3.1743 \(\int \frac {(A+B x) (d+e x)^{5/2}}{a+b x} \, dx\)

Optimal. Leaf size=164 \[ -\frac {2 (A b-a B) (b d-a e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{9/2}}+\frac {2 \sqrt {d+e x} (A b-a B) (b d-a e)^2}{b^4}+\frac {2 (d+e x)^{3/2} (A b-a B) (b d-a e)}{3 b^3}+\frac {2 (d+e x)^{5/2} (A b-a B)}{5 b^2}+\frac {2 B (d+e x)^{7/2}}{7 b e} \]

[Out]

2/3*(A*b-B*a)*(-a*e+b*d)*(e*x+d)^(3/2)/b^3+2/5*(A*b-B*a)*(e*x+d)^(5/2)/b^2+2/7*B*(e*x+d)^(7/2)/b/e-2*(A*b-B*a)
*(-a*e+b*d)^(5/2)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(9/2)+2*(A*b-B*a)*(-a*e+b*d)^2*(e*x+d)^(1/
2)/b^4

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Rubi [A]  time = 0.10, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {80, 50, 63, 208} \[ \frac {2 (d+e x)^{5/2} (A b-a B)}{5 b^2}+\frac {2 (d+e x)^{3/2} (A b-a B) (b d-a e)}{3 b^3}+\frac {2 \sqrt {d+e x} (A b-a B) (b d-a e)^2}{b^4}-\frac {2 (A b-a B) (b d-a e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{9/2}}+\frac {2 B (d+e x)^{7/2}}{7 b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A*b - a*B)*(b*d - a*e)^2*Sqrt[d + e*x])/b^4 + (2*(A*b - a*B)*(b*d - a*e)*(d + e*x)^(3/2))/(3*b^3) + (2*(A*
b - a*B)*(d + e*x)^(5/2))/(5*b^2) + (2*B*(d + e*x)^(7/2))/(7*b*e) - (2*(A*b - a*B)*(b*d - a*e)^(5/2)*ArcTanh[(
Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/b^(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 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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

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Mathematica [A]  time = 0.27, size = 136, normalized size = 0.83 \[ \frac {2 (A b-a B) \left (5 (b d-a e) \left (\sqrt {b} \sqrt {d+e x} (-3 a e+4 b d+b e x)-3 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )\right )+3 b^{5/2} (d+e x)^{5/2}\right )}{15 b^{9/2}}+\frac {2 B (d+e x)^{7/2}}{7 b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.05, size = 591, normalized size = 3.60 \[ \left [\frac {105 \, {\left ({\left (B a b^{2} - A b^{3}\right )} d^{2} e - 2 \, {\left (B a^{2} b - A a b^{2}\right )} d e^{2} + {\left (B a^{3} - A a^{2} b\right )} e^{3}\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) + 2 \, {\left (15 \, B b^{3} e^{3} x^{3} + 15 \, B b^{3} d^{3} - 161 \, {\left (B a b^{2} - A b^{3}\right )} d^{2} e + 245 \, {\left (B a^{2} b - A a b^{2}\right )} d e^{2} - 105 \, {\left (B a^{3} - A a^{2} b\right )} e^{3} + 3 \, {\left (15 \, B b^{3} d e^{2} - 7 \, {\left (B a b^{2} - A b^{3}\right )} e^{3}\right )} x^{2} + {\left (45 \, B b^{3} d^{2} e - 77 \, {\left (B a b^{2} - A b^{3}\right )} d e^{2} + 35 \, {\left (B a^{2} b - A a b^{2}\right )} e^{3}\right )} x\right )} \sqrt {e x + d}}{105 \, b^{4} e}, \frac {2 \, {\left (105 \, {\left ({\left (B a b^{2} - A b^{3}\right )} d^{2} e - 2 \, {\left (B a^{2} b - A a b^{2}\right )} d e^{2} + {\left (B a^{3} - A a^{2} b\right )} e^{3}\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) + {\left (15 \, B b^{3} e^{3} x^{3} + 15 \, B b^{3} d^{3} - 161 \, {\left (B a b^{2} - A b^{3}\right )} d^{2} e + 245 \, {\left (B a^{2} b - A a b^{2}\right )} d e^{2} - 105 \, {\left (B a^{3} - A a^{2} b\right )} e^{3} + 3 \, {\left (15 \, B b^{3} d e^{2} - 7 \, {\left (B a b^{2} - A b^{3}\right )} e^{3}\right )} x^{2} + {\left (45 \, B b^{3} d^{2} e - 77 \, {\left (B a b^{2} - A b^{3}\right )} d e^{2} + 35 \, {\left (B a^{2} b - A a b^{2}\right )} e^{3}\right )} x\right )} \sqrt {e x + d}\right )}}{105 \, b^{4} e}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/105*(105*((B*a*b^2 - A*b^3)*d^2*e - 2*(B*a^2*b - A*a*b^2)*d*e^2 + (B*a^3 - A*a^2*b)*e^3)*sqrt((b*d - a*e)/b
)*log((b*e*x + 2*b*d - a*e + 2*sqrt(e*x + d)*b*sqrt((b*d - a*e)/b))/(b*x + a)) + 2*(15*B*b^3*e^3*x^3 + 15*B*b^
3*d^3 - 161*(B*a*b^2 - A*b^3)*d^2*e + 245*(B*a^2*b - A*a*b^2)*d*e^2 - 105*(B*a^3 - A*a^2*b)*e^3 + 3*(15*B*b^3*
d*e^2 - 7*(B*a*b^2 - A*b^3)*e^3)*x^2 + (45*B*b^3*d^2*e - 77*(B*a*b^2 - A*b^3)*d*e^2 + 35*(B*a^2*b - A*a*b^2)*e
^3)*x)*sqrt(e*x + d))/(b^4*e), 2/105*(105*((B*a*b^2 - A*b^3)*d^2*e - 2*(B*a^2*b - A*a*b^2)*d*e^2 + (B*a^3 - A*
a^2*b)*e^3)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) + (15*B*b^3*e^3*x^3
+ 15*B*b^3*d^3 - 161*(B*a*b^2 - A*b^3)*d^2*e + 245*(B*a^2*b - A*a*b^2)*d*e^2 - 105*(B*a^3 - A*a^2*b)*e^3 + 3*
(15*B*b^3*d*e^2 - 7*(B*a*b^2 - A*b^3)*e^3)*x^2 + (45*B*b^3*d^2*e - 77*(B*a*b^2 - A*b^3)*d*e^2 + 35*(B*a^2*b -
A*a*b^2)*e^3)*x)*sqrt(e*x + d))/(b^4*e)]

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giac [B]  time = 1.30, size = 371, normalized size = 2.26 \[ -\frac {2 \, {\left (B a b^{3} d^{3} - A b^{4} d^{3} - 3 \, B a^{2} b^{2} d^{2} e + 3 \, A a b^{3} d^{2} e + 3 \, B a^{3} b d e^{2} - 3 \, A a^{2} b^{2} d e^{2} - B a^{4} e^{3} + A a^{3} b e^{3}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{4}} + \frac {2 \, {\left (15 \, {\left (x e + d\right )}^{\frac {7}{2}} B b^{6} e^{6} - 21 \, {\left (x e + d\right )}^{\frac {5}{2}} B a b^{5} e^{7} + 21 \, {\left (x e + d\right )}^{\frac {5}{2}} A b^{6} e^{7} - 35 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b^{5} d e^{7} + 35 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{6} d e^{7} - 105 \, \sqrt {x e + d} B a b^{5} d^{2} e^{7} + 105 \, \sqrt {x e + d} A b^{6} d^{2} e^{7} + 35 \, {\left (x e + d\right )}^{\frac {3}{2}} B a^{2} b^{4} e^{8} - 35 \, {\left (x e + d\right )}^{\frac {3}{2}} A a b^{5} e^{8} + 210 \, \sqrt {x e + d} B a^{2} b^{4} d e^{8} - 210 \, \sqrt {x e + d} A a b^{5} d e^{8} - 105 \, \sqrt {x e + d} B a^{3} b^{3} e^{9} + 105 \, \sqrt {x e + d} A a^{2} b^{4} e^{9}\right )} e^{\left (-7\right )}}{105 \, b^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(B*a*b^3*d^3 - A*b^4*d^3 - 3*B*a^2*b^2*d^2*e + 3*A*a*b^3*d^2*e + 3*B*a^3*b*d*e^2 - 3*A*a^2*b^2*d*e^2 - B*a^
4*e^3 + A*a^3*b*e^3)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^4) + 2/105*(15*(x*e
+ d)^(7/2)*B*b^6*e^6 - 21*(x*e + d)^(5/2)*B*a*b^5*e^7 + 21*(x*e + d)^(5/2)*A*b^6*e^7 - 35*(x*e + d)^(3/2)*B*a*
b^5*d*e^7 + 35*(x*e + d)^(3/2)*A*b^6*d*e^7 - 105*sqrt(x*e + d)*B*a*b^5*d^2*e^7 + 105*sqrt(x*e + d)*A*b^6*d^2*e
^7 + 35*(x*e + d)^(3/2)*B*a^2*b^4*e^8 - 35*(x*e + d)^(3/2)*A*a*b^5*e^8 + 210*sqrt(x*e + d)*B*a^2*b^4*d*e^8 - 2
10*sqrt(x*e + d)*A*a*b^5*d*e^8 - 105*sqrt(x*e + d)*B*a^3*b^3*e^9 + 105*sqrt(x*e + d)*A*a^2*b^4*e^9)*e^(-7)/b^7

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maple [B]  time = 0.01, size = 573, normalized size = 3.49 \[ -\frac {2 A \,a^{3} e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{3}}+\frac {6 A \,a^{2} d \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{2}}-\frac {6 A a \,d^{2} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b}+\frac {2 A \,d^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}}+\frac {2 B \,a^{4} e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{4}}-\frac {6 B \,a^{3} d \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{3}}+\frac {6 B \,a^{2} d^{2} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{2}}-\frac {2 B a \,d^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b}+\frac {2 \sqrt {e x +d}\, A \,a^{2} e^{2}}{b^{3}}-\frac {4 \sqrt {e x +d}\, A a d e}{b^{2}}+\frac {2 \sqrt {e x +d}\, A \,d^{2}}{b}-\frac {2 \sqrt {e x +d}\, B \,a^{3} e^{2}}{b^{4}}+\frac {4 \sqrt {e x +d}\, B \,a^{2} d e}{b^{3}}-\frac {2 \sqrt {e x +d}\, B a \,d^{2}}{b^{2}}-\frac {2 \left (e x +d \right )^{\frac {3}{2}} A a e}{3 b^{2}}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} A d}{3 b}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} B \,a^{2} e}{3 b^{3}}-\frac {2 \left (e x +d \right )^{\frac {3}{2}} B a d}{3 b^{2}}+\frac {2 \left (e x +d \right )^{\frac {5}{2}} A}{5 b}-\frac {2 \left (e x +d \right )^{\frac {5}{2}} B a}{5 b^{2}}+\frac {2 \left (e x +d \right )^{\frac {7}{2}} B}{7 b e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*B*(e*x+d)^(7/2)/b/e+2/5/b*A*(e*x+d)^(5/2)-2/5/b^2*B*(e*x+d)^(5/2)*a-2/3*e/b^2*A*(e*x+d)^(3/2)*a+2/3/b*A*(e
*x+d)^(3/2)*d+2/3*e/b^3*B*(e*x+d)^(3/2)*a^2-2/3/b^2*B*(e*x+d)^(3/2)*a*d+2*e^2/b^3*A*(e*x+d)^(1/2)*a^2-4*e/b^2*
A*(e*x+d)^(1/2)*a*d+2/b*A*(e*x+d)^(1/2)*d^2-2*e^2/b^4*B*(e*x+d)^(1/2)*a^3+4*e/b^3*B*(e*x+d)^(1/2)*a^2*d-2/b^2*
B*(e*x+d)^(1/2)*a*d^2-2*e^3/b^3/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*A*a^3+6*e^2/b^
2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*A*a^2*d-6*e/b/((a*e-b*d)*b)^(1/2)*arctan((e*
x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*A*a*d^2+2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*A*
d^3+2*e^3/b^4/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*B*a^4-6*e^2/b^3/((a*e-b*d)*b)^(1
/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*B*a^3*d+6*e/b^2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e
-b*d)*b)^(1/2)*b)*B*a^2*d^2-2/b/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*B*a*d^3

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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((B*x+A)*(e*x+d)^(5/2)/(b*x+a),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 positive or negative?

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mupad [B]  time = 1.26, size = 330, normalized size = 2.01 \[ \left (\frac {2\,A\,e-2\,B\,d}{5\,b\,e}-\frac {2\,B\,\left (a\,e^2-b\,d\,e\right )}{5\,b^2\,e^2}\right )\,{\left (d+e\,x\right )}^{5/2}+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\left (A\,b-B\,a\right )\,{\left (a\,e-b\,d\right )}^{5/2}\,\sqrt {d+e\,x}}{B\,a^4\,e^3-3\,B\,a^3\,b\,d\,e^2-A\,a^3\,b\,e^3+3\,B\,a^2\,b^2\,d^2\,e+3\,A\,a^2\,b^2\,d\,e^2-B\,a\,b^3\,d^3-3\,A\,a\,b^3\,d^2\,e+A\,b^4\,d^3}\right )\,\left (A\,b-B\,a\right )\,{\left (a\,e-b\,d\right )}^{5/2}}{b^{9/2}}+\frac {2\,B\,{\left (d+e\,x\right )}^{7/2}}{7\,b\,e}+\frac {\left (\frac {2\,A\,e-2\,B\,d}{b\,e}-\frac {2\,B\,\left (a\,e^2-b\,d\,e\right )}{b^2\,e^2}\right )\,{\left (a\,e^2-b\,d\,e\right )}^2\,\sqrt {d+e\,x}}{b^2\,e^2}-\frac {\left (\frac {2\,A\,e-2\,B\,d}{b\,e}-\frac {2\,B\,\left (a\,e^2-b\,d\,e\right )}{b^2\,e^2}\right )\,\left (a\,e^2-b\,d\,e\right )\,{\left (d+e\,x\right )}^{3/2}}{3\,b\,e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*A*e - 2*B*d)/(5*b*e) - (2*B*(a*e^2 - b*d*e))/(5*b^2*e^2))*(d + e*x)^(5/2) + (2*atan((b^(1/2)*(A*b - B*a)*(
a*e - b*d)^(5/2)*(d + e*x)^(1/2))/(A*b^4*d^3 + B*a^4*e^3 - A*a^3*b*e^3 - B*a*b^3*d^3 + 3*A*a^2*b^2*d*e^2 + 3*B
*a^2*b^2*d^2*e - 3*A*a*b^3*d^2*e - 3*B*a^3*b*d*e^2))*(A*b - B*a)*(a*e - b*d)^(5/2))/b^(9/2) + (2*B*(d + e*x)^(
7/2))/(7*b*e) + (((2*A*e - 2*B*d)/(b*e) - (2*B*(a*e^2 - b*d*e))/(b^2*e^2))*(a*e^2 - b*d*e)^2*(d + e*x)^(1/2))/
(b^2*e^2) - (((2*A*e - 2*B*d)/(b*e) - (2*B*(a*e^2 - b*d*e))/(b^2*e^2))*(a*e^2 - b*d*e)*(d + e*x)^(3/2))/(3*b*e
)

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sympy [A]  time = 70.74, size = 221, normalized size = 1.35 \[ \frac {2 B \left (d + e x\right )^{\frac {7}{2}}}{7 b e} + \frac {\left (d + e x\right )^{\frac {5}{2}} \left (2 A b - 2 B a\right )}{5 b^{2}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (- 2 A a b e + 2 A b^{2} d + 2 B a^{2} e - 2 B a b d\right )}{3 b^{3}} + \frac {\sqrt {d + e x} \left (2 A a^{2} b e^{2} - 4 A a b^{2} d e + 2 A b^{3} d^{2} - 2 B a^{3} e^{2} + 4 B a^{2} b d e - 2 B a b^{2} d^{2}\right )}{b^{4}} + \frac {2 \left (- A b + B a\right ) \left (a e - b d\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e - b d}{b}}} \right )}}{b^{5} \sqrt {\frac {a e - b d}{b}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**(5/2)/(b*x+a),x)

[Out]

2*B*(d + e*x)**(7/2)/(7*b*e) + (d + e*x)**(5/2)*(2*A*b - 2*B*a)/(5*b**2) + (d + e*x)**(3/2)*(-2*A*a*b*e + 2*A*
b**2*d + 2*B*a**2*e - 2*B*a*b*d)/(3*b**3) + sqrt(d + e*x)*(2*A*a**2*b*e**2 - 4*A*a*b**2*d*e + 2*A*b**3*d**2 -
2*B*a**3*e**2 + 4*B*a**2*b*d*e - 2*B*a*b**2*d**2)/b**4 + 2*(-A*b + B*a)*(a*e - b*d)**3*atan(sqrt(d + e*x)/sqrt
((a*e - b*d)/b))/(b**5*sqrt((a*e - b*d)/b))

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