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

Optimal. Leaf size=110 \[ \frac{5 d \sqrt{c+d x} (b c-a d)}{b^3}-\frac{5 d (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{7/2}}-\frac{(c+d x)^{5/2}}{b (a+b x)}+\frac{5 d (c+d x)^{3/2}}{3 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0569942, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {47, 50, 63, 208} \[ \frac{5 d \sqrt{c+d x} (b c-a d)}{b^3}-\frac{5 d (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{7/2}}-\frac{(c+d x)^{5/2}}{b (a+b x)}+\frac{5 d (c+d x)^{3/2}}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/2)/(a + b*x)^2,x]

[Out]

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

Rule 47

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},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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

Mathematica [C]  time = 0.0144742, size = 50, normalized size = 0.45 \[ \frac{2 d (c+d x)^{7/2} \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};-\frac{b (c+d x)}{a d-b c}\right )}{7 (a d-b c)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*(c + d*x)^(7/2)*Hypergeometric2F1[2, 7/2, 9/2, -((b*(c + d*x))/(-(b*c) + a*d))])/(7*(-(b*c) + a*d)^2)

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Maple [B]  time = 0.013, size = 258, normalized size = 2.4 \begin{align*}{\frac{2\,d}{3\,{b}^{2}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}-4\,{\frac{a{d}^{2}\sqrt{dx+c}}{{b}^{3}}}+4\,{\frac{d\sqrt{dx+c}c}{{b}^{2}}}-{\frac{{a}^{2}{d}^{3}}{{b}^{3} \left ( bdx+ad \right ) }\sqrt{dx+c}}+2\,{\frac{\sqrt{dx+c}ac{d}^{2}}{{b}^{2} \left ( bdx+ad \right ) }}-{\frac{d{c}^{2}}{b \left ( bdx+ad \right ) }\sqrt{dx+c}}+5\,{\frac{{a}^{2}{d}^{3}}{{b}^{3}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }-10\,{\frac{ac{d}^{2}}{{b}^{2}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+5\,{\frac{d{c}^{2}}{b\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)^2,x)

[Out]

2/3*d*(d*x+c)^(3/2)/b^2-4/b^3*a*d^2*(d*x+c)^(1/2)+4*d/b^2*(d*x+c)^(1/2)*c-1/b^3*(d*x+c)^(1/2)/(b*d*x+a*d)*a^2*
d^3+2/b^2*(d*x+c)^(1/2)/(b*d*x+a*d)*a*c*d^2-d/b*(d*x+c)^(1/2)/(b*d*x+a*d)*c^2+5/b^3/((a*d-b*c)*b)^(1/2)*arctan
(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*a^2*d^3-10/b^2/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^
(1/2))*a*c*d^2+5*d/b/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*c^2

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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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.67534, size = 707, normalized size = 6.43 \begin{align*} \left [-\frac{15 \,{\left (a b c d - a^{2} d^{2} +{\left (b^{2} c d - a b d^{2}\right )} x\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^{2} d^{2} x^{2} - 3 \, b^{2} c^{2} + 20 \, a b c d - 15 \, a^{2} d^{2} + 2 \,{\left (7 \, b^{2} c d - 5 \, a b d^{2}\right )} x\right )} \sqrt{d x + c}}{6 \,{\left (b^{4} x + a b^{3}\right )}}, -\frac{15 \,{\left (a b c d - a^{2} d^{2} +{\left (b^{2} c d - a b d^{2}\right )} x\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^{2} d^{2} x^{2} - 3 \, b^{2} c^{2} + 20 \, a b c d - 15 \, a^{2} d^{2} + 2 \,{\left (7 \, b^{2} c d - 5 \, a b d^{2}\right )} x\right )} \sqrt{d x + c}}{3 \,{\left (b^{4} x + a b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(15*(a*b*c*d - a^2*d^2 + (b^2*c*d - a*b*d^2)*x)*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^2*d^2*x^2 - 3*b^2*c^2 + 20*a*b*c*d - 15*a^2*d^2 + 2*(7*b^2*c
*d - 5*a*b*d^2)*x)*sqrt(d*x + c))/(b^4*x + a*b^3), -1/3*(15*(a*b*c*d - a^2*d^2 + (b^2*c*d - a*b*d^2)*x)*sqrt(-
(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) - (2*b^2*d^2*x^2 - 3*b^2*c^2 + 20*a*b
*c*d - 15*a^2*d^2 + 2*(7*b^2*c*d - 5*a*b*d^2)*x)*sqrt(d*x + c))/(b^4*x + a*b^3)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.196, size = 244, normalized size = 2.22 \begin{align*} \frac{5 \,{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} 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{\sqrt{d x + c} b^{2} c^{2} d - 2 \, \sqrt{d x + c} a b c d^{2} + \sqrt{d x + c} a^{2} d^{3}}{{\left ({\left (d x + c\right )} b - b c + a d\right )} b^{3}} + \frac{2 \,{\left ({\left (d x + c\right )}^{\frac{3}{2}} b^{4} d + 6 \, \sqrt{d x + c} b^{4} c d - 6 \, \sqrt{d x + c} a b^{3} d^{2}\right )}}{3 \, b^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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