\(\int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}} \, dx\) [708]

Optimal result

Integrand size = 19, antiderivative size = 113 \[ \int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=\frac {3 (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^2}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}+\frac {3 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2} \sqrt {d}} \]

[Out]

3/4*(-a*d+b*c)^2*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(5/2)/d^(1/2)+1/2*(d*x+c)^(3/2)*(b*x+a
)^(1/2)/b+3/4*(-a*d+b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^2

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {52, 65, 223, 212} \[ \int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=\frac {3 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2} \sqrt {d}}+\frac {3 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)}{4 b^2}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b} \]

[In]

Int[(c + d*x)^(3/2)/Sqrt[a + b*x],x]

[Out]

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

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*} \text {integral}& = \frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}+\frac {(3 (b c-a d)) \int \frac {\sqrt {c+d x}}{\sqrt {a+b x}} \, dx}{4 b} \\ & = \frac {3 (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^2}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}+\frac {\left (3 (b c-a d)^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{8 b^2} \\ & = \frac {3 (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^2}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}+\frac {\left (3 (b c-a 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 )}{4 b^3} \\ & = \frac {3 (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^2}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}+\frac {\left (3 (b c-a 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 )}{4 b^3} \\ & = \frac {3 (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^2}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}+\frac {3 (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2} \sqrt {d}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.83 \[ \int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} (5 b c-3 a d+2 b d x)}{4 b^2}+\frac {3 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{4 b^{5/2} \sqrt {d}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.24

method	result	size
default	\(\frac {\left (d x +c \right )^{\frac {3}{2}} \sqrt {b x +a}}{2 b}-\frac {3 \left (a d -b c \right ) \left (\frac {\sqrt {b x +a}\, \sqrt {d x +c}}{b}-\frac {\left (a d -b c \right ) \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d x}{\sqrt {b d}}+\sqrt {b d \,x^{2}+\left (a d +b c \right ) x +a c}\right )}{2 b \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}}\right )}{4 b}\)	\(140\)

[In]

int((d*x+c)^(3/2)/(b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/2*(d*x+c)^(3/2)*(b*x+a)^(1/2)/b-3/4*(a*d-b*c)/b*((b*x+a)^(1/2)*(d*x+c)^(1/2)/b-1/2*(a*d-b*c)/b*((b*x+a)*(d*x
+c))^(1/2)/(d*x+c)^(1/2)/(b*x+a)^(1/2)*ln((1/2*a*d+1/2*b*c+b*d*x)/(b*d)^(1/2)+(b*d*x^2+(a*d+b*c)*x+a*c)^(1/2))
/(b*d)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 306, normalized size of antiderivative = 2.71 \[ \int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=\left [\frac {3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (2 \, b^{2} d^{2} x + 5 \, b^{2} c d - 3 \, a b d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, b^{3} d}, -\frac {3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (2 \, b^{2} d^{2} x + 5 \, b^{2} c d - 3 \, a b d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, b^{3} d}\right ] \]

[In]

integrate((d*x+c)^(3/2)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(3*(b^2*c^2 - 2*a*b*c*d + a^2*d^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*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(2*b^2*d^2*x + 5*b^2*c*d
- 3*a*b*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*d), -1/8*(3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(-b*d)*arctan(
1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*
x)) - 2*(2*b^2*d^2*x + 5*b^2*c*d - 3*a*b*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*d)]

Sympy [F]

\[ \int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=\int \frac {\left (c + d x\right )^{\frac {3}{2}}}{\sqrt {a + b x}}\, dx \]

[In]

integrate((d*x+c)**(3/2)/(b*x+a)**(1/2),x)

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((d*x+c)^(3/2)/(b*x+a)^(1/2),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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (87) = 174\).

Time = 0.30 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.06 \[ \int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=-\frac {\frac {4 \, {\left (\frac {{\left (b^{2} c - a b d\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d}} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a}\right )} c {\left | b \right |}}{b^{2}} - \frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, b x + 2 \, a + \frac {b c d - 5 \, a d^{2}}{d^{2}}\right )} \sqrt {b x + a} + \frac {{\left (b^{3} c^{2} + 2 \, a b^{2} c d - 3 \, a^{2} b d^{2}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d}\right )} d {\left | b \right |}}{b^{3}}}{4 \, b} \]

[In]

integrate((d*x+c)^(3/2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

-1/4*(4*((b^2*c - a*b*d)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/sqrt(b*d) -
sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a))*c*abs(b)/b^2 - (sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*b*x
+ 2*a + (b*c*d - 5*a*d^2)/d^2)*sqrt(b*x + a) + (b^3*c^2 + 2*a*b^2*c*d - 3*a^2*b*d^2)*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))*d*abs(b)/b^3)/b

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=\int \frac {{\left (c+d\,x\right )}^{3/2}}{\sqrt {a+b\,x}} \,d x \]

[In]

int((c + d*x)^(3/2)/(a + b*x)^(1/2),x)

[Out]

int((c + d*x)^(3/2)/(a + b*x)^(1/2), x)