3.22.23 \(\int \frac {(a+b x) (d+e x)^{5/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) [2123]

Optimal. Leaf size=41 \[ \frac {2 (a+b x) (d+e x)^{7/2}}{7 e \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

2/7*(b*x+a)*(e*x+d)^(7/2)/e/((b*x+a)^2)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {784, 21, 32} \begin {gather*} \frac {2 (a+b x) (d+e x)^{7/2}}{7 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)^(5/2))/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

(2*(a + b*x)*(d + e*x)^(7/2))/(7*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 784

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)^{5/2}}{\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)^{5/2}}{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)^{5/2} \, dx}{b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (a+b x) (d+e x)^{7/2}}{7 e \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 32, normalized size = 0.78 \begin {gather*} \frac {2 (a+b x) (d+e x)^{7/2}}{7 e \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)*(d + e*x)^(7/2))/(7*e*Sqrt[(a + b*x)^2])

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Maple [A]
time = 0.02, size = 27, normalized size = 0.66

method	result	size
gosper	\(\frac {2 \left (b x +a \right ) \left (e x +d \right )^{\frac {7}{2}}}{7 e \sqrt {\left (b x +a \right )^{2}}}\)	\(27\)
default	\(\frac {2 \left (b x +a \right ) \left (e x +d \right )^{\frac {7}{2}}}{7 e \sqrt {\left (b x +a \right )^{2}}}\)	\(27\)
risch	\(\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (e^{3} x^{3}+3 d \,e^{2} x^{2}+3 d^{2} x e +d^{3}\right ) \sqrt {e x +d}}{7 \left (b x +a \right ) e}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(e*x+d)^(5/2)/((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/7*(b*x+a)*(e*x+d)^(7/2)/e/((b*x+a)^2)^(1/2)

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Maxima [A]
time = 0.37, size = 38, normalized size = 0.93 \begin {gather*} \frac {2}{7} \, {\left (x^{3} e^{3} + 3 \, d x^{2} e^{2} + 3 \, d^{2} x e + d^{3}\right )} \sqrt {x e + d} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^(5/2)/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

2/7*(x^3*e^3 + 3*d*x^2*e^2 + 3*d^2*x*e + d^3)*sqrt(x*e + d)*e^(-1)

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Fricas [A]
time = 2.92, size = 38, normalized size = 0.93 \begin {gather*} \frac {2}{7} \, {\left (x^{3} e^{3} + 3 \, d x^{2} e^{2} + 3 \, d^{2} x e + d^{3}\right )} \sqrt {x e + d} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^(5/2)/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

2/7*(x^3*e^3 + 3*d*x^2*e^2 + 3*d^2*x*e + d^3)*sqrt(x*e + d)*e^(-1)

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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)*(e*x+d)**(5/2)/((b*x+a)**2)**(1/2),x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 154 vs. \(2 (26) = 52\).
time = 1.91, size = 154, normalized size = 3.76 \begin {gather*} \frac {2}{35} \, {\left (35 \, \sqrt {x e + d} d^{3} \mathrm {sgn}\left (b x + a\right ) + 35 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {x e + d} d\right )} d^{2} \mathrm {sgn}\left (b x + a\right ) + 7 \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} - 10 \, {\left (x e + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {x e + d} d^{2}\right )} d \mathrm {sgn}\left (b x + a\right ) + {\left (5 \, {\left (x e + d\right )}^{\frac {7}{2}} - 21 \, {\left (x e + d\right )}^{\frac {5}{2}} d + 35 \, {\left (x e + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {x e + d} d^{3}\right )} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^(5/2)/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

2/35*(35*sqrt(x*e + d)*d^3*sgn(b*x + a) + 35*((x*e + d)^(3/2) - 3*sqrt(x*e + d)*d)*d^2*sgn(b*x + a) + 7*(3*(x*
e + d)^(5/2) - 10*(x*e + d)^(3/2)*d + 15*sqrt(x*e + d)*d^2)*d*sgn(b*x + a) + (5*(x*e + d)^(7/2) - 21*(x*e + d)
^(5/2)*d + 35*(x*e + d)^(3/2)*d^2 - 35*sqrt(x*e + d)*d^3)*sgn(b*x + a))*e^(-1)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\left (a+b\,x\right )\,{\left (d+e\,x\right )}^{5/2}}{\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)^(5/2))/((a + b*x)^2)^(1/2),x)

[Out]

int(((a + b*x)*(d + e*x)^(5/2))/((a + b*x)^2)^(1/2), x)

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