3.22.16 \(\int \frac {(A+B x) (d+e x)^{3/2}}{(a+b x)^{5/2}} \, dx\)

Optimal. Leaf size=201 \[ \frac {\sqrt {e} (-5 a B e+2 A b e+3 b B d) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{b^{7/2}}+\frac {e \sqrt {a+b x} \sqrt {d+e x} (-5 a B e+2 A b e+3 b B d)}{b^3 (b d-a e)}-\frac {2 (d+e x)^{3/2} (-5 a B e+2 A b e+3 b B d)}{3 b^2 \sqrt {a+b x} (b d-a e)}-\frac {2 (d+e x)^{5/2} (A b-a B)}{3 b (a+b x)^{3/2} (b d-a e)} \]

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Rubi [A]  time = 0.15, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {78, 47, 50, 63, 217, 206} \begin {gather*} -\frac {2 (d+e x)^{3/2} (-5 a B e+2 A b e+3 b B d)}{3 b^2 \sqrt {a+b x} (b d-a e)}+\frac {e \sqrt {a+b x} \sqrt {d+e x} (-5 a B e+2 A b e+3 b B d)}{b^3 (b d-a e)}+\frac {\sqrt {e} (-5 a B e+2 A b e+3 b B d) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{b^{7/2}}-\frac {2 (d+e x)^{5/2} (A b-a B)}{3 b (a+b x)^{3/2} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(3*b*B*d + 2*A*b*e - 5*a*B*e)*Sqrt[a + b*x]*Sqrt[d + e*x])/(b^3*(b*d - a*e)) - (2*(3*b*B*d + 2*A*b*e - 5*a*
B*e)*(d + e*x)^(3/2))/(3*b^2*(b*d - a*e)*Sqrt[a + b*x]) - (2*(A*b - a*B)*(d + e*x)^(5/2))/(3*b*(b*d - a*e)*(a
+ b*x)^(3/2)) + (Sqrt[e]*(3*b*B*d + 2*A*b*e - 5*a*B*e)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])
])/b^(7/2)

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

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Mathematica [C]  time = 0.14, size = 127, normalized size = 0.63 \begin {gather*} \frac {2 \sqrt {d+e x} \left (-\frac {b^2 (d+e x)^2 (A b-a B)}{b d-a e}-\frac {(a+b x) (-5 a B e+2 A b e+3 b B d) \, _2F_1\left (-\frac {3}{2},-\frac {1}{2};\frac {1}{2};\frac {e (a+b x)}{a e-b d}\right )}{\sqrt {\frac {b (d+e x)}{b d-a e}}}\right )}{3 b^3 (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d + e*x]*(-((b^2*(A*b - a*B)*(d + e*x)^2)/(b*d - a*e)) - ((3*b*B*d + 2*A*b*e - 5*a*B*e)*(a + b*x)*Hype
rgeometric2F1[-3/2, -1/2, 1/2, (e*(a + b*x))/(-(b*d) + a*e)])/Sqrt[(b*(d + e*x))/(b*d - a*e)]))/(3*b^3*(a + b*
x)^(3/2))

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IntegrateAlgebraic [A]  time = 0.31, size = 243, normalized size = 1.21 \begin {gather*} \frac {\left (-5 a B e^{3/2}+2 A b e^{3/2}+3 b B d \sqrt {e}\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{b^{7/2}}+\frac {(d+e x)^{3/2} \left (\frac {4 A b^2 e (a+b x)}{d+e x}-\frac {6 A b e^2 (a+b x)^2}{(d+e x)^2}+\frac {6 b^2 B d (a+b x)}{d+e x}-2 a b^2 B+\frac {15 a B e^2 (a+b x)^2}{(d+e x)^2}-\frac {10 a b B e (a+b x)}{d+e x}-\frac {9 b B d e (a+b x)^2}{(d+e x)^2}+2 A b^3\right )}{3 b^3 (a+b x)^{3/2} \left (\frac {e (a+b x)}{d+e x}-b\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)^(3/2)*(2*A*b^3 - 2*a*b^2*B - (9*b*B*d*e*(a + b*x)^2)/(d + e*x)^2 - (6*A*b*e^2*(a + b*x)^2)/(d + e*x
)^2 + (15*a*B*e^2*(a + b*x)^2)/(d + e*x)^2 + (6*b^2*B*d*(a + b*x))/(d + e*x) + (4*A*b^2*e*(a + b*x))/(d + e*x)
- (10*a*b*B*e*(a + b*x))/(d + e*x)))/(3*b^3*(a + b*x)^(3/2)*(-b + (e*(a + b*x))/(d + e*x))) + ((3*b*B*d*Sqrt[
e] + 2*A*b*e^(3/2) - 5*a*B*e^(3/2))*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/b^(7/2)

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fricas [A]  time = 6.17, size = 561, normalized size = 2.79 \begin {gather*} \left [\frac {3 \, {\left (3 \, B a^{2} b d + {\left (3 \, B b^{3} d - {\left (5 \, B a b^{2} - 2 \, A b^{3}\right )} e\right )} x^{2} - {\left (5 \, B a^{3} - 2 \, A a^{2} b\right )} e + 2 \, {\left (3 \, B a b^{2} d - {\left (5 \, B a^{2} b - 2 \, A a b^{2}\right )} e\right )} x\right )} \sqrt {\frac {e}{b}} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \, {\left (2 \, b^{2} e x + b^{2} d + a b e\right )} \sqrt {b x + a} \sqrt {e x + d} \sqrt {\frac {e}{b}} + 8 \, {\left (b^{2} d e + a b e^{2}\right )} x\right ) + 4 \, {\left (3 \, B b^{2} e x^{2} - 2 \, {\left (2 \, B a b + A b^{2}\right )} d + 3 \, {\left (5 \, B a^{2} - 2 \, A a b\right )} e - 2 \, {\left (3 \, B b^{2} d - 2 \, {\left (5 \, B a b - 2 \, A b^{2}\right )} e\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{12 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac {3 \, {\left (3 \, B a^{2} b d + {\left (3 \, B b^{3} d - {\left (5 \, B a b^{2} - 2 \, A b^{3}\right )} e\right )} x^{2} - {\left (5 \, B a^{3} - 2 \, A a^{2} b\right )} e + 2 \, {\left (3 \, B a b^{2} d - {\left (5 \, B a^{2} b - 2 \, A a b^{2}\right )} e\right )} x\right )} \sqrt {-\frac {e}{b}} \arctan \left (\frac {{\left (2 \, b e x + b d + a e\right )} \sqrt {b x + a} \sqrt {e x + d} \sqrt {-\frac {e}{b}}}{2 \, {\left (b e^{2} x^{2} + a d e + {\left (b d e + a e^{2}\right )} x\right )}}\right ) - 2 \, {\left (3 \, B b^{2} e x^{2} - 2 \, {\left (2 \, B a b + A b^{2}\right )} d + 3 \, {\left (5 \, B a^{2} - 2 \, A a b\right )} e - 2 \, {\left (3 \, B b^{2} d - 2 \, {\left (5 \, B a b - 2 \, A b^{2}\right )} e\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{6 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(3*(3*B*a^2*b*d + (3*B*b^3*d - (5*B*a*b^2 - 2*A*b^3)*e)*x^2 - (5*B*a^3 - 2*A*a^2*b)*e + 2*(3*B*a*b^2*d -
(5*B*a^2*b - 2*A*a*b^2)*e)*x)*sqrt(e/b)*log(8*b^2*e^2*x^2 + b^2*d^2 + 6*a*b*d*e + a^2*e^2 + 4*(2*b^2*e*x + b^
2*d + a*b*e)*sqrt(b*x + a)*sqrt(e*x + d)*sqrt(e/b) + 8*(b^2*d*e + a*b*e^2)*x) + 4*(3*B*b^2*e*x^2 - 2*(2*B*a*b
+ A*b^2)*d + 3*(5*B*a^2 - 2*A*a*b)*e - 2*(3*B*b^2*d - 2*(5*B*a*b - 2*A*b^2)*e)*x)*sqrt(b*x + a)*sqrt(e*x + d))
/(b^5*x^2 + 2*a*b^4*x + a^2*b^3), -1/6*(3*(3*B*a^2*b*d + (3*B*b^3*d - (5*B*a*b^2 - 2*A*b^3)*e)*x^2 - (5*B*a^3
- 2*A*a^2*b)*e + 2*(3*B*a*b^2*d - (5*B*a^2*b - 2*A*a*b^2)*e)*x)*sqrt(-e/b)*arctan(1/2*(2*b*e*x + b*d + a*e)*sq
rt(b*x + a)*sqrt(e*x + d)*sqrt(-e/b)/(b*e^2*x^2 + a*d*e + (b*d*e + a*e^2)*x)) - 2*(3*B*b^2*e*x^2 - 2*(2*B*a*b
+ A*b^2)*d + 3*(5*B*a^2 - 2*A*a*b)*e - 2*(3*B*b^2*d - 2*(5*B*a*b - 2*A*b^2)*e)*x)*sqrt(b*x + a)*sqrt(e*x + d))
/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)]

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giac [B]  time = 3.07, size = 951, normalized size = 4.73

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*sqrt(b*x + a)*B*abs(b)*e/b^5 - 1/2*(3*B*b^(3/2)*d*abs(b)*e^(1/2) - 5*B*a*s
qrt(b)*abs(b)*e^(3/2) + 2*A*b^(3/2)*abs(b)*e^(3/2))*log((sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a
)*b*e - a*b*e))^2)/b^5 - 4/3*(3*B*b^(13/2)*d^4*abs(b)*e^(1/2) - 16*B*a*b^(11/2)*d^3*abs(b)*e^(3/2) + 4*A*b^(13
/2)*d^3*abs(b)*e^(3/2) - 6*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2*B*b^(9/2)*d
^3*abs(b)*e^(1/2) + 30*B*a^2*b^(9/2)*d^2*abs(b)*e^(5/2) - 12*A*a*b^(11/2)*d^2*abs(b)*e^(5/2) + 24*(sqrt(b*x +
a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2*B*a*b^(7/2)*d^2*abs(b)*e^(3/2) - 6*(sqrt(b*x + a)*
sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2*A*b^(9/2)*d^2*abs(b)*e^(3/2) + 3*(sqrt(b*x + a)*sqrt(
b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^4*B*b^(5/2)*d^2*abs(b)*e^(1/2) - 24*B*a^3*b^(7/2)*d*abs(b)*e
^(7/2) + 12*A*a^2*b^(9/2)*d*abs(b)*e^(7/2) - 30*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e -
a*b*e))^2*B*a^2*b^(5/2)*d*abs(b)*e^(5/2) + 12*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*
b*e))^2*A*a*b^(7/2)*d*abs(b)*e^(5/2) - 12*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e)
)^4*B*a*b^(3/2)*d*abs(b)*e^(3/2) + 6*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^4*A
*b^(5/2)*d*abs(b)*e^(3/2) + 7*B*a^4*b^(5/2)*abs(b)*e^(9/2) - 4*A*a^3*b^(7/2)*abs(b)*e^(9/2) + 12*(sqrt(b*x + a
)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2*B*a^3*b^(3/2)*abs(b)*e^(7/2) - 6*(sqrt(b*x + a)*sqr
t(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2*A*a^2*b^(5/2)*abs(b)*e^(7/2) + 9*(sqrt(b*x + a)*sqrt(b)*
e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^4*B*a^2*sqrt(b)*abs(b)*e^(5/2) - 6*(sqrt(b*x + a)*sqrt(b)*e^(1/
2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^4*A*a*b^(3/2)*abs(b)*e^(5/2))/((b^2*d - a*b*e - (sqrt(b*x + a)*sqrt(
b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2)^3*b^4)

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maple [B]  time = 0.02, size = 698, normalized size = 3.47 \begin {gather*} \frac {\sqrt {e x +d}\, \left (6 A \,b^{3} e^{2} x^{2} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-15 B a \,b^{2} e^{2} x^{2} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+9 B \,b^{3} d e \,x^{2} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+12 A a \,b^{2} e^{2} x \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-30 B \,a^{2} b \,e^{2} x \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+18 B a \,b^{2} d e x \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+6 A \,a^{2} b \,e^{2} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )-15 B \,a^{3} e^{2} \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+9 B \,a^{2} b d e \ln \left (\frac {2 b e x +a e +b d +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}}{2 \sqrt {b e}}\right )+6 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B \,b^{2} e \,x^{2}-16 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, A \,b^{2} e x +40 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B a b e x -12 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B \,b^{2} d x -12 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, A a b e -4 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, A \,b^{2} d +30 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B \,a^{2} e -8 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B a b d \right )}{6 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \left (b x +a \right )^{\frac {3}{2}} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^(3/2)/(b*x+a)^(5/2),x)

[Out]

1/6*(e*x+d)^(1/2)*(6*A*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*x^2*b^3*e^2
-15*B*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*x^2*a*b^2*e^2+9*B*ln(1/2*(2*
b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))*x^2*b^3*d*e+12*A*a*b^2*e^2*x*ln(1/2*(2*b*e*x
+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))-30*B*a^2*b*e^2*x*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+
a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))+18*B*a*b^2*d*e*x*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)
*(b*e)^(1/2))/(b*e)^(1/2))+6*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*B*b^2*e*x^2+6*A*a^2*b*e^2*ln(1/2*(2*b*e*x+a*e
+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))-16*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*A*b^2*e*x-15*B
*a^3*e^2*ln(1/2*(2*b*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))+9*B*a^2*b*d*e*ln(1/2*(2*b
*e*x+a*e+b*d+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/(b*e)^(1/2))+40*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*B*a*b*
e*x-12*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*B*b^2*d*x-12*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*A*a*b*e-4*((b*x+a)
*(e*x+d))^(1/2)*(b*e)^(1/2)*A*b^2*d+30*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*B*a^2*e-8*((b*x+a)*(e*x+d))^(1/2)*(
b*e)^(1/2)*B*a*b*d)/(b*e)^(1/2)/((b*x+a)*(e*x+d))^(1/2)/b^3/(b*x+a)^(3/2)

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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)*(e*x+d)^(3/2)/(b*x+a)^(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*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^{3/2}}{{\left (a+b\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(d + e*x)^(3/2))/(a + b*x)^(5/2),x)

[Out]

int(((A + B*x)*(d + e*x)^(3/2))/(a + b*x)^(5/2), x)

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sympy [F(-1)]  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((B*x+A)*(e*x+d)**(3/2)/(b*x+a)**(5/2),x)

[Out]

Timed out

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