs(b) - 330*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x +
a)*b*d - a*b*d))^6*b^8*c^5*abs(b) - 338*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^7*c^4*d*abs(b) - 252*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^6*c^3*d^2*abs(b) - 396*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^3*b^5
*c^2*d^3*abs(b) - 122*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^4*b^4*c*d^
4*abs(b) + 30*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^5*b^3*d^5*abs(b) +
165*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*b^6*c^4*abs(b) + 372*sqrt(b*d
)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a*b^5*c^3*d*abs(b) + 282*sqrt(b*d)*(sqrt(b
*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a^2*b^4*c^2*d^2*abs(b) + 60*sqrt(b*d)*(sqrt(b*d)*sq
rt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a^3*b^3*c*d^3*abs(b) - 15*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x +
a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a^4*b^2*d^4*abs(b) - 33*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt
(b^2*c + (b*x + a)*b*d - a*b*d))^10*b^4*c^3*abs(b) - 99*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x
+ a)*b*d - a*b*d))^10*a*b^3*c^2*d*abs(b) - 15*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d
- a*b*d))^10*a^2*b^2*c*d^2*abs(b) + 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d
))^10*a^3*b*d^3*abs(b))/((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 + (sqr
t(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4)^3*c^2))/b

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maple [B]  time = 0.02, size = 601, normalized size = 2.64 \begin {gather*} \frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (-3 \sqrt {b d}\, a^{3} d^{3} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )+15 \sqrt {b d}\, a^{2} b c \,d^{2} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )-45 \sqrt {b d}\, a \,b^{2} c^{2} d \,x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )-15 \sqrt {b d}\, b^{3} c^{3} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )+48 \sqrt {a c}\, b^{3} c^{2} d \,x^{3} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+6 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a^{2} d^{2} x^{2}-28 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a b c d \,x^{2}-66 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b^{2} c^{2} x^{2}-4 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a^{2} c d x -52 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a b \,c^{2} x -16 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a^{2} c^{2}\right )}{48 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, c^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)*(d*x+c)^(1/2)/x^4,x)

[Out]

1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)/c^2*(48*ln(1/2*(2*b*d*x+a*d+b*c+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)
)/(b*d)^(1/2))*x^3*b^3*c^2*d*(a*c)^(1/2)-3*(b*d)^(1/2)*a^3*d^3*x^3*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*(b*d*x^
2+a*d*x+b*c*x+a*c)^(1/2))/x)+15*(b*d)^(1/2)*a^2*b*c*d^2*x^3*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*(b*d*x^2+a*d*x
+b*c*x+a*c)^(1/2))/x)-45*(b*d)^(1/2)*a*b^2*c^2*d*x^3*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+
a*c)^(1/2))/x)-15*(b*d)^(1/2)*b^3*c^3*x^3*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2))
/x)+6*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a^2*d^2*x^2-28*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(
b*d)^(1/2)*(a*c)^(1/2)*a*b*c*d*x^2-66*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*b^2*c^2*x^2-4*(b
*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a^2*c*d*x-52*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)
*(a*c)^(1/2)*a*b*c^2*x-16*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a^2*c^2)/(b*d*x^2+a*d*x+b*c*
x+a*c)^(1/2)/x^3/(b*d)^(1/2)/(a*c)^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(d*x+c)^(1/2)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^{5/2}\,\sqrt {c+d\,x}}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^(5/2)*(c + d*x)^(1/2))/x^4,x)

[Out]

int(((a + b*x)^(5/2)*(c + d*x)^(1/2))/x^4, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(5/2)*(d*x+c)**(1/2)/x**4,x)

[Out]

Timed out

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