3.7.77 \(\int \frac {(a+b x)^{5/2}}{x \sqrt {c+d x}} \, dx\) [677]

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

[Out]

1/4*(15*a^2*d^2-10*a*b*c*d+3*b^2*c^2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))*b^(1/2)/d^(5/2)-2*a
^(5/2)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/c^(1/2)+1/2*b*(b*x+a)^(3/2)*(d*x+c)^(1/2)/d-1/4*b*
(-7*a*d+3*b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/d^2

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Rubi [A]
time = 0.10, 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 = {104, 159, 163, 65, 223, 212, 95, 214} \begin {gather*} -\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {c}}+\frac {\sqrt {b} \left (15 a^2 d^2-10 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 d^{5/2}}-\frac {b \sqrt {a+b x} \sqrt {c+d x} (3 b c-7 a d)}{4 d^2}+\frac {b (a+b x)^{3/2} \sqrt {c+d x}}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(b*(3*b*c - 7*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/d^2 + (b*(a + b*x)^(3/2)*Sqrt[c + d*x])/(2*d) - (2*a^(5/2
)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/Sqrt[c] + (Sqrt[b]*(3*b^2*c^2 - 10*a*b*c*d + 15*a^
2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*d^(5/2))

Rule 65

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 95

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 104

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 159

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 163

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 212

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

Rule 214

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

Rule 223

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

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Mathematica [A]
time = 0.57, size = 148, normalized size = 0.87 \begin {gather*} \frac {1}{4} \left (\frac {b \sqrt {a+b x} \sqrt {c+d x} (-3 b c+9 a d+2 b d x)}{d^2}-\frac {8 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{\sqrt {c}}+\frac {\sqrt {b} \left (3 b^2 c^2-10 a b c d+15 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{d^{5/2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(341\) vs. \(2(133)=266\).
time = 0.08, size = 342, normalized size = 2.00

method	result	size
default	\(-\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (8 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) \sqrt {b d}\, a^{3} d^{2}-4 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {a c}\, b^{2} d x -15 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, a^{2} b \,d^{2}+10 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, a \,b^{2} c d -3 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, b^{3} c^{2}-18 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {a c}\, a b d +6 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {a c}\, b^{2} c \right )}{8 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, d^{2} \sqrt {b d}\, \sqrt {a c}}\)	\(342\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(8*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)+2*a*c)/x)*(b*d)^(1/2
)*a^3*d^2-4*(b*d)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*(a*c)^(1/2)*b^2*d*x-15*ln(1/2*(2*b*d*x+2*((d*x+c)*(b*x+a))^(1/
2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*(a*c)^(1/2)*a^2*b*d^2+10*ln(1/2*(2*b*d*x+2*((d*x+c)*(b*x+a))^(1/2)*(b*d)^
(1/2)+a*d+b*c)/(b*d)^(1/2))*(a*c)^(1/2)*a*b^2*c*d-3*ln(1/2*(2*b*d*x+2*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)+a*d+
b*c)/(b*d)^(1/2))*(a*c)^(1/2)*b^3*c^2-18*(b*d)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*(a*c)^(1/2)*a*b*d+6*(b*d)^(1/2)*(
(d*x+c)*(b*x+a))^(1/2)*(a*c)^(1/2)*b^2*c)/((d*x+c)*(b*x+a))^(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)/x/(d*x+c)^(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 detail

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Fricas [A]
time = 4.19, size = 987, normalized size = 5.77 \begin {gather*} \left [\frac {8 \, a^{2} d^{2} \sqrt {\frac {a}{c}} \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 c^{2} + {\left (b c^{2} + a c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {a}{c}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + {\left (3 \, b^{2} c^{2} - 10 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {\frac {b}{d}} \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 d^{2} x + b c d + a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {b}{d}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (2 \, b^{2} d x - 3 \, b^{2} c + 9 \, a b d\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, d^{2}}, \frac {4 \, a^{2} d^{2} \sqrt {\frac {a}{c}} \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 c^{2} + {\left (b c^{2} + a c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {a}{c}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - {\left (3 \, b^{2} c^{2} - 10 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} d x^{2} + a b c + {\left (b^{2} c + a b d\right )} x\right )}}\right ) + 2 \, {\left (2 \, b^{2} d x - 3 \, b^{2} c + 9 \, a b d\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, d^{2}}, \frac {16 \, a^{2} d^{2} \sqrt {-\frac {a}{c}} \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 {a}{c}}}{2 \, {\left (a b d x^{2} + a^{2} c + {\left (a b c + a^{2} d\right )} x\right )}}\right ) + {\left (3 \, b^{2} c^{2} - 10 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {\frac {b}{d}} \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 d^{2} x + b c d + a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {b}{d}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (2 \, b^{2} d x - 3 \, b^{2} c + 9 \, a b d\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, d^{2}}, \frac {8 \, a^{2} d^{2} \sqrt {-\frac {a}{c}} \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 {a}{c}}}{2 \, {\left (a b d x^{2} + a^{2} c + {\left (a b c + a^{2} d\right )} x\right )}}\right ) - {\left (3 \, b^{2} c^{2} - 10 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} d x^{2} + a b c + {\left (b^{2} c + a b d\right )} x\right )}}\right ) + 2 \, {\left (2 \, b^{2} d x - 3 \, b^{2} c + 9 \, a b d\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, d^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}}}{x \sqrt {c + d x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m i_lex_is_greater Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^{5/2}}{x\,\sqrt {c+d\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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