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

Optimal. Leaf size=218 \[ -2 a^{5/2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {\left (5 a^3 d^3+15 a^2 b c d^2-5 a b^2 c^2 d+b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 \sqrt {b} d^{5/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-5 a d) (a d+b c)}{8 d^2}+\frac {1}{3} (a+b x)^{5/2} \sqrt {c+d x}+\frac {(a+b x)^{3/2} \sqrt {c+d x} (5 a d+b c)}{12 d} \]

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Rubi [A]  time = 0.25, antiderivative size = 218, 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 (15 a^2 b c d^2+5 a^3 d^3-5 a b^2 c^2 d+b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 \sqrt {b} d^{5/2}}-2 a^{5/2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-5 a d) (a d+b c)}{8 d^2}+\frac {1}{3} (a+b x)^{5/2} \sqrt {c+d x}+\frac {(a+b x)^{3/2} \sqrt {c+d x} (5 a d+b c)}{12 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.82, size = 242, normalized size = 1.11 \begin {gather*} \frac {\sqrt {d} \left (\sqrt {a+b x} (c+d x) \left (33 a^2 d^2+2 a b d (7 c+13 d x)+b^2 \left (-3 c^2+2 c d x+8 d^2 x^2\right )\right )-48 a^{5/2} \sqrt {c} d^2 \sqrt {c+d x} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )\right )+\frac {3 \sqrt {b c-a d} \left (5 a^3 d^3+15 a^2 b c d^2-5 a b^2 c^2 d+b^3 c^3\right ) \sqrt {\frac {b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{b}}{24 d^{5/2} \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((3*Sqrt[b*c - a*d]*(b^3*c^3 - 5*a*b^2*c^2*d + 15*a^2*b*c*d^2 + 5*a^3*d^3)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*Arc
Sinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/b + Sqrt[d]*(Sqrt[a + b*x]*(c + d*x)*(33*a^2*d^2 + 2*a*b*d*(7*c
+ 13*d*x) + b^2*(-3*c^2 + 2*c*d*x + 8*d^2*x^2)) - 48*a^(5/2)*Sqrt[c]*d^2*Sqrt[c + d*x]*ArcTanh[(Sqrt[c]*Sqrt[
a + b*x])/(Sqrt[a]*Sqrt[c + d*x])]))/(24*d^(5/2)*Sqrt[c + d*x])

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IntegrateAlgebraic [A]  time = 0.57, size = 396, normalized size = 1.82 \begin {gather*} -2 a^{5/2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )+\frac {\left (5 a^3 d^3+15 a^2 b c d^2-5 a b^2 c^2 d+b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{8 \sqrt {b} d^{5/2}}+\frac {\sqrt {c+d x} \left (\frac {15 a^3 b^2 d^3 (c+d x)^2}{(a+b x)^2}-\frac {40 a^3 b d^4 (c+d x)}{a+b x}+33 a^3 d^5-\frac {3 a^2 b^3 c d^2 (c+d x)^2}{(a+b x)^2}+\frac {24 a^2 b^2 c d^3 (c+d x)}{a+b x}-45 a^2 b c d^4+\frac {3 b^5 c^3 (c+d x)^2}{(a+b x)^2}-\frac {8 b^4 c^3 d (c+d x)}{a+b x}-\frac {15 a b^4 c^2 d (c+d x)^2}{(a+b x)^2}+\frac {24 a b^3 c^2 d^2 (c+d x)}{a+b x}+15 a b^2 c^2 d^3-3 b^3 c^3 d^2\right )}{24 d^2 \sqrt {a+b x} \left (d-\frac {b (c+d x)}{a+b x}\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 19.89, size = 1197, normalized size = 5.49

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,x, algorithm="fricas")

[Out]

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

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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((b*x+a)^(5/2)*(d*x+c)^(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: 0.46

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maple [B]  time = 0.02, size = 583, normalized size = 2.67 \begin {gather*} \frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (-48 \sqrt {b d}\, a^{3} c \,d^{2} \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 {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 )+45 \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 )-15 \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 )+3 \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}+52 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a b \,d^{2} x +4 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, b^{2} c d x +66 \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 -6 \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}\, d^{2}} \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,x)

[Out]

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

[Out]

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

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

[Out]

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

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