3.23.28 \(\int \frac {(a+b x)^{5/2} (A+B x)}{(d+e x)^{5/2}} \, dx\) [2228]

Optimal. Leaf size=257 \[ -\frac {2 (B d-A e) (a+b x)^{7/2}}{3 e (b d-a e) (d+e x)^{3/2}}-\frac {2 (7 b B d-4 A b e-3 a B e) (a+b x)^{5/2}}{3 e^2 (b d-a e) \sqrt {d+e x}}-\frac {5 b (7 b B d-4 A b e-3 a B e) \sqrt {a+b x} \sqrt {d+e x}}{4 e^4}+\frac {5 b (7 b B d-4 A b e-3 a B e) (a+b x)^{3/2} \sqrt {d+e x}}{6 e^3 (b d-a e)}+\frac {5 \sqrt {b} (b d-a e) (7 b B d-4 A b e-3 a B e) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{4 e^{9/2}} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {79, 49, 52, 65, 223, 212} \begin {gather*} \frac {5 \sqrt {b} (b d-a e) (-3 a B e-4 A b e+7 b B d) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{4 e^{9/2}}-\frac {5 b \sqrt {a+b x} \sqrt {d+e x} (-3 a B e-4 A b e+7 b B d)}{4 e^4}+\frac {5 b (a+b x)^{3/2} \sqrt {d+e x} (-3 a B e-4 A b e+7 b B d)}{6 e^3 (b d-a e)}-\frac {2 (a+b x)^{5/2} (-3 a B e-4 A b e+7 b B d)}{3 e^2 \sqrt {d+e x} (b d-a e)}-\frac {2 (a+b x)^{7/2} (B d-A e)}{3 e (d+e x)^{3/2} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 49

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 52

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 65

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 79

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] || L
tQ[p, n]))))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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

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Mathematica [A]
time = 3.32, size = 215, normalized size = 0.84 \begin {gather*} \frac {-\frac {\sqrt {a+b x} \left (8 a^2 e^2 (2 B d+A e+3 B e x)+a b e \left (8 A e (5 d+7 e x)-B \left (115 d^2+158 d e x+27 e^2 x^2\right )\right )+b^2 \left (-4 A e \left (15 d^2+20 d e x+3 e^2 x^2\right )+B \left (105 d^3+140 d^2 e x+21 d e^2 x^2-6 e^3 x^3\right )\right )\right )}{(d+e x)^{3/2}}+15 \sqrt {\frac {b}{e}} (b d-a e) (-7 b B d+4 A b e+3 a B e) \log \left (\sqrt {a+b x}-\sqrt {\frac {b}{e}} \sqrt {d+e x}\right )}{12 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-((Sqrt[a + b*x]*(8*a^2*e^2*(2*B*d + A*e + 3*B*e*x) + a*b*e*(8*A*e*(5*d + 7*e*x) - B*(115*d^2 + 158*d*e*x + 2
7*e^2*x^2)) + b^2*(-4*A*e*(15*d^2 + 20*d*e*x + 3*e^2*x^2) + B*(105*d^3 + 140*d^2*e*x + 21*d*e^2*x^2 - 6*e^3*x^
3))))/(d + e*x)^(3/2)) + 15*Sqrt[b/e]*(b*d - a*e)*(-7*b*B*d + 4*A*b*e + 3*a*B*e)*Log[Sqrt[a + b*x] - Sqrt[b/e]
*Sqrt[d + e*x]])/(12*e^4)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1249\) vs. \(2(219)=438\).
time = 0.09, size = 1250, normalized size = 4.86

method	result	size
default	\(\text {Expression too large to display}\)	\(1250\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(b*x+a)^(1/2)*(-120*A*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^3*d^2
*e^2*x+60*A*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b^2*d^2*e^2+45*B*ln(
1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^2*b*d^2*e^2+12*B*b^2*e^3*x^3*(b*e)^
(1/2)*((b*x+a)*(e*x+d))^(1/2)+120*A*b^2*d^2*e*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)+60*A*ln(1/2*(2*b*e*x+2*((b*x
+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b^2*e^4*x^2-60*A*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1
/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^3*d*e^3*x^2+45*B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)
+a*e+b*d)/(b*e)^(1/2))*a^2*b*e^4*x^2+105*B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e
)^(1/2))*b^3*d^2*e^2*x^2+24*A*b^2*e^3*x^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+210*B*ln(1/2*(2*b*e*x+2*((b*x+a)
*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^3*d^3*e*x-150*B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*
(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b^2*d^3*e-48*B*a^2*e^3*x*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-32*B*a^2*d*e^
2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+105*B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*
e)^(1/2))*b^3*d^4+160*A*b^2*d*e^2*x*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)-80*A*a*b*d*e^2*(b*e)^(1/2)*((b*x+a)*(e
*x+d))^(1/2)+120*A*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b^2*d*e^3*x+9
0*B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^2*b*d*e^3*x-16*A*a^2*e^3*((b
*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-210*B*b^2*d^3*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+316*B*a*b*d*e^2*x*((b*x+a)*
(e*x+d))^(1/2)*(b*e)^(1/2)-60*A*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^
3*d^3*e-150*B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b^2*d*e^3*x^2-300*
B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b^2*d^2*e^2*x+54*B*a*b*e^3*x^2
*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-42*B*b^2*d*e^2*x^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-112*A*a*b*e^3*x*((
b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-280*B*b^2*d^2*e*x*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+230*B*a*b*d^2*e*((b*x+
a)*(e*x+d))^(1/2)*(b*e)^(1/2))/((b*x+a)*(e*x+d))^(1/2)/(b*e)^(1/2)/(e*x+d)^(3/2)/e^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((b*x+a)^(5/2)*(B*x+A)/(e*x+d)^(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(b*d-%e*a>0)', see `assume?` fo
r more detai

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Fricas [A]
time = 2.75, size = 822, normalized size = 3.20 \begin {gather*} \left [\frac {15 \, {\left (7 \, B b^{2} d^{4} + {\left (3 \, B a^{2} + 4 \, A a b\right )} x^{2} e^{4} - 2 \, {\left ({\left (5 \, B a b + 2 \, A b^{2}\right )} d x^{2} - {\left (3 \, B a^{2} + 4 \, A a b\right )} d x\right )} e^{3} + {\left (7 \, B b^{2} d^{2} x^{2} - 4 \, {\left (5 \, B a b + 2 \, A b^{2}\right )} d^{2} x + {\left (3 \, B a^{2} + 4 \, A a b\right )} d^{2}\right )} e^{2} + 2 \, {\left (7 \, B b^{2} d^{3} x - {\left (5 \, B a b + 2 \, A b^{2}\right )} d^{3}\right )} e\right )} \sqrt {b} e^{\left (-\frac {1}{2}\right )} \log \left (b^{2} d^{2} + 4 \, {\left (b d e + {\left (2 \, b x + a\right )} e^{2}\right )} \sqrt {b x + a} \sqrt {x e + d} \sqrt {b} e^{\left (-\frac {1}{2}\right )} + {\left (8 \, b^{2} x^{2} + 8 \, a b x + a^{2}\right )} e^{2} + 2 \, {\left (4 \, b^{2} d x + 3 \, a b d\right )} e\right ) - 4 \, {\left (105 \, B b^{2} d^{3} - {\left (6 \, B b^{2} x^{3} - 8 \, A a^{2} + 3 \, {\left (9 \, B a b + 4 \, A b^{2}\right )} x^{2} - 8 \, {\left (3 \, B a^{2} + 7 \, A a b\right )} x\right )} e^{3} + {\left (21 \, B b^{2} d x^{2} - 2 \, {\left (79 \, B a b + 40 \, A b^{2}\right )} d x + 8 \, {\left (2 \, B a^{2} + 5 \, A a b\right )} d\right )} e^{2} + 5 \, {\left (28 \, B b^{2} d^{2} x - {\left (23 \, B a b + 12 \, A b^{2}\right )} d^{2}\right )} e\right )} \sqrt {b x + a} \sqrt {x e + d}}{48 \, {\left (x^{2} e^{6} + 2 \, d x e^{5} + d^{2} e^{4}\right )}}, -\frac {15 \, {\left (7 \, B b^{2} d^{4} + {\left (3 \, B a^{2} + 4 \, A a b\right )} x^{2} e^{4} - 2 \, {\left ({\left (5 \, B a b + 2 \, A b^{2}\right )} d x^{2} - {\left (3 \, B a^{2} + 4 \, A a b\right )} d x\right )} e^{3} + {\left (7 \, B b^{2} d^{2} x^{2} - 4 \, {\left (5 \, B a b + 2 \, A b^{2}\right )} d^{2} x + {\left (3 \, B a^{2} + 4 \, A a b\right )} d^{2}\right )} e^{2} + 2 \, {\left (7 \, B b^{2} d^{3} x - {\left (5 \, B a b + 2 \, A b^{2}\right )} d^{3}\right )} e\right )} \sqrt {-b e^{\left (-1\right )}} \arctan \left (\frac {{\left (b d + {\left (2 \, b x + a\right )} e\right )} \sqrt {b x + a} \sqrt {x e + d} \sqrt {-b e^{\left (-1\right )}}}{2 \, {\left (b^{2} d x + a b d + {\left (b^{2} x^{2} + a b x\right )} e\right )}}\right ) + 2 \, {\left (105 \, B b^{2} d^{3} - {\left (6 \, B b^{2} x^{3} - 8 \, A a^{2} + 3 \, {\left (9 \, B a b + 4 \, A b^{2}\right )} x^{2} - 8 \, {\left (3 \, B a^{2} + 7 \, A a b\right )} x\right )} e^{3} + {\left (21 \, B b^{2} d x^{2} - 2 \, {\left (79 \, B a b + 40 \, A b^{2}\right )} d x + 8 \, {\left (2 \, B a^{2} + 5 \, A a b\right )} d\right )} e^{2} + 5 \, {\left (28 \, B b^{2} d^{2} x - {\left (23 \, B a b + 12 \, A b^{2}\right )} d^{2}\right )} e\right )} \sqrt {b x + a} \sqrt {x e + d}}{24 \, {\left (x^{2} e^{6} + 2 \, d x e^{5} + d^{2} e^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(15*(7*B*b^2*d^4 + (3*B*a^2 + 4*A*a*b)*x^2*e^4 - 2*((5*B*a*b + 2*A*b^2)*d*x^2 - (3*B*a^2 + 4*A*a*b)*d*x)
*e^3 + (7*B*b^2*d^2*x^2 - 4*(5*B*a*b + 2*A*b^2)*d^2*x + (3*B*a^2 + 4*A*a*b)*d^2)*e^2 + 2*(7*B*b^2*d^3*x - (5*B
*a*b + 2*A*b^2)*d^3)*e)*sqrt(b)*e^(-1/2)*log(b^2*d^2 + 4*(b*d*e + (2*b*x + a)*e^2)*sqrt(b*x + a)*sqrt(x*e + d)
*sqrt(b)*e^(-1/2) + (8*b^2*x^2 + 8*a*b*x + a^2)*e^2 + 2*(4*b^2*d*x + 3*a*b*d)*e) - 4*(105*B*b^2*d^3 - (6*B*b^2
*x^3 - 8*A*a^2 + 3*(9*B*a*b + 4*A*b^2)*x^2 - 8*(3*B*a^2 + 7*A*a*b)*x)*e^3 + (21*B*b^2*d*x^2 - 2*(79*B*a*b + 40
*A*b^2)*d*x + 8*(2*B*a^2 + 5*A*a*b)*d)*e^2 + 5*(28*B*b^2*d^2*x - (23*B*a*b + 12*A*b^2)*d^2)*e)*sqrt(b*x + a)*s
qrt(x*e + d))/(x^2*e^6 + 2*d*x*e^5 + d^2*e^4), -1/24*(15*(7*B*b^2*d^4 + (3*B*a^2 + 4*A*a*b)*x^2*e^4 - 2*((5*B*
a*b + 2*A*b^2)*d*x^2 - (3*B*a^2 + 4*A*a*b)*d*x)*e^3 + (7*B*b^2*d^2*x^2 - 4*(5*B*a*b + 2*A*b^2)*d^2*x + (3*B*a^
2 + 4*A*a*b)*d^2)*e^2 + 2*(7*B*b^2*d^3*x - (5*B*a*b + 2*A*b^2)*d^3)*e)*sqrt(-b*e^(-1))*arctan(1/2*(b*d + (2*b*
x + a)*e)*sqrt(b*x + a)*sqrt(x*e + d)*sqrt(-b*e^(-1))/(b^2*d*x + a*b*d + (b^2*x^2 + a*b*x)*e)) + 2*(105*B*b^2*
d^3 - (6*B*b^2*x^3 - 8*A*a^2 + 3*(9*B*a*b + 4*A*b^2)*x^2 - 8*(3*B*a^2 + 7*A*a*b)*x)*e^3 + (21*B*b^2*d*x^2 - 2*
(79*B*a*b + 40*A*b^2)*d*x + 8*(2*B*a^2 + 5*A*a*b)*d)*e^2 + 5*(28*B*b^2*d^2*x - (23*B*a*b + 12*A*b^2)*d^2)*e)*s
qrt(b*x + a)*sqrt(x*e + d))/(x^2*e^6 + 2*d*x*e^5 + d^2*e^4)]

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Sympy [F(-1)] Timed out
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)**(5/2)*(B*x+A)/(e*x+d)**(5/2),x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 534 vs. \(2 (231) = 462\).
time = 1.96, size = 534, normalized size = 2.08 \begin {gather*} -\frac {5 \, {\left (7 \, B b^{2} d^{2} {\left | b \right |} - 10 \, B a b d {\left | b \right |} e - 4 \, A b^{2} d {\left | b \right |} e + 3 \, B a^{2} {\left | b \right |} e^{2} + 4 \, A a b {\left | b \right |} e^{2}\right )} e^{\left (-\frac {9}{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 )}{4 \, \sqrt {b}} + \frac {{\left ({\left (3 \, {\left (b x + a\right )} {\left (\frac {2 \, {\left (B b^{5} d {\left | b \right |} e^{6} - B a b^{4} {\left | b \right |} e^{7}\right )} {\left (b x + a\right )}}{b^{4} d e^{7} - a b^{3} e^{8}} - \frac {7 \, B b^{6} d^{2} {\left | b \right |} e^{5} - 10 \, B a b^{5} d {\left | b \right |} e^{6} - 4 \, A b^{6} d {\left | b \right |} e^{6} + 3 \, B a^{2} b^{4} {\left | b \right |} e^{7} + 4 \, A a b^{5} {\left | b \right |} e^{7}}{b^{4} d e^{7} - a b^{3} e^{8}}\right )} - \frac {20 \, {\left (7 \, B b^{7} d^{3} {\left | b \right |} e^{4} - 17 \, B a b^{6} d^{2} {\left | b \right |} e^{5} - 4 \, A b^{7} d^{2} {\left | b \right |} e^{5} + 13 \, B a^{2} b^{5} d {\left | b \right |} e^{6} + 8 \, A a b^{6} d {\left | b \right |} e^{6} - 3 \, B a^{3} b^{4} {\left | b \right |} e^{7} - 4 \, A a^{2} b^{5} {\left | b \right |} e^{7}\right )}}{b^{4} d e^{7} - a b^{3} e^{8}}\right )} {\left (b x + a\right )} - \frac {15 \, {\left (7 \, B b^{8} d^{4} {\left | b \right |} e^{3} - 24 \, B a b^{7} d^{3} {\left | b \right |} e^{4} - 4 \, A b^{8} d^{3} {\left | b \right |} e^{4} + 30 \, B a^{2} b^{6} d^{2} {\left | b \right |} e^{5} + 12 \, A a b^{7} d^{2} {\left | b \right |} e^{5} - 16 \, B a^{3} b^{5} d {\left | b \right |} e^{6} - 12 \, A a^{2} b^{6} d {\left | b \right |} e^{6} + 3 \, B a^{4} b^{4} {\left | b \right |} e^{7} + 4 \, A a^{3} b^{5} {\left | b \right |} e^{7}\right )}}{b^{4} d e^{7} - a b^{3} e^{8}}\right )} \sqrt {b x + a}}{12 \, {\left (b^{2} d + {\left (b x + a\right )} b e - a b e\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/4*(7*B*b^2*d^2*abs(b) - 10*B*a*b*d*abs(b)*e - 4*A*b^2*d*abs(b)*e + 3*B*a^2*abs(b)*e^2 + 4*A*a*b*abs(b)*e^2)
*e^(-9/2)*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) + 1/12*((3*(b
*x + a)*(2*(B*b^5*d*abs(b)*e^6 - B*a*b^4*abs(b)*e^7)*(b*x + a)/(b^4*d*e^7 - a*b^3*e^8) - (7*B*b^6*d^2*abs(b)*e
^5 - 10*B*a*b^5*d*abs(b)*e^6 - 4*A*b^6*d*abs(b)*e^6 + 3*B*a^2*b^4*abs(b)*e^7 + 4*A*a*b^5*abs(b)*e^7)/(b^4*d*e^
7 - a*b^3*e^8)) - 20*(7*B*b^7*d^3*abs(b)*e^4 - 17*B*a*b^6*d^2*abs(b)*e^5 - 4*A*b^7*d^2*abs(b)*e^5 + 13*B*a^2*b
^5*d*abs(b)*e^6 + 8*A*a*b^6*d*abs(b)*e^6 - 3*B*a^3*b^4*abs(b)*e^7 - 4*A*a^2*b^5*abs(b)*e^7)/(b^4*d*e^7 - a*b^3
*e^8))*(b*x + a) - 15*(7*B*b^8*d^4*abs(b)*e^3 - 24*B*a*b^7*d^3*abs(b)*e^4 - 4*A*b^8*d^3*abs(b)*e^4 + 30*B*a^2*
b^6*d^2*abs(b)*e^5 + 12*A*a*b^7*d^2*abs(b)*e^5 - 16*B*a^3*b^5*d*abs(b)*e^6 - 12*A*a^2*b^6*d*abs(b)*e^6 + 3*B*a
^4*b^4*abs(b)*e^7 + 4*A*a^3*b^5*abs(b)*e^7)/(b^4*d*e^7 - a*b^3*e^8))*sqrt(b*x + a)/(b^2*d + (b*x + a)*b*e - a*
b*e)^(3/2)

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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 (a+b\,x\right )}^{5/2}}{{\left (d+e\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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