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

Optimal. Leaf size=171 \[ \frac {\sqrt {d} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2}}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (7 b c-3 a d)}{4 b^2}-\frac {2 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b} \]

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Rubi [A]  time = 0.14, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {102, 154, 157, 63, 217, 206, 93, 208} \begin {gather*} \frac {\sqrt {d} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2}}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (7 b c-3 a d)}{4 b^2}-\frac {2 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(7*b*c - 3*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*b^2) + (d*Sqrt[a + b*x]*(c + d*x)^(3/2))/(2*b) - (2*c^(5/2)
*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/Sqrt[a] + (Sqrt[d]*(15*b^2*c^2 - 10*a*b*c*d + 3*a^2
*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*b^(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 102

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))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegersQ[2*m, 2*n, 2*p]

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 {(c+d x)^{5/2}}{x \sqrt {a+b x}} \, dx &=\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}+\frac {\int \frac {\sqrt {c+d x} \left (2 b c^2+\frac {1}{2} d (7 b c-3 a d) x\right )}{x \sqrt {a+b x}} \, dx}{2 b}\\ &=\frac {d (7 b c-3 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^2}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}+\frac {\int \frac {2 b^2 c^3+\frac {1}{4} d \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right ) x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{2 b^2}\\ &=\frac {d (7 b c-3 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^2}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}+c^3 \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx+\frac {\left (d \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{8 b^2}\\ &=\frac {d (7 b c-3 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^2}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}+\left (2 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 (d \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right )\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 )}{4 b^3}\\ &=\frac {d (7 b c-3 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^2}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}-\frac {2 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}+\frac {\left (d \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right )\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 )}{4 b^3}\\ &=\frac {d (7 b c-3 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^2}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}-\frac {2 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}+\frac {\sqrt {d} \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 1.13, size = 186, normalized size = 1.09 \begin {gather*} \frac {1}{4} \left (\frac {\sqrt {d} \sqrt {b c-a d} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\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^3 \sqrt {c+d x}}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (-3 a d+9 b c+2 b d x)}{b^2}-\frac {8 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d*Sqrt[a + b*x]*Sqrt[c + d*x]*(9*b*c - 3*a*d + 2*b*d*x))/b^2 + (Sqrt[d]*Sqrt[b*c - a*d]*(15*b^2*c^2 - 10*a*b
*c*d + 3*a^2*d^2)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(b^3*Sqrt[
c + d*x]) - (8*c^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/Sqrt[a])/4

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IntegrateAlgebraic [A]  time = 0.47, size = 248, normalized size = 1.45 \begin {gather*} \frac {\left (3 a^2 d^{5/2}-10 a b c d^{3/2}+15 b^2 c^2 \sqrt {d}\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2}}+\frac {d \sqrt {a+b x} \left (-\frac {3 a^2 d^3 (a+b x)}{c+d x}+5 a^2 b d^2-\frac {7 b^2 c^2 d (a+b x)}{c+d x}-14 a b^2 c d+\frac {10 a b c d^2 (a+b x)}{c+d x}+9 b^3 c^2\right )}{4 b^2 \sqrt {c+d x} \left (b-\frac {d (a+b x)}{c+d x}\right )^2}-\frac {2 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*Sqrt[a + b*x]*(9*b^3*c^2 - 14*a*b^2*c*d + 5*a^2*b*d^2 - (7*b^2*c^2*d*(a + b*x))/(c + d*x) + (10*a*b*c*d^2*(
a + b*x))/(c + d*x) - (3*a^2*d^3*(a + b*x))/(c + d*x)))/(4*b^2*Sqrt[c + d*x]*(b - (d*(a + b*x))/(c + d*x))^2)
- (2*c^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/Sqrt[a] + ((15*b^2*c^2*Sqrt[d] - 10*a*b
*c*d^(3/2) + 3*a^2*d^(5/2))*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*b^(5/2))

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fricas [A]  time = 8.49, size = 987, normalized size = 5.77 \begin {gather*} \left [\frac {8 \, b^{2} c^{2} \sqrt {\frac {c}{a}} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a^{2} c + {\left (a b c + a^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {c}{a}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + {\left (15 \, b^{2} c^{2} - 10 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {\frac {d}{b}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b^{2} d x + b^{2} c + a b d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {d}{b}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (2 \, b d^{2} x + 9 \, b c d - 3 \, a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, b^{2}}, \frac {4 \, b^{2} c^{2} \sqrt {\frac {c}{a}} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a^{2} c + {\left (a b c + a^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {c}{a}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - {\left (15 \, b^{2} c^{2} - 10 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {-\frac {d}{b}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {d}{b}}}{2 \, {\left (b d^{2} x^{2} + a c d + {\left (b c d + a d^{2}\right )} x\right )}}\right ) + 2 \, {\left (2 \, b d^{2} x + 9 \, b c d - 3 \, a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, b^{2}}, \frac {16 \, b^{2} c^{2} \sqrt {-\frac {c}{a}} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {c}{a}}}{2 \, {\left (b c d x^{2} + a c^{2} + {\left (b c^{2} + a c d\right )} x\right )}}\right ) + {\left (15 \, b^{2} c^{2} - 10 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {\frac {d}{b}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b^{2} d x + b^{2} c + a b d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {d}{b}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (2 \, b d^{2} x + 9 \, b c d - 3 \, a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, b^{2}}, \frac {8 \, b^{2} c^{2} \sqrt {-\frac {c}{a}} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {c}{a}}}{2 \, {\left (b c d x^{2} + a c^{2} + {\left (b c^{2} + a c d\right )} x\right )}}\right ) - {\left (15 \, b^{2} c^{2} - 10 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {-\frac {d}{b}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {d}{b}}}{2 \, {\left (b d^{2} x^{2} + a c d + {\left (b c d + a d^{2}\right )} x\right )}}\right ) + 2 \, {\left (2 \, b d^{2} x + 9 \, b c d - 3 \, a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(8*b^2*c^2*sqrt(c/a)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a^2*c + (a*b*c + a^2*d)
*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + (15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d
^2)*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b^2*d*x + b^2*c + a*b*d)*sqrt(b*x + a)*
sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(2*b*d^2*x + 9*b*c*d - 3*a*d^2)*sqrt(b*x + a)*sqrt(d*x
+ c))/b^2, 1/8*(4*b^2*c^2*sqrt(c/a)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a^2*c + (a*b*c
+ a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - (15*b^2*c^2 - 10*a*b*c*d
+ 3*a^2*d^2)*sqrt(-d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-d/b)/(b*d^2*x^2 + a
*c*d + (b*c*d + a*d^2)*x)) + 2*(2*b*d^2*x + 9*b*c*d - 3*a*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/b^2, 1/16*(16*b^2*
c^2*sqrt(-c/a)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-c/a)/(b*c*d*x^2 + a*c^2 +
(b*c^2 + a*c*d)*x)) + (15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d
+ a^2*d^2 + 4*(2*b^2*d*x + b^2*c + a*b*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2)*x) + 4
*(2*b*d^2*x + 9*b*c*d - 3*a*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/b^2, 1/8*(8*b^2*c^2*sqrt(-c/a)*arctan(1/2*(2*a*c
+ (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-c/a)/(b*c*d*x^2 + a*c^2 + (b*c^2 + a*c*d)*x)) - (15*b^2*c^
2 - 10*a*b*c*d + 3*a^2*d^2)*sqrt(-d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-d/b)
/(b*d^2*x^2 + a*c*d + (b*c*d + a*d^2)*x)) + 2*(2*b*d^2*x + 9*b*c*d - 3*a*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/b^2
]

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

[Out]

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

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maple [B]  time = 0.02, size = 342, normalized size = 2.00 \begin {gather*} \frac {\sqrt {d x +c}\, \sqrt {b x +a}\, \left (3 \sqrt {a c}\, a^{2} d^{3} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-10 \sqrt {a c}\, a b c \,d^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-8 \sqrt {b d}\, b^{2} c^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+15 \sqrt {a c}\, b^{2} c^{2} d \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+4 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b \,d^{2} x -6 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,d^{2}+18 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b c d \right )}{8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \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)/x/(b*x+a)^(1/2),x)

[Out]

1/8*(d*x+c)^(1/2)*(b*x+a)^(1/2)*(3*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))
*a^2*d^3*(a*c)^(1/2)-10*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*a*b*c*d^2*
(a*c)^(1/2)+15*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*b^2*c^2*d*(a*c)^(1/
2)-8*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*b^2*c^3*(b*d)^(1/2)+4*x*b*d^2*(a*c)^(1/2)
*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)-6*a*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+18*b*c*d*(a*c)^(1
/2)*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/((b*x+a)*(d*x+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)/x/(b*x+a)^(1/2),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 = -0.01 \begin {gather*} \int \frac {{\left (c+d\,x\right )}^{5/2}}{x\,\sqrt {a+b\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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