3.7.75 \(\int \frac {x (a+b x)^{5/2}}{\sqrt {c+d x}} \, dx\) [675]

Optimal. Leaf size=217 \[ -\frac {5 (b c-a d)^2 (7 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{64 b d^4}+\frac {5 (b c-a d) (7 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{96 b d^3}-\frac {(7 b c+a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 b d^2}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b d}+\frac {5 (b c-a d)^3 (7 b c+a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{3/2} d^{9/2}} \]

[Out]

5/64*(-a*d+b*c)^3*(a*d+7*b*c)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(3/2)/d^(9/2)+5/96*(-a*d+
b*c)*(a*d+7*b*c)*(b*x+a)^(3/2)*(d*x+c)^(1/2)/b/d^3-1/24*(a*d+7*b*c)*(b*x+a)^(5/2)*(d*x+c)^(1/2)/b/d^2+1/4*(b*x
+a)^(7/2)*(d*x+c)^(1/2)/b/d-5/64*(-a*d+b*c)^2*(a*d+7*b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b/d^4

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Rubi [A]
time = 0.08, antiderivative size = 217, 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 = {81, 52, 65, 223, 212} \begin {gather*} \frac {5 (b c-a d)^3 (a d+7 b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{3/2} d^{9/2}}-\frac {5 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^2 (a d+7 b c)}{64 b d^4}+\frac {5 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d) (a d+7 b c)}{96 b d^3}-\frac {(a+b x)^{5/2} \sqrt {c+d x} (a d+7 b c)}{24 b d^2}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*(b*c - a*d)^2*(7*b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*b*d^4) + (5*(b*c - a*d)*(7*b*c + a*d)*(a + b*
x)^(3/2)*Sqrt[c + d*x])/(96*b*d^3) - ((7*b*c + a*d)*(a + b*x)^(5/2)*Sqrt[c + d*x])/(24*b*d^2) + ((a + b*x)^(7/
2)*Sqrt[c + d*x])/(4*b*d) + (5*(b*c - a*d)^3*(7*b*c + a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d
*x])])/(64*b^(3/2)*d^(9/2))

Rule 52

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 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 81

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

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 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 {x (a+b x)^{5/2}}{\sqrt {c+d x}} \, dx &=\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b d}-\frac {(7 b c+a d) \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}} \, dx}{8 b d}\\ &=-\frac {(7 b c+a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 b d^2}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b d}+\frac {(5 (b c-a d) (7 b c+a d)) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx}{48 b d^2}\\ &=\frac {5 (b c-a d) (7 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{96 b d^3}-\frac {(7 b c+a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 b d^2}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b d}-\frac {\left (5 (b c-a d)^2 (7 b c+a d)\right ) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx}{64 b d^3}\\ &=-\frac {5 (b c-a d)^2 (7 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{64 b d^4}+\frac {5 (b c-a d) (7 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{96 b d^3}-\frac {(7 b c+a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 b d^2}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b d}+\frac {\left (5 (b c-a d)^3 (7 b c+a d)\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{128 b d^4}\\ &=-\frac {5 (b c-a d)^2 (7 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{64 b d^4}+\frac {5 (b c-a d) (7 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{96 b d^3}-\frac {(7 b c+a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 b d^2}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b d}+\frac {\left (5 (b c-a d)^3 (7 b c+a d)\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 )}{64 b^2 d^4}\\ &=-\frac {5 (b c-a d)^2 (7 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{64 b d^4}+\frac {5 (b c-a d) (7 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{96 b d^3}-\frac {(7 b c+a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 b d^2}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b d}+\frac {\left (5 (b c-a d)^3 (7 b c+a d)\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 )}{64 b^2 d^4}\\ &=-\frac {5 (b c-a d)^2 (7 b c+a d) \sqrt {a+b x} \sqrt {c+d x}}{64 b d^4}+\frac {5 (b c-a d) (7 b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{96 b d^3}-\frac {(7 b c+a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 b d^2}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b d}+\frac {5 (b c-a d)^3 (7 b c+a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{3/2} d^{9/2}}\\ \end {align*}

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Mathematica [A]
time = 0.42, size = 175, normalized size = 0.81 \begin {gather*} \frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 a^3 d^3+a^2 b d^2 (-191 c+118 d x)+a b^2 d \left (265 c^2-172 c d x+136 d^2 x^2\right )+b^3 \left (-105 c^3+70 c^2 d x-56 c d^2 x^2+48 d^3 x^3\right )\right )}{192 b d^4}+\frac {5 (b c-a d)^3 (7 b c+a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{64 b^{3/2} d^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*Sqrt[c + d*x]*(15*a^3*d^3 + a^2*b*d^2*(-191*c + 118*d*x) + a*b^2*d*(265*c^2 - 172*c*d*x + 136*d
^2*x^2) + b^3*(-105*c^3 + 70*c^2*d*x - 56*c*d^2*x^2 + 48*d^3*x^3)))/(192*b*d^4) + (5*(b*c - a*d)^3*(7*b*c + a*
d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(64*b^(3/2)*d^(9/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(573\) vs. \(2(179)=358\).
time = 0.07, size = 574, normalized size = 2.65

method	result	size
default	\(-\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (-96 b^{3} d^{3} x^{3} \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}-272 a \,b^{2} d^{3} x^{2} \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+112 b^{3} c \,d^{2} x^{2} \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+15 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{4} d^{4}+60 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3} b c \,d^{3}-270 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b^{2} c^{2} d^{2}+300 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{3} c^{3} d -105 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{4} c^{4}-236 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a^{2} b \,d^{3} x +344 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a \,b^{2} c \,d^{2} x -140 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, b^{3} c^{2} d x -30 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a^{3} d^{3}+382 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a^{2} b c \,d^{2}-530 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a \,b^{2} c^{2} d +210 \sqrt {b d}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, b^{3} c^{3}\right )}{384 b \,d^{4} \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}}\)	\(574\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(-96*b^3*d^3*x^3*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)-272*a*b^2*d^3*x^2*((d*
x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)+112*b^3*c*d^2*x^2*((d*x+c)*(b*x+a))^(1/2)*(b*d)^(1/2)+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^4*d^4+60*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^3*b*c*d^3-270*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^2*b^2*c^2*d^2+300*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*b^3*c^3*d-105*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))*b^4*c^4-236*(b*d)^
(1/2)*((d*x+c)*(b*x+a))^(1/2)*a^2*b*d^3*x+344*(b*d)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*a*b^2*c*d^2*x-140*(b*d)^(1/2
)*((d*x+c)*(b*x+a))^(1/2)*b^3*c^2*d*x-30*(b*d)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*a^3*d^3+382*(b*d)^(1/2)*((d*x+c)*
(b*x+a))^(1/2)*a^2*b*c*d^2-530*(b*d)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*a*b^2*c^2*d+210*(b*d)^(1/2)*((d*x+c)*(b*x+a
))^(1/2)*b^3*c^3)/b/d^4/((d*x+c)*(b*x+a))^(1/2)/(b*d)^(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(x*(b*x+a)^(5/2)/(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 = 1.11, size = 546, normalized size = 2.52 \begin {gather*} \left [-\frac {15 \, {\left (7 \, b^{4} c^{4} - 20 \, a b^{3} c^{3} d + 18 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} - a^{4} d^{4}\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 (48 \, b^{4} d^{4} x^{3} - 105 \, b^{4} c^{3} d + 265 \, a b^{3} c^{2} d^{2} - 191 \, a^{2} b^{2} c d^{3} + 15 \, a^{3} b d^{4} - 8 \, {\left (7 \, b^{4} c d^{3} - 17 \, a b^{3} d^{4}\right )} x^{2} + 2 \, {\left (35 \, b^{4} c^{2} d^{2} - 86 \, a b^{3} c d^{3} + 59 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{768 \, b^{2} d^{5}}, -\frac {15 \, {\left (7 \, b^{4} c^{4} - 20 \, a b^{3} c^{3} d + 18 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} - a^{4} d^{4}\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 (48 \, b^{4} d^{4} x^{3} - 105 \, b^{4} c^{3} d + 265 \, a b^{3} c^{2} d^{2} - 191 \, a^{2} b^{2} c d^{3} + 15 \, a^{3} b d^{4} - 8 \, {\left (7 \, b^{4} c d^{3} - 17 \, a b^{3} d^{4}\right )} x^{2} + 2 \, {\left (35 \, b^{4} c^{2} d^{2} - 86 \, a b^{3} c d^{3} + 59 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{384 \, b^{2} d^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/768*(15*(7*b^4*c^4 - 20*a*b^3*c^3*d + 18*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 - a^4*d^4)*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*(48*b^4*d^4*x^3 - 105*b^4*c^3*d + 265*a*b^3*c^2*d^2 - 191*a^2*b^2*c*d^3 + 15*a^3*b*d^4
- 8*(7*b^4*c*d^3 - 17*a*b^3*d^4)*x^2 + 2*(35*b^4*c^2*d^2 - 86*a*b^3*c*d^3 + 59*a^2*b^2*d^4)*x)*sqrt(b*x + a)*s
qrt(d*x + c))/(b^2*d^5), -1/384*(15*(7*b^4*c^4 - 20*a*b^3*c^3*d + 18*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 - a^4*d^4
)*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*(48*b^4*d^4*x^3 - 105*b^4*c^3*d + 265*a*b^3*c^2*d^2 - 191*a^2*b^2*c*d^3 + 15*a^3*b
*d^4 - 8*(7*b^4*c*d^3 - 17*a*b^3*d^4)*x^2 + 2*(35*b^4*c^2*d^2 - 86*a*b^3*c*d^3 + 59*a^2*b^2*d^4)*x)*sqrt(b*x +
a)*sqrt(d*x + c))/(b^2*d^5)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 2.15, size = 290, normalized size = 1.34 \begin {gather*} \frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, {\left (b x + a\right )} {\left (4 \, {\left (b x + a\right )} {\left (\frac {6 \, {\left (b x + a\right )}}{b^{2} d} - \frac {7 \, b^{3} c d^{5} + a b^{2} d^{6}}{b^{4} d^{7}}\right )} + \frac {5 \, {\left (7 \, b^{4} c^{2} d^{4} - 6 \, a b^{3} c d^{5} - a^{2} b^{2} d^{6}\right )}}{b^{4} d^{7}}\right )} - \frac {15 \, {\left (7 \, b^{5} c^{3} d^{3} - 13 \, a b^{4} c^{2} d^{4} + 5 \, a^{2} b^{3} c d^{5} + a^{3} b^{2} d^{6}\right )}}{b^{4} d^{7}}\right )} \sqrt {b x + a} - \frac {15 \, {\left (7 \, b^{4} c^{4} - 20 \, a b^{3} c^{3} d + 18 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} - a^{4} d^{4}\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 |}\right )}{\sqrt {b d} b d^{4}}\right )} b}{192 \, {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(b*x + a)*(4*(b*x + a)*(6*(b*x + a)/(b^2*d) - (7*b^3*c*d^5 + a*b
^2*d^6)/(b^4*d^7)) + 5*(7*b^4*c^2*d^4 - 6*a*b^3*c*d^5 - a^2*b^2*d^6)/(b^4*d^7)) - 15*(7*b^5*c^3*d^3 - 13*a*b^4
*c^2*d^4 + 5*a^2*b^3*c*d^5 + a^3*b^2*d^6)/(b^4*d^7))*sqrt(b*x + a) - 15*(7*b^4*c^4 - 20*a*b^3*c^3*d + 18*a^2*b
^2*c^2*d^2 - 4*a^3*b*c*d^3 - a^4*d^4)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))
/(sqrt(b*d)*b*d^4))*b/abs(b)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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