3.14.70 \(\int \frac {(c+d x)^{3/2}}{(a+b x)^{5/2}} \, dx\)

Optimal. Leaf size=92 \[ \frac {2 d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{5/2}}-\frac {2 d \sqrt {c+d x}}{b^2 \sqrt {a+b x}}-\frac {2 (c+d x)^{3/2}}{3 b (a+b x)^{3/2}} \]

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Rubi [A]  time = 0.04, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {47, 63, 217, 206} \begin {gather*} \frac {2 d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{5/2}}-\frac {2 d \sqrt {c+d x}}{b^2 \sqrt {a+b x}}-\frac {2 (c+d x)^{3/2}}{3 b (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(3/2)/(a + b*x)^(5/2),x]

[Out]

(-2*d*Sqrt[c + d*x])/(b^2*Sqrt[a + b*x]) - (2*(c + d*x)^(3/2))/(3*b*(a + b*x)^(3/2)) + (2*d^(3/2)*ArcTanh[(Sqr
t[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/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 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 206

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

Rule 217

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

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Mathematica [C]  time = 0.05, size = 73, normalized size = 0.79 \begin {gather*} -\frac {2 (c+d x)^{3/2} \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};\frac {d (a+b x)}{a d-b c}\right )}{3 b (a+b x)^{3/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(3/2)/(a + b*x)^(5/2),x]

[Out]

(-2*(c + d*x)^(3/2)*Hypergeometric2F1[-3/2, -3/2, -1/2, (d*(a + b*x))/(-(b*c) + a*d)])/(3*b*(a + b*x)^(3/2)*((
b*(c + d*x))/(b*c - a*d))^(3/2))

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IntegrateAlgebraic [A]  time = 0.14, size = 85, normalized size = 0.92 \begin {gather*} \frac {2 d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{5/2}}-\frac {2 (c+d x)^{3/2} \left (\frac {3 d (a+b x)}{c+d x}+b\right )}{3 b^2 (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(c + d*x)^(3/2)/(a + b*x)^(5/2),x]

[Out]

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

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fricas [B]  time = 2.06, size = 325, normalized size = 3.53 \begin {gather*} \left [\frac {3 \, {\left (b^{2} d x^{2} + 2 \, a b d x + a^{2} d\right )} \sqrt {\frac {d}{b}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b^{2} d x + b^{2} c + a b d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {d}{b}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (4 \, b d x + b c + 3 \, a d\right )} \sqrt {b x + a} \sqrt {d x + c}}{6 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}}, -\frac {3 \, {\left (b^{2} d x^{2} + 2 \, a b d x + a^{2} d\right )} \sqrt {-\frac {d}{b}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {d}{b}}}{2 \, {\left (b d^{2} x^{2} + a c d + {\left (b c d + a d^{2}\right )} x\right )}}\right ) + 2 \, {\left (4 \, b d x + b c + 3 \, a d\right )} \sqrt {b x + a} \sqrt {d x + c}}{3 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(3/2)/(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(b^2*d*x^2 + 2*a*b*d*x + a^2*d)*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b^2
*d*x + b^2*c + a*b*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(4*b*d*x + b*c + 3*
a*d)*sqrt(b*x + a)*sqrt(d*x + c))/(b^4*x^2 + 2*a*b^3*x + a^2*b^2), -1/3*(3*(b^2*d*x^2 + 2*a*b*d*x + a^2*d)*sqr
t(-d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-d/b)/(b*d^2*x^2 + a*c*d + (b*c*d +
a*d^2)*x)) + 2*(4*b*d*x + b*c + 3*a*d)*sqrt(b*x + a)*sqrt(d*x + c))/(b^4*x^2 + 2*a*b^3*x + a^2*b^2)]

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giac [B]  time = 1.74, size = 455, normalized size = 4.95 \begin {gather*} -\frac {\sqrt {b d} d {\left | b \right |} \log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{b^{4}} - \frac {8 \, {\left (2 \, \sqrt {b d} b^{5} c^{3} d {\left | b \right |} - 6 \, \sqrt {b d} a b^{4} c^{2} d^{2} {\left | b \right |} + 6 \, \sqrt {b d} a^{2} b^{3} c d^{3} {\left | b \right |} - 2 \, \sqrt {b d} a^{3} b^{2} d^{4} {\left | b \right |} - 3 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{3} c^{2} d {\left | b \right |} + 6 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b^{2} c d^{2} {\left | b \right |} - 3 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{2} b d^{3} {\left | b \right |} + 3 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} b c d {\left | b \right |} - 3 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} a d^{2} {\left | b \right |}\right )}}{3 \, {\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}^{3} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(3/2)/(b*x+a)^(5/2),x, algorithm="giac")

[Out]

-sqrt(b*d)*d*abs(b)*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/b^4 - 8/3*(2*sqrt(b
*d)*b^5*c^3*d*abs(b) - 6*sqrt(b*d)*a*b^4*c^2*d^2*abs(b) + 6*sqrt(b*d)*a^2*b^3*c*d^3*abs(b) - 2*sqrt(b*d)*a^3*b
^2*d^4*abs(b) - 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^3*c^2*d*abs(b)
+ 6*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^2*c*d^2*abs(b) - 3*sqrt(b
*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b*d^3*abs(b) + 3*sqrt(b*d)*(sqrt(b*d
)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b*c*d*abs(b) - 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) -
sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a*d^2*abs(b))/((b^2*c - a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c
+ (b*x + a)*b*d - a*b*d))^2)^3*b^3)

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d x +c \right )^{\frac {3}{2}}}{\left (b x +a \right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(3/2)/(b*x+a)^(5/2),x)

[Out]

int((d*x+c)^(3/2)/(b*x+a)^(5/2),x)

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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)^(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(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c+d\,x\right )}^{3/2}}{{\left (a+b\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(3/2)/(a + b*x)^(5/2),x)

[Out]

int((c + d*x)^(3/2)/(a + b*x)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c + d x\right )^{\frac {3}{2}}}{\left (a + b x\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(3/2)/(b*x+a)**(5/2),x)

[Out]

Integral((c + d*x)**(3/2)/(a + b*x)**(5/2), x)

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