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

Optimal. Leaf size=219 \[ \frac {\left (a^3 d^3-5 a^2 b c d^2+15 a b^2 c^2 d+5 b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{5/2} \sqrt {d}}+\frac {\sqrt {a+b x} \sqrt {c+d x} (5 b c-a d) (a d+b c)}{8 b^2}-2 \sqrt {a} c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {1}{3} \sqrt {a+b x} (c+d x)^{5/2}+\frac {\sqrt {a+b x} (c+d x)^{3/2} (a d+5 b c)}{12 b} \]

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Rubi [A]  time = 0.22, antiderivative size = 219, 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 = {101, 154, 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 {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{5/2} \sqrt {d}}+\frac {\sqrt {a+b x} \sqrt {c+d x} (5 b c-a d) (a d+b c)}{8 b^2}-2 \sqrt {a} c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {1}{3} \sqrt {a+b x} (c+d x)^{5/2}+\frac {\sqrt {a+b x} (c+d x)^{3/2} (a d+5 b c)}{12 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 101

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

Rule 154

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

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

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Mathematica [A]  time = 1.01, size = 274, normalized size = 1.25 \begin {gather*} \frac {\sqrt {c+d x} \left (-\frac {b \sqrt {d} \left (\sqrt {a+b x} (c+d x) \left (-3 a^2 d^2+2 a b d (7 c+d x)+b^2 \left (33 c^2+26 c d x+8 d^2 x^2\right )\right )-48 \sqrt {a} b^2 c^{5/2} \sqrt {c+d x} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )\right )}{\sqrt {\frac {b (c+d x)}{b c-a d}}}-3 \sqrt {b c-a d} \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 ) \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )\right )}{24 b^2 \sqrt {d} (a d-b c) \sqrt {\frac {b (c+d x)}{b c-a d}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x]*(-3*Sqrt[b*c - a*d]*(5*b^3*c^3 + 15*a*b^2*c^2*d - 5*a^2*b*c*d^2 + a^3*d^3)*ArcSinh[(Sqrt[d]*Sqr
t[a + b*x])/Sqrt[b*c - a*d]] - (b*Sqrt[d]*(Sqrt[a + b*x]*(c + d*x)*(-3*a^2*d^2 + 2*a*b*d*(7*c + d*x) + b^2*(33
*c^2 + 26*c*d*x + 8*d^2*x^2)) - 48*Sqrt[a]*b^2*c^(5/2)*Sqrt[c + d*x]*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*
Sqrt[c + d*x])]))/Sqrt[(b*(c + d*x))/(b*c - a*d)]))/(24*b^2*Sqrt[d]*(-(b*c) + a*d)*Sqrt[(b*(c + d*x))/(b*c - a
*d)])

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IntegrateAlgebraic [A]  time = 0.63, size = 396, normalized size = 1.81 \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 {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{5/2} \sqrt {d}}+\frac {\sqrt {a+b x} \left (-3 a^3 b^2 d^3+\frac {3 a^3 d^5 (a+b x)^2}{(c+d x)^2}-\frac {8 a^3 b d^4 (a+b x)}{c+d x}+15 a^2 b^3 c d^2+\frac {24 a^2 b^2 c d^3 (a+b x)}{c+d x}-\frac {15 a^2 b c d^4 (a+b x)^2}{(c+d x)^2}-\frac {40 b^4 c^3 d (a+b x)}{c+d x}-45 a b^4 c^2 d+\frac {15 b^3 c^3 d^2 (a+b x)^2}{(c+d x)^2}+\frac {24 a b^3 c^2 d^2 (a+b x)}{c+d x}-\frac {3 a b^2 c^2 d^3 (a+b x)^2}{(c+d x)^2}+33 b^5 c^3\right )}{24 b^2 \sqrt {c+d x} \left (b-\frac {d (a+b x)}{c+d x}\right )^3}-2 \sqrt {a} c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 19.51, size = 1197, normalized size = 5.47

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Eval
uation time: 0.73

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maple [B]  time = 0.02, size = 583, normalized size = 2.66 \begin {gather*} \frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (3 \sqrt {a c}\, a^{3} d^{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 )-15 \sqrt {a c}\, a^{2} b c \,d^{2} \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 )-48 \sqrt {b d}\, a \,b^{2} c^{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 {a c}\, a \,b^{2} c^{2} d \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 )+15 \sqrt {a c}\, b^{3} c^{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 )+16 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b^{2} d^{2} x^{2}+4 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a b \,d^{2} x +52 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, b^{2} c d x -6 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a^{2} d^{2}+28 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a b c d +66 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, b^{2} c^{2}\right )}{48 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(16*x^2*b^2*d^2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)+3*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))*(a*c)^(1/2)*a^3*d^3-15*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))*(a*c)^(1/2)*a^2*b*c*d^2+45*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))*(a*c)^(1/2)*a*b^2*c^2*d+15*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))*(a*c)^(1/2)*b^3*c^3-48*(b*d)^(
1/2)*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)*a*b^2*c^3+4*(b*d)^(1/2)*(a*c)^(1/
2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*a*b*d^2+52*(b*d)^(1/2)*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*b^2*
c*d-6*(b*d)^(1/2)*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^2*d^2+28*(b*d)^(1/2)*(a*c)^(1/2)*(b*d*x^2+a*d*
x+b*c*x+a*c)^(1/2)*a*b*c*d+66*(b*d)^(1/2)*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*b^2*c^2)/(b*d*x^2+a*d*x+
b*c*x+a*c)^(1/2)/b^2/(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((d*x+c)^(5/2)*(b*x+a)^(1/2)/x,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 {\sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{5/2}}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + b x} \left (c + d x\right )^{\frac {5}{2}}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*x)*(c + d*x)**(5/2)/x, x)

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