\(\int \frac {(c+d x)^{3/2}}{(a+b x)^2} \, dx\) [1394]

Optimal result

Integrand size = 17, antiderivative size = 85 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^2} \, dx=\frac {3 d \sqrt {c+d x}}{b^2}-\frac {(c+d x)^{3/2}}{b (a+b x)}-\frac {3 d \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {43, 52, 65, 214} \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^2} \, dx=-\frac {3 d \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{5/2}}-\frac {(c+d x)^{3/2}}{b (a+b x)}+\frac {3 d \sqrt {c+d x}}{b^2} \]

[In]

Int[(c + d*x)^(3/2)/(a + b*x)^2,x]

[Out]

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

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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 {(c+d x)^{3/2}}{b (a+b x)}+\frac {(3 d) \int \frac {\sqrt {c+d x}}{a+b x} \, dx}{2 b} \\ & = \frac {3 d \sqrt {c+d x}}{b^2}-\frac {(c+d x)^{3/2}}{b (a+b x)}+\frac {(3 d (b c-a d)) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{2 b^2} \\ & = \frac {3 d \sqrt {c+d x}}{b^2}-\frac {(c+d x)^{3/2}}{b (a+b x)}+\frac {(3 (b c-a d)) \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} \\ & = \frac {3 d \sqrt {c+d x}}{b^2}-\frac {(c+d x)^{3/2}}{b (a+b x)}-\frac {3 d \sqrt {b c-a d} \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 = 0.22 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.98 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^2} \, dx=\frac {\sqrt {c+d x} (-b c+3 a d+2 b d x)}{b^2 (a+b x)}-\frac {3 d \sqrt {-b c+a d} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{b^{5/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.09

method	result	size
risch	\(\frac {2 d \sqrt {d x +c}}{b^{2}}-\frac {\left (2 a d -2 b c \right ) d \left (-\frac {\sqrt {d x +c}}{2 \left (\left (d x +c \right ) b +a d -b c \right )}+\frac {3 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}\right )}{b^{2}}\)	\(93\)
pseudoelliptic	\(\frac {-3 d \left (b x +a \right ) \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )+3 \sqrt {\left (a d -b c \right ) b}\, \sqrt {d x +c}\, \left (\frac {\left (2 d x -c \right ) b}{3}+a d \right )}{b^{2} \left (b x +a \right ) \sqrt {\left (a d -b c \right ) b}}\)	\(99\)
derivativedivides	\(2 d \left (\frac {\sqrt {d x +c}}{b^{2}}-\frac {\frac {\left (-\frac {a d}{2}+\frac {b c}{2}\right ) \sqrt {d x +c}}{\left (d x +c \right ) b +a d -b c}+\frac {3 \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}}{b^{2}}\right )\)	\(100\)
default	\(2 d \left (\frac {\sqrt {d x +c}}{b^{2}}-\frac {\frac {\left (-\frac {a d}{2}+\frac {b c}{2}\right ) \sqrt {d x +c}}{\left (d x +c \right ) b +a d -b c}+\frac {3 \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}}{b^{2}}\right )\)	\(100\)

[In]

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

[Out]

2*d*(d*x+c)^(1/2)/b^2-1/b^2*(2*a*d-2*b*c)*d*(-1/2*(d*x+c)^(1/2)/((d*x+c)*b+a*d-b*c)+3/2/((a*d-b*c)*b)^(1/2)*ar
ctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.47 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^2} \, dx=\left [\frac {3 \, {\left (b d x + a d\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + 2 \, {\left (2 \, b d x - b c + 3 \, a d\right )} \sqrt {d x + c}}{2 \, {\left (b^{3} x + a b^{2}\right )}}, -\frac {3 \, {\left (b d x + a d\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) - {\left (2 \, b d x - b c + 3 \, a d\right )} \sqrt {d x + c}}{b^{3} x + a b^{2}}\right ] \]

[In]

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

[Out]

[1/2*(3*(b*d*x + 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*(2*b*d*x - b*c + 3*a*d)*sqrt(d*x + c))/(b^3*x + a*b^2), -(3*(b*d*x + a*d)*sqrt(-(b*c - a*d)/b)*ar
ctan(-sqrt(d*x + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) - (2*b*d*x - b*c + 3*a*d)*sqrt(d*x + c))/(b^3*x + a*b^
2)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

none

Time = 0.31 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.33 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^2} \, dx=\frac {2 \, \sqrt {d x + c} d}{b^{2}} + \frac {3 \, {\left (b c d - a d^{2}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{2}} - \frac {\sqrt {d x + c} b c d - \sqrt {d x + c} a d^{2}}{{\left ({\left (d x + c\right )} b - b c + a d\right )} b^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.28 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^2} \, dx=\frac {\left (a\,d^2-b\,c\,d\right )\,\sqrt {c+d\,x}}{b^3\,\left (c+d\,x\right )-b^3\,c+a\,b^2\,d}+\frac {2\,d\,\sqrt {c+d\,x}}{b^2}-\frac {3\,d\,\mathrm {atan}\left (\frac {\sqrt {b}\,d\,\sqrt {a\,d-b\,c}\,\sqrt {c+d\,x}}{a\,d^2-b\,c\,d}\right )\,\sqrt {a\,d-b\,c}}{b^{5/2}} \]

[In]

int((c + d*x)^(3/2)/(a + b*x)^2,x)

[Out]

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