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

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

[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)

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Rubi [A]  time = 0.0537947, antiderivative size = 112, normalized size of antiderivative = 1., 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*}

Mathematica [A]  time = 0.125669, 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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Maple [B]  time = 0.006, size = 263, normalized size = 2.4 \begin{align*}{\frac{2}{5\,b} \left ( dx+c \right ) ^{{\frac{5}{2}}}}-{\frac{2\,ad}{3\,{b}^{2}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}+{\frac{2\,c}{3\,b} \left ( dx+c \right ) ^{{\frac{3}{2}}}}+2\,{\frac{{a}^{2}{d}^{2}\sqrt{dx+c}}{{b}^{3}}}-4\,{\frac{acd\sqrt{dx+c}}{{b}^{2}}}+2\,{\frac{{c}^{2}\sqrt{dx+c}}{b}}-2\,{\frac{{a}^{3}{d}^{3}}{{b}^{3}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+6\,{\frac{{a}^{2}c{d}^{2}}{{b}^{2}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }-6\,{\frac{a{c}^{2}d}{b\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+2\,{\frac{{c}^{3}}{\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) } \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [A]  time = 2.44269, size = 644, normalized size = 5.75 \begin{align*} \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 ] \end{align*}

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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Sympy [A]  time = 33.8379, size = 121, normalized size = 1.08 \begin{align*} \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}}} \end{align*}

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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Giac [A]  time = 1.19369, size = 231, normalized size = 2.06 \begin{align*} \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}} \end{align*}

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