\(\int \frac {(c+d x)^{5/2}}{(a+b x)^3} \, dx\) [1407]

Optimal result

Integrand size = 17, antiderivative size = 119 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^3} \, dx=\frac {15 d^2 \sqrt {c+d x}}{4 b^3}-\frac {5 d (c+d x)^{3/2}}{4 b^2 (a+b x)}-\frac {(c+d x)^{5/2}}{2 b (a+b x)^2}-\frac {15 d^2 \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 b^{7/2}} \]

[Out]

-5/4*d*(d*x+c)^(3/2)/b^2/(b*x+a)-1/2*(d*x+c)^(5/2)/b/(b*x+a)^2-15/4*d^2*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*
c)^(1/2))*(-a*d+b*c)^(1/2)/b^(7/2)+15/4*d^2*(d*x+c)^(1/2)/b^3

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {43, 52, 65, 214} \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^3} \, dx=-\frac {15 d^2 \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 b^{7/2}}-\frac {5 d (c+d x)^{3/2}}{4 b^2 (a+b x)}-\frac {(c+d x)^{5/2}}{2 b (a+b x)^2}+\frac {15 d^2 \sqrt {c+d x}}{4 b^3} \]

[In]

Int[(c + d*x)^(5/2)/(a + b*x)^3,x]

[Out]

(15*d^2*Sqrt[c + d*x])/(4*b^3) - (5*d*(c + d*x)^(3/2))/(4*b^2*(a + b*x)) - (c + d*x)^(5/2)/(2*b*(a + b*x)^2) -
(15*d^2*Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(4*b^(7/2))

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)^{5/2}}{2 b (a+b x)^2}+\frac {(5 d) \int \frac {(c+d x)^{3/2}}{(a+b x)^2} \, dx}{4 b} \\ & = -\frac {5 d (c+d x)^{3/2}}{4 b^2 (a+b x)}-\frac {(c+d x)^{5/2}}{2 b (a+b x)^2}+\frac {\left (15 d^2\right ) \int \frac {\sqrt {c+d x}}{a+b x} \, dx}{8 b^2} \\ & = \frac {15 d^2 \sqrt {c+d x}}{4 b^3}-\frac {5 d (c+d x)^{3/2}}{4 b^2 (a+b x)}-\frac {(c+d x)^{5/2}}{2 b (a+b x)^2}+\frac {\left (15 d^2 (b c-a d)\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{8 b^3} \\ & = \frac {15 d^2 \sqrt {c+d x}}{4 b^3}-\frac {5 d (c+d x)^{3/2}}{4 b^2 (a+b x)}-\frac {(c+d x)^{5/2}}{2 b (a+b x)^2}+\frac {(15 d (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 )}{4 b^3} \\ & = \frac {15 d^2 \sqrt {c+d x}}{4 b^3}-\frac {5 d (c+d x)^{3/2}}{4 b^2 (a+b x)}-\frac {(c+d x)^{5/2}}{2 b (a+b x)^2}-\frac {15 d^2 \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 b^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^3} \, dx=\frac {\sqrt {c+d x} \left (15 a^2 d^2-5 a b d (c-5 d x)+b^2 \left (-2 c^2-9 c d x+8 d^2 x^2\right )\right )}{4 b^3 (a+b x)^2}-\frac {15 d^2 \sqrt {-b c+a d} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{4 b^{7/2}} \]

[In]

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

[Out]

(Sqrt[c + d*x]*(15*a^2*d^2 - 5*a*b*d*(c - 5*d*x) + b^2*(-2*c^2 - 9*c*d*x + 8*d^2*x^2)))/(4*b^3*(a + b*x)^2) -
(15*d^2*Sqrt[-(b*c) + a*d]*ArcTan[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/(4*b^(7/2))

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.98

method	result	size
risch	\(\frac {2 d^{2} \sqrt {d x +c}}{b^{3}}-\frac {\left (2 a d -2 b c \right ) d^{2} \left (\frac {-\frac {9 b \left (d x +c \right )^{\frac {3}{2}}}{8}+\left (-\frac {7 a d}{8}+\frac {7 b c}{8}\right ) \sqrt {d x +c}}{\left (\left (d x +c \right ) b +a d -b c \right )^{2}}+\frac {15 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \sqrt {\left (a d -b c \right ) b}}\right )}{b^{3}}\)	\(117\)
pseudoelliptic	\(-\frac {15 \left (d^{2} \left (b x +a \right )^{2} \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )-\sqrt {d x +c}\, \left (\left (\frac {8}{15} d^{2} x^{2}-\frac {3}{5} c d x -\frac {2}{15} c^{2}\right ) b^{2}-\frac {a d \left (-5 d x +c \right ) b}{3}+a^{2} d^{2}\right ) \sqrt {\left (a d -b c \right ) b}\right )}{4 \sqrt {\left (a d -b c \right ) b}\, b^{3} \left (b x +a \right )^{2}}\)	\(130\)
derivativedivides	\(2 d^{2} \left (\frac {\sqrt {d x +c}}{b^{3}}-\frac {\frac {\left (-\frac {9}{8} a b d +\frac {9}{8} b^{2} c \right ) \left (d x +c \right )^{\frac {3}{2}}+\left (-\frac {7}{8} a^{2} d^{2}+\frac {7}{4} a b c d -\frac {7}{8} b^{2} c^{2}\right ) \sqrt {d x +c}}{\left (\left (d x +c \right ) b +a d -b c \right )^{2}}+\frac {15 \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \sqrt {\left (a d -b c \right ) b}}}{b^{3}}\right )\)	\(138\)
default	\(2 d^{2} \left (\frac {\sqrt {d x +c}}{b^{3}}-\frac {\frac {\left (-\frac {9}{8} a b d +\frac {9}{8} b^{2} c \right ) \left (d x +c \right )^{\frac {3}{2}}+\left (-\frac {7}{8} a^{2} d^{2}+\frac {7}{4} a b c d -\frac {7}{8} b^{2} c^{2}\right ) \sqrt {d x +c}}{\left (\left (d x +c \right ) b +a d -b c \right )^{2}}+\frac {15 \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \sqrt {\left (a d -b c \right ) b}}}{b^{3}}\right )\)	\(138\)

[In]

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

[Out]

2*d^2*(d*x+c)^(1/2)/b^3-1/b^3*(2*a*d-2*b*c)*d^2*((-9/8*b*(d*x+c)^(3/2)+(-7/8*a*d+7/8*b*c)*(d*x+c)^(1/2))/((d*x
+c)*b+a*d-b*c)^2+15/8/((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 = 344, normalized size of antiderivative = 2.89 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^3} \, dx=\left [\frac {15 \, {\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + 2 \, {\left (8 \, b^{2} d^{2} x^{2} - 2 \, b^{2} c^{2} - 5 \, a b c d + 15 \, a^{2} d^{2} - {\left (9 \, b^{2} c d - 25 \, a b d^{2}\right )} x\right )} \sqrt {d x + c}}{8 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac {15 \, {\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) - {\left (8 \, b^{2} d^{2} x^{2} - 2 \, b^{2} c^{2} - 5 \, a b c d + 15 \, a^{2} d^{2} - {\left (9 \, b^{2} c d - 25 \, a b d^{2}\right )} x\right )} \sqrt {d x + c}}{4 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}\right ] \]

[In]

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

[Out]

[1/8*(15*(b^2*d^2*x^2 + 2*a*b*d^2*x + a^2*d^2)*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*(8*b^2*d^2*x^2 - 2*b^2*c^2 - 5*a*b*c*d + 15*a^2*d^2 - (9*b^2*c*d - 25*a*
b*d^2)*x)*sqrt(d*x + c))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3), -1/4*(15*(b^2*d^2*x^2 + 2*a*b*d^2*x + a^2*d^2)*sqrt(
-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) - (8*b^2*d^2*x^2 - 2*b^2*c^2 - 5*a*b
*c*d + 15*a^2*d^2 - (9*b^2*c*d - 25*a*b*d^2)*x)*sqrt(d*x + c))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)]

Sympy [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^3} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((d*x+c)^(5/2)/(b*x+a)^3,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 = 0.33 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.44 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^3} \, dx=\frac {2 \, \sqrt {d x + c} d^{2}}{b^{3}} + \frac {15 \, {\left (b c d^{2} - a d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{4 \, \sqrt {-b^{2} c + a b d} b^{3}} - \frac {9 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} c d^{2} - 7 \, \sqrt {d x + c} b^{2} c^{2} d^{2} - 9 \, {\left (d x + c\right )}^{\frac {3}{2}} a b d^{3} + 14 \, \sqrt {d x + c} a b c d^{3} - 7 \, \sqrt {d x + c} a^{2} d^{4}}{4 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}^{2} b^{3}} \]

[In]

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

[Out]

2*sqrt(d*x + c)*d^2/b^3 + 15/4*(b*c*d^2 - a*d^3)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a
*b*d)*b^3) - 1/4*(9*(d*x + c)^(3/2)*b^2*c*d^2 - 7*sqrt(d*x + c)*b^2*c^2*d^2 - 9*(d*x + c)^(3/2)*a*b*d^3 + 14*s
qrt(d*x + c)*a*b*c*d^3 - 7*sqrt(d*x + c)*a^2*d^4)/(((d*x + c)*b - b*c + a*d)^2*b^3)

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.67 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^3} \, dx=\frac {2\,d^2\,\sqrt {c+d\,x}}{b^3}-\frac {\left (\frac {9\,b^2\,c\,d^2}{4}-\frac {9\,a\,b\,d^3}{4}\right )\,{\left (c+d\,x\right )}^{3/2}-\sqrt {c+d\,x}\,\left (\frac {7\,a^2\,d^4}{4}-\frac {7\,a\,b\,c\,d^3}{2}+\frac {7\,b^2\,c^2\,d^2}{4}\right )}{b^5\,{\left (c+d\,x\right )}^2-\left (2\,b^5\,c-2\,a\,b^4\,d\right )\,\left (c+d\,x\right )+b^5\,c^2+a^2\,b^3\,d^2-2\,a\,b^4\,c\,d}-\frac {15\,d^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,d^2\,\sqrt {a\,d-b\,c}\,\sqrt {c+d\,x}}{a\,d^3-b\,c\,d^2}\right )\,\sqrt {a\,d-b\,c}}{4\,b^{7/2}} \]

[In]

int((c + d*x)^(5/2)/(a + b*x)^3,x)

[Out]

(2*d^2*(c + d*x)^(1/2))/b^3 - (((9*b^2*c*d^2)/4 - (9*a*b*d^3)/4)*(c + d*x)^(3/2) - (c + d*x)^(1/2)*((7*a^2*d^4
)/4 + (7*b^2*c^2*d^2)/4 - (7*a*b*c*d^3)/2))/(b^5*(c + d*x)^2 - (2*b^5*c - 2*a*b^4*d)*(c + d*x) + b^5*c^2 + a^2
*b^3*d^2 - 2*a*b^4*c*d) - (15*d^2*atan((b^(1/2)*d^2*(a*d - b*c)^(1/2)*(c + d*x)^(1/2))/(a*d^3 - b*c*d^2))*(a*d
- b*c)^(1/2))/(4*b^(7/2))