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

Optimal. Leaf size=222 \[ \frac {5 \sqrt {b} (7 b c-3 a d) (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 d^{9/2}}-\frac {5 b \sqrt {a+b x} \sqrt {c+d x} (7 b c-3 a d)}{4 d^4}+\frac {5 b (a+b x)^{3/2} \sqrt {c+d x} (7 b c-3 a d)}{6 d^3 (b c-a d)}-\frac {2 (a+b x)^{5/2} (7 b c-3 a d)}{3 d^2 \sqrt {c+d x} (b c-a d)}-\frac {2 c (a+b x)^{7/2}}{3 d (c+d x)^{3/2} (b c-a d)} \]

[Out]

-2/3*c*(b*x+a)^(7/2)/d/(-a*d+b*c)/(d*x+c)^(3/2)+5/4*(-3*a*d+7*b*c)*(-a*d+b*c)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^
(1/2)/(d*x+c)^(1/2))*b^(1/2)/d^(9/2)-2/3*(-3*a*d+7*b*c)*(b*x+a)^(5/2)/d^2/(-a*d+b*c)/(d*x+c)^(1/2)+5/6*b*(-3*a
*d+7*b*c)*(b*x+a)^(3/2)*(d*x+c)^(1/2)/d^3/(-a*d+b*c)-5/4*b*(-3*a*d+7*b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/d^4

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Rubi [A]  time = 0.12, antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {78, 47, 50, 63, 217, 206} \[ -\frac {2 (a+b x)^{5/2} (7 b c-3 a d)}{3 d^2 \sqrt {c+d x} (b c-a d)}+\frac {5 b (a+b x)^{3/2} \sqrt {c+d x} (7 b c-3 a d)}{6 d^3 (b c-a d)}-\frac {5 b \sqrt {a+b x} \sqrt {c+d x} (7 b c-3 a d)}{4 d^4}+\frac {5 \sqrt {b} (7 b c-3 a d) (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 d^{9/2}}-\frac {2 c (a+b x)^{7/2}}{3 d (c+d x)^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c*(a + b*x)^(7/2))/(3*d*(b*c - a*d)*(c + d*x)^(3/2)) - (2*(7*b*c - 3*a*d)*(a + b*x)^(5/2))/(3*d^2*(b*c - a
*d)*Sqrt[c + d*x]) - (5*b*(7*b*c - 3*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*d^4) + (5*b*(7*b*c - 3*a*d)*(a + b*x
)^(3/2)*Sqrt[c + d*x])/(6*d^3*(b*c - a*d)) + (5*Sqrt[b]*(7*b*c - 3*a*d)*(b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a +
b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*d^(9/2))

Rule 47

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

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

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Mathematica [C]  time = 0.16, size = 110, normalized size = 0.50 \[ \frac {2 (a+b x)^{7/2} \left ((c+d x) (7 b c-3 a d) \sqrt {\frac {b (c+d x)}{b c-a d}} \, _2F_1\left (\frac {3}{2},\frac {7}{2};\frac {9}{2};\frac {d (a+b x)}{a d-b c}\right )+7 c (a d-b c)\right )}{21 d (c+d x)^{3/2} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)^(7/2)*(7*c*(-(b*c) + a*d) + (7*b*c - 3*a*d)*(c + d*x)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*Hypergeomet
ric2F1[3/2, 7/2, 9/2, (d*(a + b*x))/(-(b*c) + a*d)]))/(21*d*(b*c - a*d)^2*(c + d*x)^(3/2))

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fricas [A]  time = 1.65, size = 619, normalized size = 2.79 \[ \left [\frac {15 \, {\left (7 \, b^{2} c^{4} - 10 \, a b c^{3} d + 3 \, a^{2} c^{2} d^{2} + {\left (7 \, b^{2} c^{2} d^{2} - 10 \, a b c d^{3} + 3 \, a^{2} d^{4}\right )} x^{2} + 2 \, {\left (7 \, b^{2} c^{3} d - 10 \, a b c^{2} d^{2} + 3 \, a^{2} c d^{3}\right )} x\right )} \sqrt {\frac {b}{d}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d^{2} x + b c d + a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {b}{d}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (6 \, b^{2} d^{3} x^{3} - 105 \, b^{2} c^{3} + 115 \, a b c^{2} d - 16 \, a^{2} c d^{2} - 3 \, {\left (7 \, b^{2} c d^{2} - 9 \, a b d^{3}\right )} x^{2} - 2 \, {\left (70 \, b^{2} c^{2} d - 79 \, a b c d^{2} + 12 \, a^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, {\left (d^{6} x^{2} + 2 \, c d^{5} x + c^{2} d^{4}\right )}}, -\frac {15 \, {\left (7 \, b^{2} c^{4} - 10 \, a b c^{3} d + 3 \, a^{2} c^{2} d^{2} + {\left (7 \, b^{2} c^{2} d^{2} - 10 \, a b c d^{3} + 3 \, a^{2} d^{4}\right )} x^{2} + 2 \, {\left (7 \, b^{2} c^{3} d - 10 \, a b c^{2} d^{2} + 3 \, a^{2} c d^{3}\right )} x\right )} \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} d x^{2} + a b c + {\left (b^{2} c + a b d\right )} x\right )}}\right ) - 2 \, {\left (6 \, b^{2} d^{3} x^{3} - 105 \, b^{2} c^{3} + 115 \, a b c^{2} d - 16 \, a^{2} c d^{2} - 3 \, {\left (7 \, b^{2} c d^{2} - 9 \, a b d^{3}\right )} x^{2} - 2 \, {\left (70 \, b^{2} c^{2} d - 79 \, a b c d^{2} + 12 \, a^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{24 \, {\left (d^{6} x^{2} + 2 \, c d^{5} x + c^{2} d^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(15*(7*b^2*c^4 - 10*a*b*c^3*d + 3*a^2*c^2*d^2 + (7*b^2*c^2*d^2 - 10*a*b*c*d^3 + 3*a^2*d^4)*x^2 + 2*(7*b^
2*c^3*d - 10*a*b*c^2*d^2 + 3*a^2*c*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^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*(6*b^2*d^3*x^3 -
105*b^2*c^3 + 115*a*b*c^2*d - 16*a^2*c*d^2 - 3*(7*b^2*c*d^2 - 9*a*b*d^3)*x^2 - 2*(70*b^2*c^2*d - 79*a*b*c*d^2
+ 12*a^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(d^6*x^2 + 2*c*d^5*x + c^2*d^4), -1/24*(15*(7*b^2*c^4 - 10*a*b*
c^3*d + 3*a^2*c^2*d^2 + (7*b^2*c^2*d^2 - 10*a*b*c*d^3 + 3*a^2*d^4)*x^2 + 2*(7*b^2*c^3*d - 10*a*b*c^2*d^2 + 3*a
^2*c*d^3)*x)*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*(6*b^2*d^3*x^3 - 105*b^2*c^3 + 115*a*b*c^2*d - 16*a^2*c*d^2 - 3*(7*b^2*c*d^2 -
9*a*b*d^3)*x^2 - 2*(70*b^2*c^2*d - 79*a*b*c*d^2 + 12*a^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(d^6*x^2 + 2*c*d
^5*x + c^2*d^4)]

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giac [B]  time = 1.61, size = 404, normalized size = 1.82 \[ \frac {{\left ({\left (3 \, {\left (b x + a\right )} {\left (\frac {2 \, {\left (b^{5} c d^{6} {\left | b \right |} - a b^{4} d^{7} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{4} c d^{7} - a b^{3} d^{8}} - \frac {7 \, b^{6} c^{2} d^{5} {\left | b \right |} - 10 \, a b^{5} c d^{6} {\left | b \right |} + 3 \, a^{2} b^{4} d^{7} {\left | b \right |}}{b^{4} c d^{7} - a b^{3} d^{8}}\right )} - \frac {20 \, {\left (7 \, b^{7} c^{3} d^{4} {\left | b \right |} - 17 \, a b^{6} c^{2} d^{5} {\left | b \right |} + 13 \, a^{2} b^{5} c d^{6} {\left | b \right |} - 3 \, a^{3} b^{4} d^{7} {\left | b \right |}\right )}}{b^{4} c d^{7} - a b^{3} d^{8}}\right )} {\left (b x + a\right )} - \frac {15 \, {\left (7 \, b^{8} c^{4} d^{3} {\left | b \right |} - 24 \, a b^{7} c^{3} d^{4} {\left | b \right |} + 30 \, a^{2} b^{6} c^{2} d^{5} {\left | b \right |} - 16 \, a^{3} b^{5} c d^{6} {\left | b \right |} + 3 \, a^{4} b^{4} d^{7} {\left | b \right |}\right )}}{b^{4} c d^{7} - a b^{3} d^{8}}\right )} \sqrt {b x + a}}{12 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} - \frac {5 \, {\left (7 \, b^{2} c^{2} {\left | b \right |} - 10 \, a b c d {\left | b \right |} + 3 \, a^{2} d^{2} {\left | b \right |}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{4 \, \sqrt {b d} d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*((3*(b*x + a)*(2*(b^5*c*d^6*abs(b) - a*b^4*d^7*abs(b))*(b*x + a)/(b^4*c*d^7 - a*b^3*d^8) - (7*b^6*c^2*d^5
*abs(b) - 10*a*b^5*c*d^6*abs(b) + 3*a^2*b^4*d^7*abs(b))/(b^4*c*d^7 - a*b^3*d^8)) - 20*(7*b^7*c^3*d^4*abs(b) -
17*a*b^6*c^2*d^5*abs(b) + 13*a^2*b^5*c*d^6*abs(b) - 3*a^3*b^4*d^7*abs(b))/(b^4*c*d^7 - a*b^3*d^8))*(b*x + a) -
15*(7*b^8*c^4*d^3*abs(b) - 24*a*b^7*c^3*d^4*abs(b) + 30*a^2*b^6*c^2*d^5*abs(b) - 16*a^3*b^5*c*d^6*abs(b) + 3*
a^4*b^4*d^7*abs(b))/(b^4*c*d^7 - a*b^3*d^8))*sqrt(b*x + a)/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) - 5/4*(7*b^2*
c^2*abs(b) - 10*a*b*c*d*abs(b) + 3*a^2*d^2*abs(b))*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b
*d - a*b*d)))/(sqrt(b*d)*d^4)

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maple [B]  time = 0.02, size = 750, normalized size = 3.38 \[ \frac {\sqrt {b x +a}\, \left (45 a^{2} b \,d^{4} x^{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 )-150 a \,b^{2} c \,d^{3} x^{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 )+105 b^{3} c^{2} d^{2} x^{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 )+90 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 )-300 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 )+210 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 )+45 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 )-150 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 )+12 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{2} d^{3} x^{3}+54 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a b \,d^{3} x^{2}-42 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{2} c \,d^{2} x^{2}-48 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} d^{3} x +316 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a b c \,d^{2} x -280 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, b^{2} c^{2} d x -32 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, a^{2} c \,d^{2}+230 \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 )}{24 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \left (d x +c \right )^{\frac {3}{2}} d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(b*x+a)^(1/2)*(45*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^2*a^2*b*d
^4-150*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^2*a*b^2*c*d^3+105*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^2*b^3*c^2*d^2+12*((b*x+a)*(d*x+c))^(1/2
)*(b*d)^(1/2)*b^2*d^3*x^3+90*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))-300*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))+210*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))+54*((b*x+a)*(d*x+c))^(1
/2)*(b*d)^(1/2)*a*b*d^3*x^2-42*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*b^2*c*d^2*x^2+45*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))-150*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))-48*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a^2*d^3*x+316*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/
2)*a*b*c*d^2*x-280*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*b^2*c^2*d*x-32*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*a^2*
c*d^2+230*((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)^(3/2)/d^4

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x+a)^(5/2)/(d*x+c)^(5/2),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 \[ \int \frac {x\,{\left (a+b\,x\right )}^{5/2}}{{\left (c+d\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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