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

Optimal. Leaf size=193 \[ -\frac {(b d-a e)^2 (5 a B e-6 A b e+b B d) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{8 b^{7/2} e^{3/2}}-\frac {\sqrt {a+b x} \sqrt {d+e x} (b d-a e) (5 a B e-6 A b e+b B d)}{8 b^3 e}-\frac {\sqrt {a+b x} (d+e x)^{3/2} (5 a B e-6 A b e+b B d)}{12 b^2 e}+\frac {B \sqrt {a+b x} (d+e x)^{5/2}}{3 b e} \]

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Rubi [A]  time = 0.15, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {80, 50, 63, 217, 206} \begin {gather*} -\frac {(b d-a e)^2 (5 a B e-6 A b e+b B d) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{8 b^{7/2} e^{3/2}}-\frac {\sqrt {a+b x} (d+e x)^{3/2} (5 a B e-6 A b e+b B d)}{12 b^2 e}-\frac {\sqrt {a+b x} \sqrt {d+e x} (b d-a e) (5 a B e-6 A b e+b B d)}{8 b^3 e}+\frac {B \sqrt {a+b x} (d+e x)^{5/2}}{3 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*d - a*e)*(b*B*d - 6*A*b*e + 5*a*B*e)*Sqrt[a + b*x]*Sqrt[d + e*x])/(8*b^3*e) - ((b*B*d - 6*A*b*e + 5*a*B*e
)*Sqrt[a + b*x]*(d + e*x)^(3/2))/(12*b^2*e) + (B*Sqrt[a + b*x]*(d + e*x)^(5/2))/(3*b*e) - ((b*d - a*e)^2*(b*B*
d - 6*A*b*e + 5*a*B*e)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(8*b^(7/2)*e^(3/2))

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 80

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

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Mathematica [A]  time = 0.70, size = 182, normalized size = 0.94 \begin {gather*} \frac {\sqrt {d+e x} \left (\sqrt {e} \sqrt {a+b x} \left (15 a^2 B e^2-2 a b e (9 A e+11 B d+5 B e x)+b^2 \left (6 A e (5 d+2 e x)+B \left (3 d^2+14 d e x+8 e^2 x^2\right )\right )\right )-\frac {3 (b d-a e)^{3/2} (5 a B e-6 A b e+b B d) \sinh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b d-a e}}\right )}{\sqrt {\frac {b (d+e x)}{b d-a e}}}\right )}{24 b^3 e^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x]*(Sqrt[e]*Sqrt[a + b*x]*(15*a^2*B*e^2 - 2*a*b*e*(11*B*d + 9*A*e + 5*B*e*x) + b^2*(6*A*e*(5*d + 2
*e*x) + B*(3*d^2 + 14*d*e*x + 8*e^2*x^2))) - (3*(b*d - a*e)^(3/2)*(b*B*d - 6*A*b*e + 5*a*B*e)*ArcSinh[(Sqrt[e]
*Sqrt[a + b*x])/Sqrt[b*d - a*e]])/Sqrt[(b*(d + e*x))/(b*d - a*e)]))/(24*b^3*e^(3/2))

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IntegrateAlgebraic [A]  time = 0.41, size = 275, normalized size = 1.42 \begin {gather*} \frac {\sqrt {a+b x} (b d-a e)^2 \left (-\frac {48 A b^2 e^2 (a+b x)}{d+e x}+\frac {18 A b e^3 (a+b x)^2}{(d+e x)^2}+\frac {8 b^2 B d e (a+b x)}{d+e x}-33 a b^2 B e-\frac {15 a B e^3 (a+b x)^2}{(d+e x)^2}+\frac {40 a b B e^2 (a+b x)}{d+e x}-\frac {3 b B d e^2 (a+b x)^2}{(d+e x)^2}+30 A b^3 e+3 b^3 B d\right )}{24 b^3 e \sqrt {d+e x} \left (b-\frac {e (a+b x)}{d+e x}\right )^3}-\frac {(b d-a e)^2 (5 a B e-6 A b e+b B d) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{8 b^{7/2} e^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*d - a*e)^2*Sqrt[a + b*x]*(3*b^3*B*d + 30*A*b^3*e - 33*a*b^2*B*e - (3*b*B*d*e^2*(a + b*x)^2)/(d + e*x)^2 +
(18*A*b*e^3*(a + b*x)^2)/(d + e*x)^2 - (15*a*B*e^3*(a + b*x)^2)/(d + e*x)^2 + (8*b^2*B*d*e*(a + b*x))/(d + e*x
) - (48*A*b^2*e^2*(a + b*x))/(d + e*x) + (40*a*b*B*e^2*(a + b*x))/(d + e*x)))/(24*b^3*e*Sqrt[d + e*x]*(b - (e*
(a + b*x))/(d + e*x))^3) - ((b*d - a*e)^2*(b*B*d - 6*A*b*e + 5*a*B*e)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]
*Sqrt[d + e*x])])/(8*b^(7/2)*e^(3/2))

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fricas [A]  time = 1.81, size = 542, normalized size = 2.81 \begin {gather*} \left [-\frac {3 \, {\left (B b^{3} d^{3} + 3 \, {\left (B a b^{2} - 2 \, A b^{3}\right )} d^{2} e - 3 \, {\left (3 \, B a^{2} b - 4 \, A a b^{2}\right )} d e^{2} + {\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} e^{3}\right )} \sqrt {b e} \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 e x + b d + a e\right )} \sqrt {b e} \sqrt {b x + a} \sqrt {e x + d} + 8 \, {\left (b^{2} d e + a b e^{2}\right )} x\right ) - 4 \, {\left (8 \, B b^{3} e^{3} x^{2} + 3 \, B b^{3} d^{2} e - 2 \, {\left (11 \, B a b^{2} - 15 \, A b^{3}\right )} d e^{2} + 3 \, {\left (5 \, B a^{2} b - 6 \, A a b^{2}\right )} e^{3} + 2 \, {\left (7 \, B b^{3} d e^{2} - {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} e^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{96 \, b^{4} e^{2}}, \frac {3 \, {\left (B b^{3} d^{3} + 3 \, {\left (B a b^{2} - 2 \, A b^{3}\right )} d^{2} e - 3 \, {\left (3 \, B a^{2} b - 4 \, A a b^{2}\right )} d e^{2} + {\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} e^{3}\right )} \sqrt {-b e} \arctan \left (\frac {{\left (2 \, b e x + b d + a e\right )} \sqrt {-b e} \sqrt {b x + a} \sqrt {e x + d}}{2 \, {\left (b^{2} e^{2} x^{2} + a b d e + {\left (b^{2} d e + a b e^{2}\right )} x\right )}}\right ) + 2 \, {\left (8 \, B b^{3} e^{3} x^{2} + 3 \, B b^{3} d^{2} e - 2 \, {\left (11 \, B a b^{2} - 15 \, A b^{3}\right )} d e^{2} + 3 \, {\left (5 \, B a^{2} b - 6 \, A a b^{2}\right )} e^{3} + 2 \, {\left (7 \, B b^{3} d e^{2} - {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} e^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{48 \, b^{4} e^{2}}\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)^(1/2),x, algorithm="fricas")

[Out]

[-1/96*(3*(B*b^3*d^3 + 3*(B*a*b^2 - 2*A*b^3)*d^2*e - 3*(3*B*a^2*b - 4*A*a*b^2)*d*e^2 + (5*B*a^3 - 6*A*a^2*b)*e
^3)*sqrt(b*e)*log(8*b^2*e^2*x^2 + b^2*d^2 + 6*a*b*d*e + a^2*e^2 + 4*(2*b*e*x + b*d + a*e)*sqrt(b*e)*sqrt(b*x +
a)*sqrt(e*x + d) + 8*(b^2*d*e + a*b*e^2)*x) - 4*(8*B*b^3*e^3*x^2 + 3*B*b^3*d^2*e - 2*(11*B*a*b^2 - 15*A*b^3)*
d*e^2 + 3*(5*B*a^2*b - 6*A*a*b^2)*e^3 + 2*(7*B*b^3*d*e^2 - (5*B*a*b^2 - 6*A*b^3)*e^3)*x)*sqrt(b*x + a)*sqrt(e*
x + d))/(b^4*e^2), 1/48*(3*(B*b^3*d^3 + 3*(B*a*b^2 - 2*A*b^3)*d^2*e - 3*(3*B*a^2*b - 4*A*a*b^2)*d*e^2 + (5*B*a
^3 - 6*A*a^2*b)*e^3)*sqrt(-b*e)*arctan(1/2*(2*b*e*x + b*d + a*e)*sqrt(-b*e)*sqrt(b*x + a)*sqrt(e*x + d)/(b^2*e
^2*x^2 + a*b*d*e + (b^2*d*e + a*b*e^2)*x)) + 2*(8*B*b^3*e^3*x^2 + 3*B*b^3*d^2*e - 2*(11*B*a*b^2 - 15*A*b^3)*d*
e^2 + 3*(5*B*a^2*b - 6*A*a*b^2)*e^3 + 2*(7*B*b^3*d*e^2 - (5*B*a*b^2 - 6*A*b^3)*e^3)*x)*sqrt(b*x + a)*sqrt(e*x
+ d))/(b^4*e^2)]

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giac [B]  time = 1.54, size = 575, normalized size = 2.98 \begin {gather*} -\frac {\frac {24 \, {\left (\frac {{\left (b^{2} d - a b e\right )} e^{\left (-\frac {1}{2}\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt {b}} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \sqrt {b x + a}\right )} A d {\left | b \right |}}{b^{2}} - \frac {{\left (\sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )}}{b^{2}} + \frac {{\left (b^{6} d e^{3} - 13 \, a b^{5} e^{4}\right )} e^{\left (-4\right )}}{b^{7}}\right )} - \frac {3 \, {\left (b^{7} d^{2} e^{2} + 2 \, a b^{6} d e^{3} - 11 \, a^{2} b^{5} e^{4}\right )} e^{\left (-4\right )}}{b^{7}}\right )} - \frac {3 \, {\left (b^{3} d^{3} + a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - 5 \, a^{3} e^{3}\right )} e^{\left (-\frac {5}{2}\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{b^{\frac {3}{2}}}\right )} B {\left | b \right |} e}{b^{2}} - \frac {6 \, {\left (\frac {{\left (b^{3} d^{2} + 2 \, a b^{2} d e - 3 \, a^{2} b e^{2}\right )} e^{\left (-\frac {3}{2}\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt {b}} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} {\left (2 \, b x + {\left (b d e - 5 \, a e^{2}\right )} e^{\left (-2\right )} + 2 \, a\right )} \sqrt {b x + a}\right )} B d {\left | b \right |}}{b^{3}} - \frac {6 \, {\left (\frac {{\left (b^{3} d^{2} + 2 \, a b^{2} d e - 3 \, a^{2} b e^{2}\right )} e^{\left (-\frac {3}{2}\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt {b}} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} {\left (2 \, b x + {\left (b d e - 5 \, a e^{2}\right )} e^{\left (-2\right )} + 2 \, a\right )} \sqrt {b x + a}\right )} A {\left | b \right |} e}{b^{3}}}{24 \, b} \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)^(1/2),x, algorithm="giac")

[Out]

-1/24*(24*((b^2*d - a*b*e)*e^(-1/2)*log(abs(-sqrt(b*x + a)*sqrt(b)*e^(1/2) + sqrt(b^2*d + (b*x + a)*b*e - a*b*
e)))/sqrt(b) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*sqrt(b*x + a))*A*d*abs(b)/b^2 - (sqrt(b^2*d + (b*x + a)*b*e
- a*b*e)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)/b^2 + (b^6*d*e^3 - 13*a*b^5*e^4)*e^(-4)/b^7) - 3*(b^7*d^2*e^
2 + 2*a*b^6*d*e^3 - 11*a^2*b^5*e^4)*e^(-4)/b^7) - 3*(b^3*d^3 + a*b^2*d^2*e + 3*a^2*b*d*e^2 - 5*a^3*e^3)*e^(-5/
2)*log(abs(-sqrt(b*x + a)*sqrt(b)*e^(1/2) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/b^(3/2))*B*abs(b)*e/b^2 - 6*
((b^3*d^2 + 2*a*b^2*d*e - 3*a^2*b*e^2)*e^(-3/2)*log(abs(-sqrt(b*x + a)*sqrt(b)*e^(1/2) + sqrt(b^2*d + (b*x + a
)*b*e - a*b*e)))/sqrt(b) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*(2*b*x + (b*d*e - 5*a*e^2)*e^(-2) + 2*a)*sqrt(b
*x + a))*B*d*abs(b)/b^3 - 6*((b^3*d^2 + 2*a*b^2*d*e - 3*a^2*b*e^2)*e^(-3/2)*log(abs(-sqrt(b*x + a)*sqrt(b)*e^(
1/2) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/sqrt(b) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*(2*b*x + (b*d*e - 5
*a*e^2)*e^(-2) + 2*a)*sqrt(b*x + a))*A*abs(b)*e/b^3)/b

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maple [B]  time = 0.02, size = 636, normalized size = 3.30 \begin {gather*} \frac {\sqrt {e x +d}\, \sqrt {b x +a}\, \left (18 A \,a^{2} b \,e^{3} \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 )-36 A a \,b^{2} d \,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 )+18 A \,b^{3} d^{2} 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 )-15 B \,a^{3} e^{3} \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 )+27 B \,a^{2} b d \,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 \,b^{2} d^{2} 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 )-3 B \,b^{3} d^{3} \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 )+16 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, B \,b^{2} e^{2} x^{2}+24 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, A \,b^{2} e^{2} x -20 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B a b \,e^{2} x +28 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B \,b^{2} d e x -36 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, A a b \,e^{2}+60 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, A \,b^{2} d e +30 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, B \,a^{2} e^{2}-44 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, B a b d e +6 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, B \,b^{2} d^{2}\right )}{48 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, b^{3} e} \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)^(1/2),x)

[Out]

1/48*(e*x+d)^(1/2)*(b*x+a)^(1/2)*(16*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*B*b^2*e^2*x^2+18*A*a^2*b*e^3*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))-36*A*a*b^2*d*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))+18*A*b^3*d^2*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))+24*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*A*b^2*e^2*x-15*B*a^3*e^3*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))+27*B*a^2*b*d*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*b^2*d^2*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))-3*B*b^3*d^3*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))-20*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*B*a*b*e^2*x+28*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*B*b^2*d*e*x-36*
(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*A*a*b*e^2+60*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*A*b^2*d*e+30*((b*x+a)*(e*
x+d))^(1/2)*(b*e)^(1/2)*B*a^2*e^2-44*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*B*a*b*d*e+6*(b*e)^(1/2)*((b*x+a)*(e*x
+d))^(1/2)*B*b^2*d^2)/b^3/e/((b*x+a)*(e*x+d))^(1/2)/(b*e)^(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((B*x+A)*(e*x+d)^(3/2)/(b*x+a)^(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*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 = -0.01 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^{3/2}}{\sqrt {a+b\,x}} \,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)^(1/2),x)

[Out]

int(((A + B*x)*(d + e*x)^(3/2))/(a + b*x)^(1/2), 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)**(1/2),x)

[Out]

Timed out

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