3.1.10 \(\int \frac {(a+b x) \sqrt {c+d x} (e+f x)}{x} \, dx\)

Optimal. Leaf size=77 \[ -\frac {2 (c+d x)^{3/2} (-5 d (a f+b e)+2 b c f-3 b d f x)}{15 d^2}+2 a e \sqrt {c+d x}-2 a \sqrt {c} e \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {147, 50, 63, 208} \begin {gather*} -\frac {2 (c+d x)^{3/2} (-5 d (a f+b e)+2 b c f-3 b d f x)}{15 d^2}+2 a e \sqrt {c+d x}-2 a \sqrt {c} e \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*Sqrt[c + d*x]*(e + f*x))/x,x]

[Out]

2*a*e*Sqrt[c + d*x] - (2*(c + d*x)^(3/2)*(2*b*c*f - 5*d*(b*e + a*f) - 3*b*d*f*x))/(15*d^2) - 2*a*Sqrt[c]*e*Arc
Tanh[Sqrt[c + d*x]/Sqrt[c]]

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 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 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) \sqrt {c+d x} (e+f x)}{x} \, dx &=-\frac {2 (c+d x)^{3/2} (2 b c f-5 d (b e+a f)-3 b d f x)}{15 d^2}+(a e) \int \frac {\sqrt {c+d x}}{x} \, dx\\ &=2 a e \sqrt {c+d x}-\frac {2 (c+d x)^{3/2} (2 b c f-5 d (b e+a f)-3 b d f x)}{15 d^2}+(a c e) \int \frac {1}{x \sqrt {c+d x}} \, dx\\ &=2 a e \sqrt {c+d x}-\frac {2 (c+d x)^{3/2} (2 b c f-5 d (b e+a f)-3 b d f x)}{15 d^2}+\frac {(2 a c e) \operatorname {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{d}\\ &=2 a e \sqrt {c+d x}-\frac {2 (c+d x)^{3/2} (2 b c f-5 d (b e+a f)-3 b d f x)}{15 d^2}-2 a \sqrt {c} e \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 81, normalized size = 1.05 \begin {gather*} \frac {2 \sqrt {c+d x} (5 a d (c f+3 d e+d f x)-b (c+d x) (2 c f-5 d e-3 d f x))}{15 d^2}-2 a \sqrt {c} e \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*Sqrt[c + d*x]*(e + f*x))/x,x]

[Out]

(2*Sqrt[c + d*x]*(-(b*(c + d*x)*(-5*d*e + 2*c*f - 3*d*f*x)) + 5*a*d*(3*d*e + c*f + d*f*x)))/(15*d^2) - 2*a*Sqr
t[c]*e*ArcTanh[Sqrt[c + d*x]/Sqrt[c]]

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IntegrateAlgebraic [A]  time = 0.06, size = 105, normalized size = 1.36 \begin {gather*} \frac {2 \left (15 a d^2 e \sqrt {c+d x}+5 a d f (c+d x)^{3/2}+5 b d e (c+d x)^{3/2}+3 b f (c+d x)^{5/2}-5 b c f (c+d x)^{3/2}\right )}{15 d^2}-2 a \sqrt {c} e \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x)*Sqrt[c + d*x]*(e + f*x))/x,x]

[Out]

(2*(15*a*d^2*e*Sqrt[c + d*x] + 5*b*d*e*(c + d*x)^(3/2) - 5*b*c*f*(c + d*x)^(3/2) + 5*a*d*f*(c + d*x)^(3/2) + 3
*b*f*(c + d*x)^(5/2)))/(15*d^2) - 2*a*Sqrt[c]*e*ArcTanh[Sqrt[c + d*x]/Sqrt[c]]

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fricas [A]  time = 0.90, size = 219, normalized size = 2.84 \begin {gather*} \left [\frac {15 \, a \sqrt {c} d^{2} e \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (3 \, b d^{2} f x^{2} + 5 \, {\left (b c d + 3 \, a d^{2}\right )} e - {\left (2 \, b c^{2} - 5 \, a c d\right )} f + {\left (5 \, b d^{2} e + {\left (b c d + 5 \, a d^{2}\right )} f\right )} x\right )} \sqrt {d x + c}}{15 \, d^{2}}, \frac {2 \, {\left (15 \, a \sqrt {-c} d^{2} e \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) + {\left (3 \, b d^{2} f x^{2} + 5 \, {\left (b c d + 3 \, a d^{2}\right )} e - {\left (2 \, b c^{2} - 5 \, a c d\right )} f + {\left (5 \, b d^{2} e + {\left (b c d + 5 \, a d^{2}\right )} f\right )} x\right )} \sqrt {d x + c}\right )}}{15 \, d^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(f*x+e)*(d*x+c)^(1/2)/x,x, algorithm="fricas")

[Out]

[1/15*(15*a*sqrt(c)*d^2*e*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 2*(3*b*d^2*f*x^2 + 5*(b*c*d + 3*a*d^2
)*e - (2*b*c^2 - 5*a*c*d)*f + (5*b*d^2*e + (b*c*d + 5*a*d^2)*f)*x)*sqrt(d*x + c))/d^2, 2/15*(15*a*sqrt(-c)*d^2
*e*arctan(sqrt(d*x + c)*sqrt(-c)/c) + (3*b*d^2*f*x^2 + 5*(b*c*d + 3*a*d^2)*e - (2*b*c^2 - 5*a*c*d)*f + (5*b*d^
2*e + (b*c*d + 5*a*d^2)*f)*x)*sqrt(d*x + c))/d^2]

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giac [A]  time = 1.33, size = 105, normalized size = 1.36 \begin {gather*} \frac {2 \, a c \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right ) e}{\sqrt {-c}} + \frac {2 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} b d^{8} f - 5 \, {\left (d x + c\right )}^{\frac {3}{2}} b c d^{8} f + 5 \, {\left (d x + c\right )}^{\frac {3}{2}} a d^{9} f + 5 \, {\left (d x + c\right )}^{\frac {3}{2}} b d^{9} e + 15 \, \sqrt {d x + c} a d^{10} e\right )}}{15 \, d^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(f*x+e)*(d*x+c)^(1/2)/x,x, algorithm="giac")

[Out]

2*a*c*arctan(sqrt(d*x + c)/sqrt(-c))*e/sqrt(-c) + 2/15*(3*(d*x + c)^(5/2)*b*d^8*f - 5*(d*x + c)^(3/2)*b*c*d^8*
f + 5*(d*x + c)^(3/2)*a*d^9*f + 5*(d*x + c)^(3/2)*b*d^9*e + 15*sqrt(d*x + c)*a*d^10*e)/d^10

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maple [A]  time = 0.01, size = 89, normalized size = 1.16 \begin {gather*} \frac {-2 a \sqrt {c}\, d^{2} e \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )+2 \sqrt {d x +c}\, a \,d^{2} e +\frac {2 \left (d x +c \right )^{\frac {3}{2}} a d f}{3}-\frac {2 \left (d x +c \right )^{\frac {3}{2}} b c f}{3}+\frac {2 \left (d x +c \right )^{\frac {3}{2}} b d e}{3}+\frac {2 \left (d x +c \right )^{\frac {5}{2}} b f}{5}}{d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(f*x+e)*(d*x+c)^(1/2)/x,x)

[Out]

2/d^2*(1/5*f*b*(d*x+c)^(5/2)+1/3*(d*x+c)^(3/2)*a*d*f-1/3*(d*x+c)^(3/2)*b*c*f+1/3*(d*x+c)^(3/2)*b*d*e+a*d^2*e*(
d*x+c)^(1/2)-a*c^(1/2)*d^2*e*arctanh((d*x+c)^(1/2)/c^(1/2)))

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maxima [A]  time = 0.98, size = 91, normalized size = 1.18 \begin {gather*} a \sqrt {c} e \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right ) + \frac {2 \, {\left (15 \, \sqrt {d x + c} a d^{2} e + 3 \, {\left (d x + c\right )}^{\frac {5}{2}} b f + 5 \, {\left (b d e - {\left (b c - a d\right )} f\right )} {\left (d x + c\right )}^{\frac {3}{2}}\right )}}{15 \, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(f*x+e)*(d*x+c)^(1/2)/x,x, algorithm="maxima")

[Out]

a*sqrt(c)*e*log((sqrt(d*x + c) - sqrt(c))/(sqrt(d*x + c) + sqrt(c))) + 2/15*(15*sqrt(d*x + c)*a*d^2*e + 3*(d*x
+ c)^(5/2)*b*f + 5*(b*d*e - (b*c - a*d)*f)*(d*x + c)^(3/2))/d^2

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mupad [B]  time = 0.09, size = 136, normalized size = 1.77 \begin {gather*} \left (c\,\left (\frac {2\,a\,d\,f-4\,b\,c\,f+2\,b\,d\,e}{d^2}+\frac {2\,b\,c\,f}{d^2}\right )-\frac {2\,\left (a\,d-b\,c\right )\,\left (c\,f-d\,e\right )}{d^2}\right )\,\sqrt {c+d\,x}+\left (\frac {2\,a\,d\,f-4\,b\,c\,f+2\,b\,d\,e}{3\,d^2}+\frac {2\,b\,c\,f}{3\,d^2}\right )\,{\left (c+d\,x\right )}^{3/2}+\frac {2\,b\,f\,{\left (c+d\,x\right )}^{5/2}}{5\,d^2}+a\,\sqrt {c}\,e\,\mathrm {atan}\left (\frac {\sqrt {c+d\,x}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,2{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e + f*x)*(a + b*x)*(c + d*x)^(1/2))/x,x)

[Out]

(c*((2*a*d*f - 4*b*c*f + 2*b*d*e)/d^2 + (2*b*c*f)/d^2) - (2*(a*d - b*c)*(c*f - d*e))/d^2)*(c + d*x)^(1/2) + ((
2*a*d*f - 4*b*c*f + 2*b*d*e)/(3*d^2) + (2*b*c*f)/(3*d^2))*(c + d*x)^(3/2) + (2*b*f*(c + d*x)^(5/2))/(5*d^2) +
a*c^(1/2)*e*atan(((c + d*x)^(1/2)*1i)/c^(1/2))*2i

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sympy [A]  time = 25.99, size = 92, normalized size = 1.19 \begin {gather*} \frac {2 a c e \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + 2 a e \sqrt {c + d x} + \frac {2 b f \left (c + d x\right )^{\frac {5}{2}}}{5 d^{2}} + \frac {2 \left (c + d x\right )^{\frac {3}{2}} \left (a d f - b c f + b d e\right )}{3 d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(f*x+e)*(d*x+c)**(1/2)/x,x)

[Out]

2*a*c*e*atan(sqrt(c + d*x)/sqrt(-c))/sqrt(-c) + 2*a*e*sqrt(c + d*x) + 2*b*f*(c + d*x)**(5/2)/(5*d**2) + 2*(c +
d*x)**(3/2)*(a*d*f - b*c*f + b*d*e)/(3*d**2)

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