3.1407 \(\int \frac{(c+d x)^{5/2}}{(a+b x)^3} \, dx\)

Optimal. Leaf size=119 \[ -\frac{15 d^2 \sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{4 b^{7/2}}-\frac{5 d (c+d x)^{3/2}}{4 b^2 (a+b x)}-\frac{(c+d x)^{5/2}}{2 b (a+b x)^2}+\frac{15 d^2 \sqrt{c+d x}}{4 b^3} \]

[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))

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Rubi [A]  time = 0.0493503, antiderivative size = 119, 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} \[ -\frac{15 d^2 \sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{4 b^{7/2}}-\frac{5 d (c+d x)^{3/2}}{4 b^2 (a+b x)}-\frac{(c+d x)^{5/2}}{2 b (a+b x)^2}+\frac{15 d^2 \sqrt{c+d x}}{4 b^3} \]

Antiderivative was successfully verified.

[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 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)^3} \, dx &=-\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)) \operatorname{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 [C]  time = 0.0161051, size = 52, normalized size = 0.44 \[ \frac{2 d^2 (c+d x)^{7/2} \, _2F_1\left (3,\frac{7}{2};\frac{9}{2};-\frac{b (c+d x)}{a d-b c}\right )}{7 (a d-b c)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d^2*(c + d*x)^(7/2)*Hypergeometric2F1[3, 7/2, 9/2, -((b*(c + d*x))/(-(b*c) + a*d))])/(7*(-(b*c) + a*d)^3)

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Maple [B]  time = 0.015, size = 238, normalized size = 2. \begin{align*} 2\,{\frac{{d}^{2}\sqrt{dx+c}}{{b}^{3}}}+{\frac{9\,{d}^{3}a}{4\,{b}^{2} \left ( bdx+ad \right ) ^{2}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}-{\frac{9\,{d}^{2}c}{4\,b \left ( bdx+ad \right ) ^{2}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}+{\frac{7\,{d}^{4}{a}^{2}}{4\,{b}^{3} \left ( bdx+ad \right ) ^{2}}\sqrt{dx+c}}-{\frac{7\,{d}^{3}ac}{2\,{b}^{2} \left ( bdx+ad \right ) ^{2}}\sqrt{dx+c}}+{\frac{7\,{d}^{2}{c}^{2}}{4\,b \left ( bdx+ad \right ) ^{2}}\sqrt{dx+c}}-{\frac{15\,{d}^{3}a}{4\,{b}^{3}}\arctan \left ({b\sqrt{dx+c}{\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}}+{\frac{15\,{d}^{2}c}{4\,{b}^{2}}\arctan \left ({b\sqrt{dx+c}{\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.8732, size = 728, normalized size = 6.12 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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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)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.28623, size = 231, normalized size = 1.94 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)