3.18.90 \(\int \frac {(a+b x) (d+e x)^4}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=39 \[ \frac {(a+b x) (d+e x)^5}{5 e \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.03, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {770, 21, 32} \begin {gather*} \frac {(a+b x) (d+e x)^5}{5 e \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(d + e*x)^4)/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

((a + b*x)*(d + e*x)^5)/(5*e*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {(a+b x) (d+e x)^4}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\left (a b+b^2 x\right ) \int \frac {(a+b x) (d+e x)^4}{a b+b^2 x} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (a b+b^2 x\right ) \int (d+e x)^4 \, dx}{b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(a+b x) (d+e x)^5}{5 e \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 30, normalized size = 0.77 \begin {gather*} \frac {(a+b x) (d+e x)^5}{5 e \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(d + e*x)^4)/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

((a + b*x)*(d + e*x)^5)/(5*e*Sqrt[(a + b*x)^2])

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IntegrateAlgebraic [F]  time = 0.81, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(a+b x) (d+e x)^4}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((a + b*x)*(d + e*x)^4)/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

Defer[IntegrateAlgebraic][((a + b*x)*(d + e*x)^4)/Sqrt[a^2 + 2*a*b*x + b^2*x^2], x]

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fricas [A]  time = 0.42, size = 42, normalized size = 1.08 \begin {gather*} \frac {1}{5} \, e^{4} x^{5} + d e^{3} x^{4} + 2 \, d^{2} e^{2} x^{3} + 2 \, d^{3} e x^{2} + d^{4} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^4/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

1/5*e^4*x^5 + d*e^3*x^4 + 2*d^2*e^2*x^3 + 2*d^3*e*x^2 + d^4*x

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giac [A]  time = 0.15, size = 18, normalized size = 0.46 \begin {gather*} \frac {1}{5} \, {\left (x e + d\right )}^{5} e^{\left (-1\right )} \mathrm {sgn}\left (b x + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^4/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/5*(x*e + d)^5*e^(-1)*sgn(b*x + a)

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maple [B]  time = 0.05, size = 58, normalized size = 1.49 \begin {gather*} \frac {\left (e^{4} x^{4}+5 d \,e^{3} x^{3}+10 d^{2} e^{2} x^{2}+10 d^{3} e x +5 d^{4}\right ) \left (b x +a \right ) x}{5 \sqrt {\left (b x +a \right )^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(e*x+d)^4/((b*x+a)^2)^(1/2),x)

[Out]

1/5*x*(e^4*x^4+5*d*e^3*x^3+10*d^2*e^2*x^2+10*d^3*e*x+5*d^4)*(b*x+a)/((b*x+a)^2)^(1/2)

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maxima [B]  time = 0.57, size = 688, normalized size = 17.64 \begin {gather*} \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} e^{4} x^{4}}{5 \, b} - \frac {9 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a e^{4} x^{3}}{20 \, b^{2}} - \frac {77 \, a^{3} e^{4} x^{2}}{60 \, b^{3}} + \frac {47 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} e^{4} x^{2}}{60 \, b^{3}} + \frac {77 \, a^{4} e^{4} x}{30 \, b^{4}} + \frac {a d^{4} \log \left (x + \frac {a}{b}\right )}{b} - \frac {a^{5} e^{4} \log \left (x + \frac {a}{b}\right )}{b^{5}} - \frac {47 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{4} e^{4}}{30 \, b^{5}} + \frac {{\left (4 \, b d e^{3} + a e^{4}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} x^{3}}{4 \, b^{2}} + \frac {13 \, {\left (4 \, b d e^{3} + a e^{4}\right )} a^{2} x^{2}}{12 \, b^{3}} - \frac {5 \, {\left (3 \, b d^{2} e^{2} + 2 \, a d e^{3}\right )} a x^{2}}{3 \, b^{2}} + \frac {{\left (2 \, b d^{3} e + 3 \, a d^{2} e^{2}\right )} x^{2}}{b} - \frac {7 \, {\left (4 \, b d e^{3} + a e^{4}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a x^{2}}{12 \, b^{3}} + \frac {2 \, {\left (3 \, b d^{2} e^{2} + 2 \, a d e^{3}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} x^{2}}{3 \, b^{2}} - \frac {13 \, {\left (4 \, b d e^{3} + a e^{4}\right )} a^{3} x}{6 \, b^{4}} + \frac {10 \, {\left (3 \, b d^{2} e^{2} + 2 \, a d e^{3}\right )} a^{2} x}{3 \, b^{3}} - \frac {2 \, {\left (2 \, b d^{3} e + 3 \, a d^{2} e^{2}\right )} a x}{b^{2}} + \frac {{\left (4 \, b d e^{3} + a e^{4}\right )} a^{4} \log \left (x + \frac {a}{b}\right )}{b^{5}} - \frac {2 \, {\left (3 \, b d^{2} e^{2} + 2 \, a d e^{3}\right )} a^{3} \log \left (x + \frac {a}{b}\right )}{b^{4}} + \frac {2 \, {\left (2 \, b d^{3} e + 3 \, a d^{2} e^{2}\right )} a^{2} \log \left (x + \frac {a}{b}\right )}{b^{3}} - \frac {{\left (b d^{4} + 4 \, a d^{3} e\right )} a \log \left (x + \frac {a}{b}\right )}{b^{2}} + \frac {7 \, {\left (4 \, b d e^{3} + a e^{4}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3}}{6 \, b^{5}} - \frac {4 \, {\left (3 \, b d^{2} e^{2} + 2 \, a d e^{3}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2}}{3 \, b^{4}} + \frac {{\left (b d^{4} + 4 \, a d^{3} e\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^4/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

1/5*sqrt(b^2*x^2 + 2*a*b*x + a^2)*e^4*x^4/b - 9/20*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a*e^4*x^3/b^2 - 77/60*a^3*e^4
*x^2/b^3 + 47/60*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^2*e^4*x^2/b^3 + 77/30*a^4*e^4*x/b^4 + a*d^4*log(x + a/b)/b -
a^5*e^4*log(x + a/b)/b^5 - 47/30*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^4*e^4/b^5 + 1/4*(4*b*d*e^3 + a*e^4)*sqrt(b^2*
x^2 + 2*a*b*x + a^2)*x^3/b^2 + 13/12*(4*b*d*e^3 + a*e^4)*a^2*x^2/b^3 - 5/3*(3*b*d^2*e^2 + 2*a*d*e^3)*a*x^2/b^2
+ (2*b*d^3*e + 3*a*d^2*e^2)*x^2/b - 7/12*(4*b*d*e^3 + a*e^4)*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a*x^2/b^3 + 2/3*(3
*b*d^2*e^2 + 2*a*d*e^3)*sqrt(b^2*x^2 + 2*a*b*x + a^2)*x^2/b^2 - 13/6*(4*b*d*e^3 + a*e^4)*a^3*x/b^4 + 10/3*(3*b
*d^2*e^2 + 2*a*d*e^3)*a^2*x/b^3 - 2*(2*b*d^3*e + 3*a*d^2*e^2)*a*x/b^2 + (4*b*d*e^3 + a*e^4)*a^4*log(x + a/b)/b
^5 - 2*(3*b*d^2*e^2 + 2*a*d*e^3)*a^3*log(x + a/b)/b^4 + 2*(2*b*d^3*e + 3*a*d^2*e^2)*a^2*log(x + a/b)/b^3 - (b*
d^4 + 4*a*d^3*e)*a*log(x + a/b)/b^2 + 7/6*(4*b*d*e^3 + a*e^4)*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^3/b^5 - 4/3*(3*b
*d^2*e^2 + 2*a*d*e^3)*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^2/b^4 + (b*d^4 + 4*a*d^3*e)*sqrt(b^2*x^2 + 2*a*b*x + a^2
)/b^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {\left (a+b\,x\right )\,{\left (d+e\,x\right )}^4}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)*(d + e*x)^4)/((a + b*x)^2)^(1/2),x)

[Out]

int(((a + b*x)*(d + e*x)^4)/((a + b*x)^2)^(1/2), x)

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sympy [A]  time = 0.12, size = 42, normalized size = 1.08 \begin {gather*} d^{4} x + 2 d^{3} e x^{2} + 2 d^{2} e^{2} x^{3} + d e^{3} x^{4} + \frac {e^{4} x^{5}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)**4/((b*x+a)**2)**(1/2),x)

[Out]

d**4*x + 2*d**3*e*x**2 + 2*d**2*e**2*x**3 + d*e**3*x**4 + e**4*x**5/5

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