3.1397 \(\int \frac{(c+d x)^{3/2}}{(a+b x)^5} \, dx\)

Optimal. Leaf size=172 \[ \frac{3 d^3 \sqrt{c+d x}}{64 b^2 (a+b x) (b c-a d)^2}-\frac{d^2 \sqrt{c+d x}}{32 b^2 (a+b x)^2 (b c-a d)}-\frac{3 d^4 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{64 b^{5/2} (b c-a d)^{5/2}}-\frac{d \sqrt{c+d x}}{8 b^2 (a+b x)^3}-\frac{(c+d x)^{3/2}}{4 b (a+b x)^4} \]

[Out]

-(d*Sqrt[c + d*x])/(8*b^2*(a + b*x)^3) - (d^2*Sqrt[c + d*x])/(32*b^2*(b*c - a*d)*(a + b*x)^2) + (3*d^3*Sqrt[c
+ d*x])/(64*b^2*(b*c - a*d)^2*(a + b*x)) - (c + d*x)^(3/2)/(4*b*(a + b*x)^4) - (3*d^4*ArcTanh[(Sqrt[b]*Sqrt[c
+ d*x])/Sqrt[b*c - a*d]])/(64*b^(5/2)*(b*c - a*d)^(5/2))

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Rubi [A]  time = 0.0737483, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {47, 51, 63, 208} \[ \frac{3 d^3 \sqrt{c+d x}}{64 b^2 (a+b x) (b c-a d)^2}-\frac{d^2 \sqrt{c+d x}}{32 b^2 (a+b x)^2 (b c-a d)}-\frac{3 d^4 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{64 b^{5/2} (b c-a d)^{5/2}}-\frac{d \sqrt{c+d x}}{8 b^2 (a+b x)^3}-\frac{(c+d x)^{3/2}}{4 b (a+b x)^4} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(3/2)/(a + b*x)^5,x]

[Out]

-(d*Sqrt[c + d*x])/(8*b^2*(a + b*x)^3) - (d^2*Sqrt[c + d*x])/(32*b^2*(b*c - a*d)*(a + b*x)^2) + (3*d^3*Sqrt[c
+ d*x])/(64*b^2*(b*c - a*d)^2*(a + b*x)) - (c + d*x)^(3/2)/(4*b*(a + b*x)^4) - (3*d^4*ArcTanh[(Sqrt[b]*Sqrt[c
+ d*x])/Sqrt[b*c - a*d]])/(64*b^(5/2)*(b*c - a*d)^(5/2))

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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)^{3/2}}{(a+b x)^5} \, dx &=-\frac{(c+d x)^{3/2}}{4 b (a+b x)^4}+\frac{(3 d) \int \frac{\sqrt{c+d x}}{(a+b x)^4} \, dx}{8 b}\\ &=-\frac{d \sqrt{c+d x}}{8 b^2 (a+b x)^3}-\frac{(c+d x)^{3/2}}{4 b (a+b x)^4}+\frac{d^2 \int \frac{1}{(a+b x)^3 \sqrt{c+d x}} \, dx}{16 b^2}\\ &=-\frac{d \sqrt{c+d x}}{8 b^2 (a+b x)^3}-\frac{d^2 \sqrt{c+d x}}{32 b^2 (b c-a d) (a+b x)^2}-\frac{(c+d x)^{3/2}}{4 b (a+b x)^4}-\frac{\left (3 d^3\right ) \int \frac{1}{(a+b x)^2 \sqrt{c+d x}} \, dx}{64 b^2 (b c-a d)}\\ &=-\frac{d \sqrt{c+d x}}{8 b^2 (a+b x)^3}-\frac{d^2 \sqrt{c+d x}}{32 b^2 (b c-a d) (a+b x)^2}+\frac{3 d^3 \sqrt{c+d x}}{64 b^2 (b c-a d)^2 (a+b x)}-\frac{(c+d x)^{3/2}}{4 b (a+b x)^4}+\frac{\left (3 d^4\right ) \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{128 b^2 (b c-a d)^2}\\ &=-\frac{d \sqrt{c+d x}}{8 b^2 (a+b x)^3}-\frac{d^2 \sqrt{c+d x}}{32 b^2 (b c-a d) (a+b x)^2}+\frac{3 d^3 \sqrt{c+d x}}{64 b^2 (b c-a d)^2 (a+b x)}-\frac{(c+d x)^{3/2}}{4 b (a+b x)^4}+\frac{\left (3 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 )}{64 b^2 (b c-a d)^2}\\ &=-\frac{d \sqrt{c+d x}}{8 b^2 (a+b x)^3}-\frac{d^2 \sqrt{c+d x}}{32 b^2 (b c-a d) (a+b x)^2}+\frac{3 d^3 \sqrt{c+d x}}{64 b^2 (b c-a d)^2 (a+b x)}-\frac{(c+d x)^{3/2}}{4 b (a+b x)^4}-\frac{3 d^4 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{64 b^{5/2} (b c-a d)^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0169324, size = 52, normalized size = 0.3 \[ \frac{2 d^4 (c+d x)^{5/2} \, _2F_1\left (\frac{5}{2},5;\frac{7}{2};-\frac{b (c+d x)}{a d-b c}\right )}{5 (a d-b c)^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d^4*(c + d*x)^(5/2)*Hypergeometric2F1[5/2, 5, 7/2, -((b*(c + d*x))/(-(b*c) + a*d))])/(5*(-(b*c) + a*d)^5)

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Maple [A]  time = 0.013, size = 222, normalized size = 1.3 \begin{align*}{\frac{3\,{d}^{4}b}{64\, \left ( bdx+ad \right ) ^{4} \left ({a}^{2}{d}^{2}-2\,abcd+{b}^{2}{c}^{2} \right ) } \left ( dx+c \right ) ^{{\frac{7}{2}}}}+{\frac{11\,{d}^{4}}{64\, \left ( bdx+ad \right ) ^{4} \left ( ad-bc \right ) } \left ( dx+c \right ) ^{{\frac{5}{2}}}}-{\frac{11\,{d}^{4}}{64\, \left ( bdx+ad \right ) ^{4}b} \left ( dx+c \right ) ^{{\frac{3}{2}}}}-{\frac{3\,{d}^{5}a}{64\, \left ( bdx+ad \right ) ^{4}{b}^{2}}\sqrt{dx+c}}+{\frac{3\,{d}^{4}c}{64\, \left ( bdx+ad \right ) ^{4}b}\sqrt{dx+c}}+{\frac{3\,{d}^{4}}{ \left ( 64\,{a}^{2}{d}^{2}-128\,abcd+64\,{b}^{2}{c}^{2} \right ){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)^(3/2)/(b*x+a)^5,x)

[Out]

3/64*d^4/(b*d*x+a*d)^4*b/(a^2*d^2-2*a*b*c*d+b^2*c^2)*(d*x+c)^(7/2)+11/64*d^4/(b*d*x+a*d)^4/(a*d-b*c)*(d*x+c)^(
5/2)-11/64*d^4/(b*d*x+a*d)^4/b*(d*x+c)^(3/2)-3/64*d^5/(b*d*x+a*d)^4/b^2*(d*x+c)^(1/2)*a+3/64*d^4/(b*d*x+a*d)^4
/b*(d*x+c)^(1/2)*c+3/64*d^4/(a^2*d^2-2*a*b*c*d+b^2*c^2)/b^2/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b
*c)*b)^(1/2))

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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)^(3/2)/(b*x+a)^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.15403, size = 2101, normalized size = 12.22 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/128*(3*(b^4*d^4*x^4 + 4*a*b^3*d^4*x^3 + 6*a^2*b^2*d^4*x^2 + 4*a^3*b*d^4*x + a^4*d^4)*sqrt(b^2*c - a*b*d)*lo
g((b*d*x + 2*b*c - a*d - 2*sqrt(b^2*c - a*b*d)*sqrt(d*x + c))/(b*x + a)) - 2*(16*b^5*c^4 - 40*a*b^4*c^3*d + 26
*a^2*b^3*c^2*d^2 + a^3*b^2*c*d^3 - 3*a^4*b*d^4 - 3*(b^5*c*d^3 - a*b^4*d^4)*x^3 + (2*b^5*c^2*d^2 - 13*a*b^4*c*d
^3 + 11*a^2*b^3*d^4)*x^2 + (24*b^5*c^3*d - 68*a*b^4*c^2*d^2 + 55*a^2*b^3*c*d^3 - 11*a^3*b^2*d^4)*x)*sqrt(d*x +
c))/(a^4*b^6*c^3 - 3*a^5*b^5*c^2*d + 3*a^6*b^4*c*d^2 - a^7*b^3*d^3 + (b^10*c^3 - 3*a*b^9*c^2*d + 3*a^2*b^8*c*
d^2 - a^3*b^7*d^3)*x^4 + 4*(a*b^9*c^3 - 3*a^2*b^8*c^2*d + 3*a^3*b^7*c*d^2 - a^4*b^6*d^3)*x^3 + 6*(a^2*b^8*c^3
- 3*a^3*b^7*c^2*d + 3*a^4*b^6*c*d^2 - a^5*b^5*d^3)*x^2 + 4*(a^3*b^7*c^3 - 3*a^4*b^6*c^2*d + 3*a^5*b^5*c*d^2 -
a^6*b^4*d^3)*x), 1/64*(3*(b^4*d^4*x^4 + 4*a*b^3*d^4*x^3 + 6*a^2*b^2*d^4*x^2 + 4*a^3*b*d^4*x + a^4*d^4)*sqrt(-b
^2*c + a*b*d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) - (16*b^5*c^4 - 40*a*b^4*c^3*d + 26*a^2
*b^3*c^2*d^2 + a^3*b^2*c*d^3 - 3*a^4*b*d^4 - 3*(b^5*c*d^3 - a*b^4*d^4)*x^3 + (2*b^5*c^2*d^2 - 13*a*b^4*c*d^3 +
11*a^2*b^3*d^4)*x^2 + (24*b^5*c^3*d - 68*a*b^4*c^2*d^2 + 55*a^2*b^3*c*d^3 - 11*a^3*b^2*d^4)*x)*sqrt(d*x + c))
/(a^4*b^6*c^3 - 3*a^5*b^5*c^2*d + 3*a^6*b^4*c*d^2 - a^7*b^3*d^3 + (b^10*c^3 - 3*a*b^9*c^2*d + 3*a^2*b^8*c*d^2
- a^3*b^7*d^3)*x^4 + 4*(a*b^9*c^3 - 3*a^2*b^8*c^2*d + 3*a^3*b^7*c*d^2 - a^4*b^6*d^3)*x^3 + 6*(a^2*b^8*c^3 - 3*
a^3*b^7*c^2*d + 3*a^4*b^6*c*d^2 - a^5*b^5*d^3)*x^2 + 4*(a^3*b^7*c^3 - 3*a^4*b^6*c^2*d + 3*a^5*b^5*c*d^2 - a^6*
b^4*d^3)*x)]

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

[Out]

Timed out

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Giac [A]  time = 1.12818, size = 385, normalized size = 2.24 \begin{align*} \frac{3 \, d^{4} \arctan \left (\frac{\sqrt{d x + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{64 \,{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} \sqrt{-b^{2} c + a b d}} + \frac{3 \,{\left (d x + c\right )}^{\frac{7}{2}} b^{3} d^{4} - 11 \,{\left (d x + c\right )}^{\frac{5}{2}} b^{3} c d^{4} - 11 \,{\left (d x + c\right )}^{\frac{3}{2}} b^{3} c^{2} d^{4} + 3 \, \sqrt{d x + c} b^{3} c^{3} d^{4} + 11 \,{\left (d x + c\right )}^{\frac{5}{2}} a b^{2} d^{5} + 22 \,{\left (d x + c\right )}^{\frac{3}{2}} a b^{2} c d^{5} - 9 \, \sqrt{d x + c} a b^{2} c^{2} d^{5} - 11 \,{\left (d x + c\right )}^{\frac{3}{2}} a^{2} b d^{6} + 9 \, \sqrt{d x + c} a^{2} b c d^{6} - 3 \, \sqrt{d x + c} a^{3} d^{7}}{64 \,{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )}{\left ({\left (d x + c\right )} b - b c + a d\right )}^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/64*d^4*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2)*sqrt(-b^2*c + a*b
*d)) + 1/64*(3*(d*x + c)^(7/2)*b^3*d^4 - 11*(d*x + c)^(5/2)*b^3*c*d^4 - 11*(d*x + c)^(3/2)*b^3*c^2*d^4 + 3*sqr
t(d*x + c)*b^3*c^3*d^4 + 11*(d*x + c)^(5/2)*a*b^2*d^5 + 22*(d*x + c)^(3/2)*a*b^2*c*d^5 - 9*sqrt(d*x + c)*a*b^2
*c^2*d^5 - 11*(d*x + c)^(3/2)*a^2*b*d^6 + 9*sqrt(d*x + c)*a^2*b*c*d^6 - 3*sqrt(d*x + c)*a^3*d^7)/((b^4*c^2 - 2
*a*b^3*c*d + a^2*b^2*d^2)*((d*x + c)*b - b*c + a*d)^4)