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

Optimal. Leaf size=218 \[ -\frac {5 (b c-a d)^2 (7 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 \sqrt {b} d^{9/2}}+\frac {5 \sqrt {a+b x} \sqrt {c+d x} (b c-a d) (7 b c-a d)}{8 d^4}-\frac {5 (a+b x)^{3/2} \sqrt {c+d x} (7 b c-a d)}{12 d^3}+\frac {(a+b x)^{5/2} \sqrt {c+d x} (7 b c-a d)}{3 d^2 (b c-a d)}-\frac {2 c (a+b x)^{7/2}}{d \sqrt {c+d x} (b c-a d)} \]

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Rubi [A]  time = 0.13, antiderivative size = 218, 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 {(a+b x)^{5/2} \sqrt {c+d x} (7 b c-a d)}{3 d^2 (b c-a d)}-\frac {5 (a+b x)^{3/2} \sqrt {c+d x} (7 b c-a d)}{12 d^3}+\frac {5 \sqrt {a+b x} \sqrt {c+d x} (b c-a d) (7 b c-a d)}{8 d^4}-\frac {5 (b c-a d)^2 (7 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 \sqrt {b} d^{9/2}}-\frac {2 c (a+b x)^{7/2}}{d \sqrt {c+d x} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.59, size = 221, normalized size = 1.01 \begin {gather*} \frac {\frac {\sqrt {d} \left (3 a^3 d^2 (27 c+11 d x)+a^2 b d \left (-190 c^2+13 c d x+59 d^2 x^2\right )+a b^2 \left (105 c^3-155 c^2 d x-82 c d^2 x^2+34 d^3 x^3\right )+b^3 x \left (105 c^3+35 c^2 d x-14 c d^2 x^2+8 d^3 x^3\right )\right )}{\sqrt {a+b x}}-\frac {15 (b c-a d)^{5/2} (7 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}}{24 d^{9/2} \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d]*(3*a^3*d^2*(27*c + 11*d*x) + a^2*b*d*(-190*c^2 + 13*c*d*x + 59*d^2*x^2) + b^3*x*(105*c^3 + 35*c^2*d*
x - 14*c*d^2*x^2 + 8*d^3*x^3) + a*b^2*(105*c^3 - 155*c^2*d*x - 82*c*d^2*x^2 + 34*d^3*x^3)))/Sqrt[a + b*x] - (1
5*(b*c - a*d)^(5/2)*(7*b*c - a*d)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a
*d]])/b)/(24*d^(9/2)*Sqrt[c + d*x])

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IntegrateAlgebraic [A]  time = 0.60, size = 232, normalized size = 1.06 \begin {gather*} \frac {\sqrt {a+b x} (a d-b c)^2 \left (\frac {280 b^2 c d (a+b x)}{c+d x}+15 a b^2 d+\frac {33 a d^3 (a+b x)^2}{(c+d x)^2}+\frac {48 c d^3 (a+b x)^3}{(c+d x)^3}-\frac {40 a b d^2 (a+b x)}{c+d x}-\frac {231 b c d^2 (a+b x)^2}{(c+d x)^2}-105 b^3 c\right )}{24 d^4 \sqrt {c+d x} \left (\frac {d (a+b x)}{c+d x}-b\right )^3}-\frac {5 (b c-a d)^2 (7 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 \sqrt {b} d^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.84, size = 598, normalized size = 2.74 \begin {gather*} \left [-\frac {15 \, {\left (7 \, b^{3} c^{4} - 15 \, a b^{2} c^{3} d + 9 \, a^{2} b c^{2} d^{2} - a^{3} c d^{3} + {\left (7 \, b^{3} c^{3} d - 15 \, a b^{2} c^{2} d^{2} + 9 \, a^{2} b c d^{3} - a^{3} d^{4}\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^{3} d^{4} x^{3} + 105 \, b^{3} c^{3} d - 190 \, a b^{2} c^{2} d^{2} + 81 \, a^{2} b c d^{3} - 2 \, {\left (7 \, b^{3} c d^{3} - 13 \, a b^{2} d^{4}\right )} x^{2} + {\left (35 \, b^{3} c^{2} d^{2} - 68 \, a b^{2} c d^{3} + 33 \, a^{2} b d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, {\left (b d^{6} x + b c d^{5}\right )}}, \frac {15 \, {\left (7 \, b^{3} c^{4} - 15 \, a b^{2} c^{3} d + 9 \, a^{2} b c^{2} d^{2} - a^{3} c d^{3} + {\left (7 \, b^{3} c^{3} d - 15 \, a b^{2} c^{2} d^{2} + 9 \, a^{2} b c d^{3} - a^{3} d^{4}\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^{3} d^{4} x^{3} + 105 \, b^{3} c^{3} d - 190 \, a b^{2} c^{2} d^{2} + 81 \, a^{2} b c d^{3} - 2 \, {\left (7 \, b^{3} c d^{3} - 13 \, a b^{2} d^{4}\right )} x^{2} + {\left (35 \, b^{3} c^{2} d^{2} - 68 \, a b^{2} c d^{3} + 33 \, a^{2} b d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, {\left (b d^{6} x + b c d^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.48, size = 292, normalized size = 1.34 \begin {gather*} \frac {{\left ({\left (b x + a\right )} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )} {\left | b \right |}}{b d} - \frac {7 \, b^{2} c d^{5} {\left | b \right |} - a b d^{6} {\left | b \right |}}{b^{2} d^{7}}\right )} + \frac {5 \, {\left (7 \, b^{3} c^{2} d^{4} {\left | b \right |} - 8 \, a b^{2} c d^{5} {\left | b \right |} + a^{2} b d^{6} {\left | b \right |}\right )}}{b^{2} d^{7}}\right )} + \frac {15 \, {\left (7 \, b^{4} c^{3} d^{3} {\left | b \right |} - 15 \, a b^{3} c^{2} d^{4} {\left | b \right |} + 9 \, a^{2} b^{2} c d^{5} {\left | b \right |} - a^{3} b d^{6} {\left | b \right |}\right )}}{b^{2} d^{7}}\right )} \sqrt {b x + a}}{24 \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} + \frac {5 \, {\left (7 \, b^{3} c^{3} {\left | b \right |} - 15 \, a b^{2} c^{2} d {\left | b \right |} + 9 \, a^{2} b c d^{2} {\left | b \right |} - 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 |}\right )}{8 \, \sqrt {b d} b d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((b*x + a)*(2*(b*x + a)*(4*(b*x + a)*abs(b)/(b*d) - (7*b^2*c*d^5*abs(b) - a*b*d^6*abs(b))/(b^2*d^7)) + 5*
(7*b^3*c^2*d^4*abs(b) - 8*a*b^2*c*d^5*abs(b) + a^2*b*d^6*abs(b))/(b^2*d^7)) + 15*(7*b^4*c^3*d^3*abs(b) - 15*a*
b^3*c^2*d^4*abs(b) + 9*a^2*b^2*c*d^5*abs(b) - a^3*b*d^6*abs(b))/(b^2*d^7))*sqrt(b*x + a)/sqrt(b^2*c + (b*x + a
)*b*d - a*b*d) + 5/8*(7*b^3*c^3*abs(b) - 15*a*b^2*c^2*d*abs(b) + 9*a^2*b*c*d^2*abs(b) - a^3*d^3*abs(b))*log(ab
s(-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)

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maple [B]  time = 0.03, size = 689, normalized size = 3.16 \begin {gather*} \frac {\sqrt {b x +a}\, \left (15 a^{3} d^{4} 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^{2} b c \,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 \,b^{2} c^{2} 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 )-105 b^{3} c^{3} 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 a^{3} c \,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 )-135 a^{2} b \,c^{2} 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 )+225 a \,b^{2} c^{3} 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 )-105 b^{3} c^{4} \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^{2} d^{3} x^{3}+52 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a b \,d^{3} x^{2}-28 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{2} c \,d^{2} x^{2}+66 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} d^{3} x -136 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a b c \,d^{2} x +70 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{2} c^{2} d x +162 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} c \,d^{2}-380 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a b \,c^{2} d +210 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{2} c^{3}\right )}{48 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {d x +c}\, d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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