3.15.76 \(\int \frac {(c+d x)^{3/2}}{(a+b x)^{5/2}} \, dx\) [1476]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.03, 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 = {49, 65, 223, 212} \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 49

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 65

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 212

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

Rule 223

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 ) \text {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 ) \text {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*}

________________________________________________________________________________________

Mathematica [A]
time = 0.14, size = 81, normalized size = 0.88 \begin {gather*} -\frac {2 \sqrt {c+d x} (3 a d+b (c+4 d x))}{3 b^2 (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 {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c + d*x]*(3*a*d + b*(c + 4*d*x)))/(3*b^2*(a + b*x)^(3/2)) + (2*d^(3/2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x]
)/(Sqrt[b]*Sqrt[c + d*x])])/b^(5/2)

________________________________________________________________________________________

Mathics [F(-1)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\left (d x +c \right )^{\frac {3}{2}}}{\left (b x +a \right )^{\frac {5}{2}}}\, dx\]

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)

________________________________________________________________________________________

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 detail

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 150 vs. \(2 (70) = 140\).
time = 0.39, 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)]

________________________________________________________________________________________

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)

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 181 vs. \(2 (70) = 140\).
time = 0.05, size = 237, normalized size = 2.58 \begin {gather*} \frac {2 \left (-\frac {\left (-12 b^{3} d^{4} c+12 b^{2} d^{5} a\right ) \sqrt {c+d x} \sqrt {c+d x}}{-9 b^{4} \left |d\right | c+9 b^{3} d \left |d\right | a}-\frac {9 b^{3} d^{4} c^{2}-18 b^{2} d^{5} a c+9 b d^{6} a^{2}}{-9 b^{4} \left |d\right | c+9 b^{3} d \left |d\right | a}\right ) \sqrt {c+d x} \sqrt {a d^{2}-b c d+b d \left (c+d x\right )}}{\left (a d^{2}-b c d+b d \left (c+d x\right )\right )^{2}}-\frac {2 d^{3} \ln \left |\sqrt {a d^{2}-b c d+b d \left (c+d x\right )}-\sqrt {b d} \sqrt {c+d x}\right |}{b^{2} \sqrt {b d} \left |d\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)

[Out]

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

________________________________________________________________________________________

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)

________________________________________________________________________________________