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

Optimal. Leaf size=112 \[ -\frac {2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{7/2}}+\frac {2 \sqrt {c+d x} (b c-a d)^2}{b^3}+\frac {2 (c+d x)^{3/2} (b c-a d)}{3 b^2}+\frac {2 (c+d x)^{5/2}}{5 b} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {50, 63, 208} \[ \frac {2 \sqrt {c+d x} (b c-a d)^2}{b^3}+\frac {2 (c+d x)^{3/2} (b c-a d)}{3 b^2}-\frac {2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{7/2}}+\frac {2 (c+d x)^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/2)/(a + b*x),x]

[Out]

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

Rule 50

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 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 208

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

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 105, normalized size = 0.94 \[ \frac {2 (b c-a d) \left (\sqrt {b} \sqrt {c+d x} (-3 a d+4 b c+b d x)-3 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )\right )}{3 b^{7/2}}+\frac {2 (c+d x)^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.64, size = 290, normalized size = 2.59 \[ \left [\frac {15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + 2 \, {\left (3 \, b^{2} d^{2} x^{2} + 23 \, b^{2} c^{2} - 35 \, a b c d + 15 \, a^{2} d^{2} + {\left (11 \, b^{2} c d - 5 \, a b d^{2}\right )} x\right )} \sqrt {d x + c}}{15 \, b^{3}}, -\frac {2 \, {\left (15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) - {\left (3 \, b^{2} d^{2} x^{2} + 23 \, b^{2} c^{2} - 35 \, a b c d + 15 \, a^{2} d^{2} + {\left (11 \, b^{2} c d - 5 \, a b d^{2}\right )} x\right )} \sqrt {d x + c}\right )}}{15 \, b^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/15*(15*(b^2*c^2 - 2*a*b*c*d + 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*sqr
t((b*c - a*d)/b))/(b*x + a)) + 2*(3*b^2*d^2*x^2 + 23*b^2*c^2 - 35*a*b*c*d + 15*a^2*d^2 + (11*b^2*c*d - 5*a*b*d
^2)*x)*sqrt(d*x + c))/b^3, -2/15*(15*(b^2*c^2 - 2*a*b*c*d + 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)) - (3*b^2*d^2*x^2 + 23*b^2*c^2 - 35*a*b*c*d + 15*a^2*d^2 + (11*b^2*c*d -
5*a*b*d^2)*x)*sqrt(d*x + c))/b^3]

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giac [A]  time = 1.30, size = 171, normalized size = 1.53 \[ \frac {2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{3}} + \frac {2 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{4} + 5 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{4} c + 15 \, \sqrt {d x + c} b^{4} c^{2} - 5 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{3} d - 30 \, \sqrt {d x + c} a b^{3} c d + 15 \, \sqrt {d x + c} a^{2} b^{2} d^{2}\right )}}{15 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*
c + a*b*d)*b^3) + 2/15*(3*(d*x + c)^(5/2)*b^4 + 5*(d*x + c)^(3/2)*b^4*c + 15*sqrt(d*x + c)*b^4*c^2 - 5*(d*x +
c)^(3/2)*a*b^3*d - 30*sqrt(d*x + c)*a*b^3*c*d + 15*sqrt(d*x + c)*a^2*b^2*d^2)/b^5

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maple [B]  time = 0.01, size = 263, normalized size = 2.35 \[ -\frac {2 a^{3} d^{3} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, b^{3}}+\frac {6 a^{2} c \,d^{2} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, b^{2}}-\frac {6 a \,c^{2} d \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, b}+\frac {2 c^{3} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}+\frac {2 \sqrt {d x +c}\, a^{2} d^{2}}{b^{3}}-\frac {4 \sqrt {d x +c}\, a c d}{b^{2}}+\frac {2 \sqrt {d x +c}\, c^{2}}{b}-\frac {2 \left (d x +c \right )^{\frac {3}{2}} a d}{3 b^{2}}+\frac {2 \left (d x +c \right )^{\frac {3}{2}} c}{3 b}+\frac {2 \left (d x +c \right )^{\frac {5}{2}}}{5 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(d*x+c)^(5/2)/b-2/3/b^2*(d*x+c)^(3/2)*a*d+2/3/b*(d*x+c)^(3/2)*c+2/b^3*(d*x+c)^(1/2)*a^2*d^2-4/b^2*(d*x+c)^
(1/2)*a*c*d+2/b*(d*x+c)^(1/2)*c^2-2/b^3/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)*a^3*d^
3+6/b^2/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)*a^2*c*d^2-6/b/((a*d-b*c)*b)^(1/2)*arct
an((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)*a*c^2*d+2/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)
*b)*c^3

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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((d*x+c)^(5/2)/(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 details)Is a*d-b*c positive or negative?

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mupad [B]  time = 0.40, size = 130, normalized size = 1.16 \[ \frac {2\,{\left (c+d\,x\right )}^{5/2}}{5\,b}-\frac {2\,\left (a\,d-b\,c\right )\,{\left (c+d\,x\right )}^{3/2}}{3\,b^2}+\frac {2\,{\left (a\,d-b\,c\right )}^2\,\sqrt {c+d\,x}}{b^3}-\frac {2\,\mathrm {atan}\left (\frac {\sqrt {b}\,{\left (a\,d-b\,c\right )}^{5/2}\,\sqrt {c+d\,x}}{a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3}\right )\,{\left (a\,d-b\,c\right )}^{5/2}}{b^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(5/2)/(a + b*x),x)

[Out]

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

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sympy [A]  time = 26.84, size = 121, normalized size = 1.08 \[ \frac {2 \left (c + d x\right )^{\frac {5}{2}}}{5 b} + \frac {\left (c + d x\right )^{\frac {3}{2}} \left (- 2 a d + 2 b c\right )}{3 b^{2}} + \frac {\sqrt {c + d x} \left (2 a^{2} d^{2} - 4 a b c d + 2 b^{2} c^{2}\right )}{b^{3}} - \frac {2 \left (a d - b c\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{b^{4} \sqrt {\frac {a d - b c}{b}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(c + d*x)**(5/2)/(5*b) + (c + d*x)**(3/2)*(-2*a*d + 2*b*c)/(3*b**2) + sqrt(c + d*x)*(2*a**2*d**2 - 4*a*b*c*d
+ 2*b**2*c**2)/b**3 - 2*(a*d - b*c)**3*atan(sqrt(c + d*x)/sqrt((a*d - b*c)/b))/(b**4*sqrt((a*d - b*c)/b))

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