3.13.88 \(\int \frac {(c+d x)^{3/2}}{(a+b x)^2} \, dx\)

Optimal. Leaf size=85 \[ -\frac {3 d \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{5/2}}-\frac {(c+d x)^{3/2}}{b (a+b x)}+\frac {3 d \sqrt {c+d x}}{b^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {47, 50, 63, 208} \begin {gather*} -\frac {3 d \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{5/2}}-\frac {(c+d x)^{3/2}}{b (a+b x)}+\frac {3 d \sqrt {c+d x}}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(3/2)/(a + b*x)^2,x]

[Out]

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

Rule 47

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

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Mathematica [C]  time = 0.01, size = 50, normalized size = 0.59 \begin {gather*} \frac {2 d (c+d x)^{5/2} \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};-\frac {b (c+d x)}{a d-b c}\right )}{5 (a d-b c)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(3/2)/(a + b*x)^2,x]

[Out]

(2*d*(c + d*x)^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, -((b*(c + d*x))/(-(b*c) + a*d))])/(5*(-(b*c) + a*d)^2)

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IntegrateAlgebraic [A]  time = 0.26, size = 107, normalized size = 1.26 \begin {gather*} \frac {3 d \sqrt {a d-b c} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x} \sqrt {a d-b c}}{b c-a d}\right )}{b^{5/2}}+\frac {d \sqrt {c+d x} (3 a d+2 b (c+d x)-3 b c)}{b^2 (a d+b (c+d x)-b c)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(c + d*x)^(3/2)/(a + b*x)^2,x]

[Out]

(d*Sqrt[c + d*x]*(-3*b*c + 3*a*d + 2*b*(c + d*x)))/(b^2*(-(b*c) + a*d + b*(c + d*x))) + (3*d*Sqrt[-(b*c) + a*d
]*ArcTan[(Sqrt[b]*Sqrt[-(b*c) + a*d]*Sqrt[c + d*x])/(b*c - a*d)])/b^(5/2)

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fricas [A]  time = 1.74, size = 210, normalized size = 2.47 \begin {gather*} \left [\frac {3 \, {\left (b d x + a d\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + 2 \, {\left (2 \, b d x - b c + 3 \, a d\right )} \sqrt {d x + c}}{2 \, {\left (b^{3} x + a b^{2}\right )}}, -\frac {3 \, {\left (b d x + a d\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) - {\left (2 \, b d x - b c + 3 \, a d\right )} \sqrt {d x + c}}{b^{3} x + a b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(3/2)/(b*x+a)^2,x, algorithm="fricas")

[Out]

[1/2*(3*(b*d*x + a*d)*sqrt((b*c - a*d)/b)*log((b*d*x + 2*b*c - a*d - 2*sqrt(d*x + c)*b*sqrt((b*c - a*d)/b))/(b
*x + a)) + 2*(2*b*d*x - b*c + 3*a*d)*sqrt(d*x + c))/(b^3*x + a*b^2), -(3*(b*d*x + a*d)*sqrt(-(b*c - a*d)/b)*ar
ctan(-sqrt(d*x + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) - (2*b*d*x - b*c + 3*a*d)*sqrt(d*x + c))/(b^3*x + a*b^
2)]

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giac [A]  time = 1.30, size = 113, normalized size = 1.33 \begin {gather*} \frac {2 \, \sqrt {d x + c} d}{b^{2}} + \frac {3 \, {\left (b c d - a d^{2}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{2}} - \frac {\sqrt {d x + c} b c d - \sqrt {d x + c} a d^{2}}{{\left ({\left (d x + c\right )} b - b c + a d\right )} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(3/2)/(b*x+a)^2,x, algorithm="giac")

[Out]

2*sqrt(d*x + c)*d/b^2 + 3*(b*c*d - a*d^2)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b
^2) - (sqrt(d*x + c)*b*c*d - sqrt(d*x + c)*a*d^2)/(((d*x + c)*b - b*c + a*d)*b^2)

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maple [B]  time = 0.01, size = 148, normalized size = 1.74 \begin {gather*} -\frac {3 a \,d^{2} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, b^{2}}+\frac {3 c d \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, b}+\frac {\sqrt {d x +c}\, a \,d^{2}}{\left (b d x +a d \right ) b^{2}}-\frac {\sqrt {d x +c}\, c d}{\left (b d x +a d \right ) b}+\frac {2 \sqrt {d x +c}\, d}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(3/2)/(b*x+a)^2,x)

[Out]

2*d*(d*x+c)^(1/2)/b^2+1/b^2*(d*x+c)^(1/2)/(b*d*x+a*d)*a*d^2-d/b*(d*x+c)^(1/2)/(b*d*x+a*d)*c-3/b^2/((a*d-b*c)*b
)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)*a*d^2+3*d/b/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d
-b*c)*b)^(1/2)*b)*c

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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((d*x+c)^(3/2)/(b*x+a)^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(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c positive or negative?

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mupad [B]  time = 0.11, size = 109, normalized size = 1.28 \begin {gather*} \frac {\left (a\,d^2-b\,c\,d\right )\,\sqrt {c+d\,x}}{b^3\,\left (c+d\,x\right )-b^3\,c+a\,b^2\,d}+\frac {2\,d\,\sqrt {c+d\,x}}{b^2}-\frac {3\,d\,\mathrm {atan}\left (\frac {\sqrt {b}\,d\,\sqrt {a\,d-b\,c}\,\sqrt {c+d\,x}}{a\,d^2-b\,c\,d}\right )\,\sqrt {a\,d-b\,c}}{b^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(3/2)/(a + b*x)^2,x)

[Out]

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

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sympy [F(-1)]  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((d*x+c)**(3/2)/(b*x+a)**2,x)

[Out]

Timed out

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