3.8.44 \(\int \frac {x (c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx\)

Optimal. Leaf size=214 \[ \frac {5 (b c-7 a d) (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{9/2} \sqrt {d}}+\frac {5 \sqrt {a+b x} \sqrt {c+d x} (b c-7 a d) (b c-a d)}{8 b^4}+\frac {5 \sqrt {a+b x} (c+d x)^{3/2} (b c-7 a d)}{12 b^3}+\frac {\sqrt {a+b x} (c+d x)^{5/2} (b c-7 a d)}{3 b^2 (b c-a d)}+\frac {2 a (c+d x)^{7/2}}{b \sqrt {a+b x} (b c-a d)} \]

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Rubi [A]  time = 0.12, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {78, 50, 63, 217, 206} \begin {gather*} \frac {\sqrt {a+b x} (c+d x)^{5/2} (b c-7 a d)}{3 b^2 (b c-a d)}+\frac {5 \sqrt {a+b x} (c+d x)^{3/2} (b c-7 a d)}{12 b^3}+\frac {5 \sqrt {a+b x} \sqrt {c+d x} (b c-7 a d) (b c-a d)}{8 b^4}+\frac {5 (b c-7 a d) (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{9/2} \sqrt {d}}+\frac {2 a (c+d x)^{7/2}}{b \sqrt {a+b x} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(c + d*x)^(5/2))/(a + b*x)^(3/2),x]

[Out]

(5*(b*c - 7*a*d)*(b*c - a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*b^4) + (5*(b*c - 7*a*d)*Sqrt[a + b*x]*(c + d*x)^(
3/2))/(12*b^3) + ((b*c - 7*a*d)*Sqrt[a + b*x]*(c + d*x)^(5/2))/(3*b^2*(b*c - a*d)) + (2*a*(c + d*x)^(7/2))/(b*
(b*c - a*d)*Sqrt[a + b*x]) + (5*(b*c - 7*a*d)*(b*c - a*d)^2*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c +
d*x])])/(8*b^(9/2)*Sqrt[d])

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 78

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

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 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 {x (c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx &=\frac {2 a (c+d x)^{7/2}}{b (b c-a d) \sqrt {a+b x}}+\frac {(b c-7 a d) \int \frac {(c+d x)^{5/2}}{\sqrt {a+b x}} \, dx}{b (b c-a d)}\\ &=\frac {(b c-7 a d) \sqrt {a+b x} (c+d x)^{5/2}}{3 b^2 (b c-a d)}+\frac {2 a (c+d x)^{7/2}}{b (b c-a d) \sqrt {a+b x}}+\frac {(5 (b c-7 a d)) \int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}} \, dx}{6 b^2}\\ &=\frac {5 (b c-7 a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^3}+\frac {(b c-7 a d) \sqrt {a+b x} (c+d x)^{5/2}}{3 b^2 (b c-a d)}+\frac {2 a (c+d x)^{7/2}}{b (b c-a d) \sqrt {a+b x}}+\frac {(5 (b c-7 a d) (b c-a d)) \int \frac {\sqrt {c+d x}}{\sqrt {a+b x}} \, dx}{8 b^3}\\ &=\frac {5 (b c-7 a d) (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{8 b^4}+\frac {5 (b c-7 a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^3}+\frac {(b c-7 a d) \sqrt {a+b x} (c+d x)^{5/2}}{3 b^2 (b c-a d)}+\frac {2 a (c+d x)^{7/2}}{b (b c-a d) \sqrt {a+b x}}+\frac {\left (5 (b c-7 a d) (b c-a d)^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{16 b^4}\\ &=\frac {5 (b c-7 a d) (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{8 b^4}+\frac {5 (b c-7 a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^3}+\frac {(b c-7 a d) \sqrt {a+b x} (c+d x)^{5/2}}{3 b^2 (b c-a d)}+\frac {2 a (c+d x)^{7/2}}{b (b c-a d) \sqrt {a+b x}}+\frac {\left (5 (b c-7 a d) (b c-a d)^2\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^5}\\ &=\frac {5 (b c-7 a d) (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{8 b^4}+\frac {5 (b c-7 a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^3}+\frac {(b c-7 a d) \sqrt {a+b x} (c+d x)^{5/2}}{3 b^2 (b c-a d)}+\frac {2 a (c+d x)^{7/2}}{b (b c-a d) \sqrt {a+b x}}+\frac {\left (5 (b c-7 a d) (b c-a d)^2\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^5}\\ &=\frac {5 (b c-7 a d) (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{8 b^4}+\frac {5 (b c-7 a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^3}+\frac {(b c-7 a d) \sqrt {a+b x} (c+d x)^{5/2}}{3 b^2 (b c-a d)}+\frac {2 a (c+d x)^{7/2}}{b (b c-a d) \sqrt {a+b x}}+\frac {5 (b c-7 a d) (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{9/2} \sqrt {d}}\\ \end {align*}

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Mathematica [A]  time = 0.65, size = 175, normalized size = 0.82 \begin {gather*} \frac {\sqrt {c+d x} \left (\frac {105 a^3 d^2+5 a^2 b d (7 d x-38 c)+a b^2 \left (81 c^2-68 c d x-14 d^2 x^2\right )+b^3 x \left (33 c^2+26 c d x+8 d^2 x^2\right )}{\sqrt {a+b x}}+\frac {15 (b c-7 a d) (b c-a d)^{3/2} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{\sqrt {d} \sqrt {\frac {b (c+d x)}{b c-a d}}}\right )}{24 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*(c + d*x)^(5/2))/(a + b*x)^(3/2),x]

[Out]

(Sqrt[c + d*x]*((105*a^3*d^2 + 5*a^2*b*d*(-38*c + 7*d*x) + a*b^2*(81*c^2 - 68*c*d*x - 14*d^2*x^2) + b^3*x*(33*
c^2 + 26*c*d*x + 8*d^2*x^2))/Sqrt[a + b*x] + (15*(b*c - 7*a*d)*(b*c - a*d)^(3/2)*ArcSinh[(Sqrt[d]*Sqrt[a + b*x
])/Sqrt[b*c - a*d]])/(Sqrt[d]*Sqrt[(b*(c + d*x))/(b*c - a*d)])))/(24*b^4)

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IntegrateAlgebraic [A]  time = 0.53, size = 231, normalized size = 1.08 \begin {gather*} \frac {5 (b c-7 a d) (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{8 b^{9/2} \sqrt {d}}+\frac {\sqrt {c+d x} (b c-a d)^2 \left (\frac {48 a b^3 (c+d x)^3}{(a+b x)^3}+\frac {33 b^3 c (c+d x)^2}{(a+b x)^2}-\frac {231 a b^2 d (c+d x)^2}{(a+b x)^2}-\frac {40 b^2 c d (c+d x)}{a+b x}+\frac {280 a b d^2 (c+d x)}{a+b x}-105 a d^3+15 b c d^2\right )}{24 b^4 \sqrt {a+b x} \left (\frac {b (c+d x)}{a+b x}-d\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x*(c + d*x)^(5/2))/(a + b*x)^(3/2),x]

[Out]

((b*c - a*d)^2*Sqrt[c + d*x]*(15*b*c*d^2 - 105*a*d^3 - (40*b^2*c*d*(c + d*x))/(a + b*x) + (280*a*b*d^2*(c + d*
x))/(a + b*x) + (33*b^3*c*(c + d*x)^2)/(a + b*x)^2 - (231*a*b^2*d*(c + d*x)^2)/(a + b*x)^2 + (48*a*b^3*(c + d*
x)^3)/(a + b*x)^3))/(24*b^4*Sqrt[a + b*x]*(-d + (b*(c + d*x))/(a + b*x))^3) + (5*(b*c - 7*a*d)*(b*c - a*d)^2*A
rcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(8*b^(9/2)*Sqrt[d])

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fricas [A]  time = 1.47, size = 594, normalized size = 2.78 \begin {gather*} \left [-\frac {15 \, {\left (a b^{3} c^{3} - 9 \, a^{2} b^{2} c^{2} d + 15 \, a^{3} b c d^{2} - 7 \, a^{4} d^{3} + {\left (b^{4} c^{3} - 9 \, a b^{3} c^{2} d + 15 \, a^{2} b^{2} c d^{2} - 7 \, a^{3} b d^{3}\right )} x\right )} \sqrt {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 x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (8 \, b^{4} d^{3} x^{3} + 81 \, a b^{3} c^{2} d - 190 \, a^{2} b^{2} c d^{2} + 105 \, a^{3} b d^{3} + 2 \, {\left (13 \, b^{4} c d^{2} - 7 \, a b^{3} d^{3}\right )} x^{2} + {\left (33 \, b^{4} c^{2} d - 68 \, a b^{3} c d^{2} + 35 \, a^{2} b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, {\left (b^{6} d x + a b^{5} d\right )}}, -\frac {15 \, {\left (a b^{3} c^{3} - 9 \, a^{2} b^{2} c^{2} d + 15 \, a^{3} b c d^{2} - 7 \, a^{4} d^{3} + {\left (b^{4} c^{3} - 9 \, a b^{3} c^{2} d + 15 \, a^{2} b^{2} c d^{2} - 7 \, a^{3} b d^{3}\right )} x\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (8 \, b^{4} d^{3} x^{3} + 81 \, a b^{3} c^{2} d - 190 \, a^{2} b^{2} c d^{2} + 105 \, a^{3} b d^{3} + 2 \, {\left (13 \, b^{4} c d^{2} - 7 \, a b^{3} d^{3}\right )} x^{2} + {\left (33 \, b^{4} c^{2} d - 68 \, a b^{3} c d^{2} + 35 \, a^{2} b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, {\left (b^{6} d x + a b^{5} d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(d*x+c)^(5/2)/(b*x+a)^(3/2),x, algorithm="fricas")

[Out]

[-1/96*(15*(a*b^3*c^3 - 9*a^2*b^2*c^2*d + 15*a^3*b*c*d^2 - 7*a^4*d^3 + (b^4*c^3 - 9*a*b^3*c^2*d + 15*a^2*b^2*c
*d^2 - 7*a^3*b*d^3)*x)*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)*s
qrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(8*b^4*d^3*x^3 + 81*a*b^3*c^2*d - 190*a^2*
b^2*c*d^2 + 105*a^3*b*d^3 + 2*(13*b^4*c*d^2 - 7*a*b^3*d^3)*x^2 + (33*b^4*c^2*d - 68*a*b^3*c*d^2 + 35*a^2*b^2*d
^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^6*d*x + a*b^5*d), -1/48*(15*(a*b^3*c^3 - 9*a^2*b^2*c^2*d + 15*a^3*b*c*d
^2 - 7*a^4*d^3 + (b^4*c^3 - 9*a*b^3*c^2*d + 15*a^2*b^2*c*d^2 - 7*a^3*b*d^3)*x)*sqrt(-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^
4*d^3*x^3 + 81*a*b^3*c^2*d - 190*a^2*b^2*c*d^2 + 105*a^3*b*d^3 + 2*(13*b^4*c*d^2 - 7*a*b^3*d^3)*x^2 + (33*b^4*
c^2*d - 68*a*b^3*c*d^2 + 35*a^2*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^6*d*x + a*b^5*d)]

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giac [B]  time = 2.09, size = 370, normalized size = 1.73 \begin {gather*} \frac {1}{24} \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )} d^{2} {\left | b \right |}}{b^{6}} + \frac {13 \, b^{18} c d^{5} {\left | b \right |} - 19 \, a b^{17} d^{6} {\left | b \right |}}{b^{23} d^{4}}\right )} + \frac {3 \, {\left (11 \, b^{19} c^{2} d^{4} {\left | b \right |} - 40 \, a b^{18} c d^{5} {\left | b \right |} + 29 \, a^{2} b^{17} d^{6} {\left | b \right |}\right )}}{b^{23} d^{4}}\right )} + \frac {4 \, {\left (\sqrt {b d} a b^{3} c^{3} {\left | b \right |} - 3 \, \sqrt {b d} a^{2} b^{2} c^{2} d {\left | b \right |} + 3 \, \sqrt {b d} a^{3} b c d^{2} {\left | b \right |} - \sqrt {b d} a^{4} d^{3} {\left | b \right |}\right )}}{{\left (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}\right )} b^{5}} - \frac {5 \, {\left (\sqrt {b d} b^{3} c^{3} {\left | b \right |} - 9 \, \sqrt {b d} a b^{2} c^{2} d {\left | b \right |} + 15 \, \sqrt {b d} a^{2} b c d^{2} {\left | b \right |} - 7 \, \sqrt {b d} a^{3} d^{3} {\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 )}{16 \, b^{6} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(d*x+c)^(5/2)/(b*x+a)^(3/2),x, algorithm="giac")

[Out]

1/24*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)*d^2*abs(b)/b^6 + (13*b^18*c*d
^5*abs(b) - 19*a*b^17*d^6*abs(b))/(b^23*d^4)) + 3*(11*b^19*c^2*d^4*abs(b) - 40*a*b^18*c*d^5*abs(b) + 29*a^2*b^
17*d^6*abs(b))/(b^23*d^4)) + 4*(sqrt(b*d)*a*b^3*c^3*abs(b) - 3*sqrt(b*d)*a^2*b^2*c^2*d*abs(b) + 3*sqrt(b*d)*a^
3*b*c*d^2*abs(b) - sqrt(b*d)*a^4*d^3*abs(b))/((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)*b^5) - 5/16*(sqrt(b*d)*b^3*c^3*abs(b) - 9*sqrt(b*d)*a*b^2*c^2*d*abs(b) + 15*sqrt(b*d)*a^2*
b*c*d^2*abs(b) - 7*sqrt(b*d)*a^3*d^3*abs(b))*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d
))^2)/(b^6*d)

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maple [B]  time = 0.03, size = 689, normalized size = 3.22 \begin {gather*} -\frac {\sqrt {d x +c}\, \left (105 a^{3} b \,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 )-225 a^{2} b^{2} 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 )+135 a \,b^{3} c^{2} d 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 )-15 b^{4} c^{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 )+105 a^{4} 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 )-225 a^{3} 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 )+135 a^{2} 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 )-15 a \,b^{3} c^{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 )-16 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{3} d^{2} x^{3}+28 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,b^{2} d^{2} x^{2}-52 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{3} c d \,x^{2}-70 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} b \,d^{2} x +136 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,b^{2} c d x -66 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{3} c^{2} x -210 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{3} d^{2}+380 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} b c d -162 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a \,b^{2} c^{2}\right )}{48 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {b x +a}\, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(d*x+c)^(5/2)/(b*x+a)^(3/2),x)

[Out]

-1/48*(d*x+c)^(1/2)*(-16*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*b^3*d^2*x^3+105*a^3*b*d^3*x*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))-225*a^2*b^2*c*d^2*x*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))+135*a*b^3*c^2*d*x*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))-15*b^4*c^3*x*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))+28*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a*b^2*d^2*x^2-52*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*b^3*c*d*x^2+10
5*a^4*d^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))-225*a^3*b*c*d^2*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))+135*a^2*b^2*c^2*d*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))-15*a*b^3*c^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))-70*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a^2*b*d^2*x+136*((b*x+a)*(d*x+c))^(1/2)*
(b*d)^(1/2)*a*b^2*c*d*x-66*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*b^3*c^2*x-210*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/
2)*a^3*d^2+380*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a^2*b*c*d-162*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a*b^2*c^2
)/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)/(b*x+a)^(1/2)/b^4

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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(x*(d*x+c)^(5/2)/(b*x+a)^(3/2),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 = -0.00 \begin {gather*} \int \frac {x\,{\left (c+d\,x\right )}^{5/2}}{{\left (a+b\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(c + d*x)^(5/2))/(a + b*x)^(3/2),x)

[Out]

int((x*(c + d*x)^(5/2))/(a + b*x)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \left (c + d x\right )^{\frac {5}{2}}}{\left (a + b x\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(d*x+c)**(5/2)/(b*x+a)**(3/2),x)

[Out]

Integral(x*(c + d*x)**(5/2)/(a + b*x)**(3/2), x)

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