\(\int \frac {(c+d x)^{3/2}}{a+b x} \, dx\) [1393]

Optimal result

Integrand size = 17, antiderivative size = 86 \[ \int \frac {(c+d x)^{3/2}}{a+b x} \, dx=\frac {2 (b c-a d) \sqrt {c+d x}}{b^2}+\frac {2 (c+d x)^{3/2}}{3 b}-\frac {2 (b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {52, 65, 214} \[ \int \frac {(c+d x)^{3/2}}{a+b x} \, dx=-\frac {2 (b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{5/2}}+\frac {2 \sqrt {c+d x} (b c-a d)}{b^2}+\frac {2 (c+d x)^{3/2}}{3 b} \]

[In]

Int[(c + d*x)^(3/2)/(a + b*x),x]

[Out]

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

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 214

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

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.90 \[ \int \frac {(c+d x)^{3/2}}{a+b x} \, dx=\frac {2 \sqrt {c+d x} (4 b c-3 a d+b d x)}{3 b^2}+\frac {2 (-b c+a d)^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{b^{5/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.73 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.02

method	result	size
risch	\(-\frac {2 \left (-b d x +3 a d -4 b c \right ) \sqrt {d x +c}}{3 b^{2}}+\frac {2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b^{2} \sqrt {\left (a d -b c \right ) b}}\)	\(88\)
pseudoelliptic	\(-\frac {2 \left (-\left (a d -b c \right )^{2} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\sqrt {\left (a d -b c \right ) b}\, \sqrt {d x +c}\, \left (\frac {\left (-d x -4 c \right ) b}{3}+a d \right )\right )}{\sqrt {\left (a d -b c \right ) b}\, b^{2}}\)	\(88\)
derivativedivides	\(-\frac {2 \left (-\frac {b \left (d x +c \right )^{\frac {3}{2}}}{3}+\sqrt {d x +c}\, a d -\sqrt {d x +c}\, b c \right )}{b^{2}}+\frac {2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b^{2} \sqrt {\left (a d -b c \right ) b}}\)	\(99\)
default	\(-\frac {2 \left (-\frac {b \left (d x +c \right )^{\frac {3}{2}}}{3}+\sqrt {d x +c}\, a d -\sqrt {d x +c}\, b c \right )}{b^{2}}+\frac {2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b^{2} \sqrt {\left (a d -b c \right ) b}}\)	\(99\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.19 \[ \int \frac {(c+d x)^{3/2}}{a+b x} \, dx=\left [-\frac {3 \, {\left (b c - a d\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d + 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) - 2 \, {\left (b d x + 4 \, b c - 3 \, a d\right )} \sqrt {d x + c}}{3 \, b^{2}}, -\frac {2 \, {\left (3 \, {\left (b c - a d\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) - {\left (b d x + 4 \, b c - 3 \, a d\right )} \sqrt {d x + c}\right )}}{3 \, b^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.32 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.22 \[ \int \frac {(c+d x)^{3/2}}{a+b x} \, dx=\begin {cases} \frac {2 \left (\frac {d \left (c + d x\right )^{\frac {3}{2}}}{3 b} + \frac {\sqrt {c + d x} \left (- a d^{2} + b c d\right )}{b^{2}} + \frac {d \left (a d - b c\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{b^{3} \sqrt {\frac {a d - b c}{b}}}\right )}{d} & \text {for}\: d \neq 0 \\c^{\frac {3}{2}} \left (\begin {cases} \frac {x}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x \right )}}{b} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(d*(c + d*x)**(3/2)/(3*b) + sqrt(c + d*x)*(-a*d**2 + b*c*d)/b**2 + d*(a*d - b*c)**2*atan(sqrt(c +
d*x)/sqrt((a*d - b*c)/b))/(b**3*sqrt((a*d - b*c)/b)))/d, Ne(d, 0)), (c**(3/2)*Piecewise((x/a, Eq(b, 0)), (log
(a + b*x)/b, True)), True))

Maxima [F(-2)]

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

[In]

integrate((d*x+c)^(3/2)/(b*x+a),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 [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.22 \[ \int \frac {(c+d x)^{3/2}}{a+b x} \, dx=\frac {2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{2}} + \frac {2 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} b^{2} + 3 \, \sqrt {d x + c} b^{2} c - 3 \, \sqrt {d x + c} a b d\right )}}{3 \, b^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.08 \[ \int \frac {(c+d x)^{3/2}}{a+b x} \, dx=\frac {2\,{\left (c+d\,x\right )}^{3/2}}{3\,b}-\frac {2\,\left (a\,d-b\,c\right )\,\sqrt {c+d\,x}}{b^2}+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {b}\,{\left (a\,d-b\,c\right )}^{3/2}\,\sqrt {c+d\,x}}{a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2}\right )\,{\left (a\,d-b\,c\right )}^{3/2}}{b^{5/2}} \]

[In]

int((c + d*x)^(3/2)/(a + b*x),x)

[Out]

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