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

Optimal. Leaf size=160 \[ \frac {c^{3/2} (b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{3/2}}+\frac {d^{3/2} (5 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2}}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{a x}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (a d+b c)}{a b} \]

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Rubi [A]  time = 0.16, antiderivative size = 160, 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 = {98, 154, 157, 63, 217, 206, 93, 208} \begin {gather*} \frac {c^{3/2} (b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{3/2}}+\frac {d^{3/2} (5 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2}}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{a x}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (a d+b c)}{a b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(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 - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[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 {(c+d x)^{5/2}}{x^2 \sqrt {a+b x}} \, dx &=-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{a x}-\frac {\int \frac {\sqrt {c+d x} \left (\frac {1}{2} c (b c-5 a d)-d (b c+a d) x\right )}{x \sqrt {a+b x}} \, dx}{a}\\ &=\frac {d (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{a b}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{a x}-\frac {\int \frac {\frac {1}{2} b c^2 (b c-5 a d)-\frac {1}{2} a d^2 (5 b c-a d) x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{a b}\\ &=\frac {d (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{a b}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{a x}-\frac {\left (c^2 (b c-5 a d)\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{2 a}+\frac {\left (d^2 (5 b c-a d)\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{2 b}\\ &=\frac {d (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{a b}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{a x}-\frac {\left (c^2 (b c-5 a d)\right ) \operatorname {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{a}+\frac {\left (d^2 (5 b c-a 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 )}{b^2}\\ &=\frac {d (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{a b}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{a x}+\frac {c^{3/2} (b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{3/2}}+\frac {\left (d^2 (5 b c-a 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 )}{b^2}\\ &=\frac {d (b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{a b}-\frac {c \sqrt {a+b x} (c+d x)^{3/2}}{a x}+\frac {c^{3/2} (b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{3/2}}+\frac {d^{3/2} (5 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 1.01, size = 195, normalized size = 1.22 \begin {gather*} \frac {a^{3/2} d^{3/2} x \sqrt {b c-a d} (5 b c-a d) \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 \left (b c^{3/2} x \sqrt {c+d x} (b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\sqrt {a} \sqrt {a+b x} (c+d x) \left (a d^2 x-b c^2\right )\right )}{a^{3/2} b^2 x \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.42, size = 233, normalized size = 1.46 \begin {gather*} \frac {\left (b c^{5/2}-5 a c^{3/2} d\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{3/2}}-\frac {\sqrt {a+b x} (a d-b c) \left (a^2 d^2-\frac {b c^2 d (a+b x)}{c+d x}-\frac {a c d^2 (a+b x)}{c+d x}+b^2 c^2\right )}{a b \sqrt {c+d x} \left (a-\frac {c (a+b x)}{c+d x}\right ) \left (b-\frac {d (a+b x)}{c+d x}\right )}+\frac {\left (5 b c d^{3/2}-a d^{5/2}\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 6.90, size = 991, normalized size = 6.19 \begin {gather*} \left [-\frac {{\left (5 \, a b c d - a^{2} d^{2}\right )} x \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 ) + {\left (b^{2} c^{2} - 5 \, a b c d\right )} x \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 ) - 4 \, {\left (a d^{2} x - b c^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{4 \, a b x}, -\frac {2 \, {\left (5 \, a b c d - a^{2} d^{2}\right )} x \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 ) + {\left (b^{2} c^{2} - 5 \, a b c d\right )} x \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 ) - 4 \, {\left (a d^{2} x - b c^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{4 \, a b x}, -\frac {2 \, {\left (b^{2} c^{2} - 5 \, a b c d\right )} x \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 (5 \, a b c d - a^{2} d^{2}\right )} x \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 (a d^{2} x - b c^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{4 \, a b x}, -\frac {{\left (b^{2} c^{2} - 5 \, a b c d\right )} x \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 (5 \, a b c d - a^{2} d^{2}\right )} x \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 (a d^{2} x - b c^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{2 \, a b x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((5*a*b*c*d - a^2*d^2)*x*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) + (b^2*c^2 - 5*a*b*c*d)*x*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) - 4*(a*d^2*x - b*c^2)*sqrt(b*x + a)*sqrt(d*x + c))/(a*b*x),
-1/4*(2*(5*a*b*c*d - a^2*d^2)*x*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)) + (b^2*c^2 - 5*a*b*c*d)*x*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) - 4*(a*d^2*x - b*c^2)*sqrt(b*x + a)*sqrt(d*x + c))/(a*b*x), -1/4*(2*(b^2*c^2 - 5*a*b*c*d)*
x*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)) + (5*a*b*c*d - a^2*d^2)*x*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*(a*d^2*x - b*c^
2)*sqrt(b*x + a)*sqrt(d*x + c))/(a*b*x), -1/2*((b^2*c^2 - 5*a*b*c*d)*x*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)) + (5*a*b*c*d - a^2*d^2)
*x*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*(a*d^2*x - b*c^2)*sqrt(b*x + a)*sqrt(d*x + c))/(a*b*x)]

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giac [B]  time = 2.87, size = 547, normalized size = 3.42 \begin {gather*} \frac {\frac {2 \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} d^{2} {\left | b \right |}}{b^{2}} - \frac {{\left (5 \, \sqrt {b d} b c d {\left | b \right |} - \sqrt {b d} a d^{2} {\left | b \right |}\right )} \log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{b^{2}} + \frac {2 \, {\left (\sqrt {b d} b^{2} c^{3} {\left | b \right |} - 5 \, \sqrt {b d} a b c^{2} d {\left | b \right |}\right )} \arctan \left (-\frac {b^{2} c + a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} a b} - \frac {4 \, {\left (\sqrt {b d} b^{4} c^{4} {\left | b \right |} - 2 \, \sqrt {b d} a b^{3} c^{3} d {\left | b \right |} + \sqrt {b d} a^{2} b^{2} c^{2} d^{2} {\left | b \right |} - \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{2} c^{3} {\left | b \right |} - \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b c^{2} d {\left | b \right |}\right )}}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b d + {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4}\right )} a}}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*d^2*abs(b)/b^2 - (5*sqrt(b*d)*b*c*d*abs(b) - sqrt(b*d
)*a*d^2*abs(b))*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/b^2 + 2*(sqrt(b*d)*b^2*
c^3*abs(b) - 5*sqrt(b*d)*a*b*c^2*d*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)*a*b) - 4*(sqrt(b*d)*b^4*c^4*abs(b) - 2*sqrt(b
*d)*a*b^3*c^3*d*abs(b) + sqrt(b*d)*a^2*b^2*c^2*d^2*abs(b) - sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c +
(b*x + a)*b*d - a*b*d))^2*b^2*c^3*abs(b) - 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*c^2*d*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 + (
sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4)*a))/b

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maple [B]  time = 0.02, size = 320, normalized size = 2.00 \begin {gather*} -\frac {\sqrt {d x +c}\, \sqrt {b x +a}\, \left (\sqrt {a c}\, a^{2} d^{3} x \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 )+5 \sqrt {b d}\, a b \,c^{2} d x \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 )-5 \sqrt {a c}\, a b c \,d^{2} x \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 )-\sqrt {b d}\, b^{2} c^{3} x \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 )-2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,d^{2} x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b \,c^{2}\right )}{2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, a b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(d*x+c)^(1/2)*(b*x+a)^(1/2)/a*(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)
)*x*a^2*d^3*(a*c)^(1/2)-5*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))*x*a*b*c*
d^2*(a*c)^(1/2)+5*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*a*b*c^2*d*(b*d)^(1/2)-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*b^2*c^3*(b*d)^(1/2)-2*x*a*d^2*(a*c)^(1/2)*((b*x+
a)*(d*x+c))^(1/2)*(b*d)^(1/2)+2*b*c^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/((b*x+a)*(d*x+c))^(1/2)
/x/(b*d)^(1/2)/(a*c)^(1/2)/b

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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^2/(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^2\,\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^2*(a + b*x)^(1/2)),x)

[Out]

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

[Out]

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

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