3.15.75 \(\int \frac {(c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx\) [1475]

Optimal. Leaf size=98 \[ \frac {3 d \sqrt {a+b x} \sqrt {c+d x}}{b^2}-\frac {2 (c+d x)^{3/2}}{b \sqrt {a+b x}}+\frac {3 \sqrt {d} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{5/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {49, 52, 65, 223, 212} \begin {gather*} \frac {3 \sqrt {d} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{5/2}}+\frac {3 d \sqrt {a+b x} \sqrt {c+d x}}{b^2}-\frac {2 (c+d x)^{3/2}}{b \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*d*Sqrt[a + b*x]*Sqrt[c + d*x])/b^2 - (2*(c + d*x)^(3/2))/(b*Sqrt[a + b*x]) + (3*Sqrt[d]*(b*c - a*d)*ArcTanh
[(Sqrt[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 52

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 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)^{3/2}} \, dx &=-\frac {2 (c+d x)^{3/2}}{b \sqrt {a+b x}}+\frac {(3 d) \int \frac {\sqrt {c+d x}}{\sqrt {a+b x}} \, dx}{b}\\ &=\frac {3 d \sqrt {a+b x} \sqrt {c+d x}}{b^2}-\frac {2 (c+d x)^{3/2}}{b \sqrt {a+b x}}+\frac {(3 d (b c-a d)) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{2 b^2}\\ &=\frac {3 d \sqrt {a+b x} \sqrt {c+d x}}{b^2}-\frac {2 (c+d x)^{3/2}}{b \sqrt {a+b x}}+\frac {(3 d (b c-a d)) \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 {3 d \sqrt {a+b x} \sqrt {c+d x}}{b^2}-\frac {2 (c+d x)^{3/2}}{b \sqrt {a+b x}}+\frac {(3 d (b c-a d)) \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 {3 d \sqrt {a+b x} \sqrt {c+d x}}{b^2}-\frac {2 (c+d x)^{3/2}}{b \sqrt {a+b x}}+\frac {3 \sqrt {d} (b c-a d) \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 = 0.43, size = 89, normalized size = 0.91 \begin {gather*} \frac {\frac {b \sqrt {c+d x} (-2 b c+3 a d+b d x)}{\sqrt {a+b x}}+3 \sqrt {\frac {b}{d}} d (-b c+a d) \log \left (\sqrt {a+b x}-\sqrt {\frac {b}{d}} \sqrt {c+d x}\right )}{b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Sqrt[c + d*x]*(-2*b*c + 3*a*d + b*d*x))/Sqrt[a + b*x] + 3*Sqrt[b/d]*d*(-(b*c) + a*d)*Log[Sqrt[a + b*x] - S
qrt[b/d]*Sqrt[c + d*x]])/b^3

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((d*x+c)^(3/2)/(b*x+a)^(3/2),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)^(3/2),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 [A]
time = 1.29, size = 311, normalized size = 3.17 \begin {gather*} \left [-\frac {3 \, {\left (a b c - a^{2} d + {\left (b^{2} c - a b d\right )} x\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 (b d x - 2 \, b c + 3 \, a d\right )} \sqrt {b x + a} \sqrt {d x + c}}{4 \, {\left (b^{3} x + a b^{2}\right )}}, -\frac {3 \, {\left (a b c - a^{2} d + {\left (b^{2} c - a b d\right )} x\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 (b d x - 2 \, b c + 3 \, a d\right )} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{3} x + a 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)^(3/2),x, algorithm="fricas")

[Out]

[-1/4*(3*(a*b*c - a^2*d + (b^2*c - a*b*d)*x)*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*(b*d*x - 2*b*c
+ 3*a*d)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*x + a*b^2), -1/2*(3*(a*b*c - a^2*d + (b^2*c - a*b*d)*x)*sqrt(-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*(b*d*x - 2*b*c + 3*a*d)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*x + a*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 {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (78) = 156\).
time = 1.36, size = 204, normalized size = 2.08 \begin {gather*} \frac {\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} d {\left | b \right |}}{b^{4}} - \frac {3 \, {\left (\sqrt {b d} b c {\left | b \right |} - \sqrt {b d} a d {\left | b \right |}\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 )}{2 \, b^{4}} - \frac {4 \, {\left (\sqrt {b d} b^{2} c^{2} {\left | b \right |} - 2 \, \sqrt {b d} a b c d {\left | b \right |} + \sqrt {b d} a^{2} d^{2} {\left | b \right |}\right )}}{{\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 )} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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 )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________