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

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

[Out]

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

________________________________________________________________________________________

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

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.54, size = 111, normalized size = 1.21 \[ \frac {6 (b c-a d)^{3/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{3/2} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )-2 \sqrt {d} \sqrt {a+b x} (a d+3 b c+4 b d x)}{3 d^{5/2} (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[d]*Sqrt[a + b*x]*(3*b*c + a*d + 4*b*d*x) + 6*(b*c - a*d)^(3/2)*((b*(c + d*x))/(b*c - a*d))^(3/2)*ArcS
inh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(3*d^(5/2)*(c + d*x)^(3/2))

________________________________________________________________________________________

fricas [B]  time = 1.08, size = 325, normalized size = 3.53 \[ \left [\frac {3 \, {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \sqrt {\frac {b}{d}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d^{2} x + b c d + a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {b}{d}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (4 \, b d x + 3 \, b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c}}{6 \, {\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}}, -\frac {3 \, {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} d x^{2} + a b c + {\left (b^{2} c + a b d\right )} x\right )}}\right ) + 2 \, {\left (4 \, b d x + 3 \, b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c}}{3 \, {\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [B]  time = 1.75, size = 181, normalized size = 1.97 \[ -\frac {2 \, b^{3} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d^{2} {\left | b \right |}} - \frac {2 \, \sqrt {b x + a} {\left (\frac {4 \, {\left (b^{5} c d^{2} - a b^{4} d^{3}\right )} {\left (b x + a\right )}}{b c d^{3} {\left | b \right |} - a d^{4} {\left | b \right |}} + \frac {3 \, {\left (b^{6} c^{2} d - 2 \, a b^{5} c d^{2} + a^{2} b^{4} d^{3}\right )}}{b c d^{3} {\left | b \right |} - a d^{4} {\left | b \right |}}\right )}}{3 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________