3.1396 \(\int \frac {(c+d x)^{3/2}}{(a+b x)^4} \, dx\)

Optimal. Leaf size=136 \[ \frac {d^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{8 b^{5/2} (b c-a d)^{3/2}}-\frac {d^2 \sqrt {c+d x}}{8 b^2 (a+b x) (b c-a d)}-\frac {d \sqrt {c+d x}}{4 b^2 (a+b x)^2}-\frac {(c+d x)^{3/2}}{3 b (a+b x)^3} \]

[Out]

-1/3*(d*x+c)^(3/2)/b/(b*x+a)^3+1/8*d^3*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(5/2)/(-a*d+b*c)^(3/2
)-1/4*d*(d*x+c)^(1/2)/b^2/(b*x+a)^2-1/8*d^2*(d*x+c)^(1/2)/b^2/(-a*d+b*c)/(b*x+a)

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Rubi [A]  time = 0.06, antiderivative size = 136, normalized size of antiderivative = 1.00, 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, 51, 63, 208} \[ -\frac {d^2 \sqrt {c+d x}}{8 b^2 (a+b x) (b c-a d)}+\frac {d^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{8 b^{5/2} (b c-a d)^{3/2}}-\frac {d \sqrt {c+d x}}{4 b^2 (a+b x)^2}-\frac {(c+d x)^{3/2}}{3 b (a+b x)^3} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(3/2)/(a + b*x)^4,x]

[Out]

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

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)^4} \, dx &=-\frac {(c+d x)^{3/2}}{3 b (a+b x)^3}+\frac {d \int \frac {\sqrt {c+d x}}{(a+b x)^3} \, dx}{2 b}\\ &=-\frac {d \sqrt {c+d x}}{4 b^2 (a+b x)^2}-\frac {(c+d x)^{3/2}}{3 b (a+b x)^3}+\frac {d^2 \int \frac {1}{(a+b x)^2 \sqrt {c+d x}} \, dx}{8 b^2}\\ &=-\frac {d \sqrt {c+d x}}{4 b^2 (a+b x)^2}-\frac {d^2 \sqrt {c+d x}}{8 b^2 (b c-a d) (a+b x)}-\frac {(c+d x)^{3/2}}{3 b (a+b x)^3}-\frac {d^3 \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{16 b^2 (b c-a d)}\\ &=-\frac {d \sqrt {c+d x}}{4 b^2 (a+b x)^2}-\frac {d^2 \sqrt {c+d x}}{8 b^2 (b c-a d) (a+b x)}-\frac {(c+d x)^{3/2}}{3 b (a+b x)^3}-\frac {d^2 \operatorname {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{8 b^2 (b c-a d)}\\ &=-\frac {d \sqrt {c+d x}}{4 b^2 (a+b x)^2}-\frac {d^2 \sqrt {c+d x}}{8 b^2 (b c-a d) (a+b x)}-\frac {(c+d x)^{3/2}}{3 b (a+b x)^3}+\frac {d^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{8 b^{5/2} (b c-a d)^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 52, normalized size = 0.38 \[ \frac {2 d^3 (c+d x)^{5/2} \, _2F_1\left (\frac {5}{2},4;\frac {7}{2};-\frac {b (c+d x)}{a d-b c}\right )}{5 (a d-b c)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(3/2)/(a + b*x)^4,x]

[Out]

(2*d^3*(c + d*x)^(5/2)*Hypergeometric2F1[5/2, 4, 7/2, -((b*(c + d*x))/(-(b*c) + a*d))])/(5*(-(b*c) + a*d)^4)

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fricas [B]  time = 0.50, size = 666, normalized size = 4.90 \[ \left [-\frac {3 \, {\left (b^{3} d^{3} x^{3} + 3 \, a b^{2} d^{3} x^{2} + 3 \, a^{2} b d^{3} x + a^{3} d^{3}\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {b^{2} c - a b d} \sqrt {d x + c}}{b x + a}\right ) + 2 \, {\left (8 \, b^{4} c^{3} - 10 \, a b^{3} c^{2} d - a^{2} b^{2} c d^{2} + 3 \, a^{3} b d^{3} + 3 \, {\left (b^{4} c d^{2} - a b^{3} d^{3}\right )} x^{2} + 2 \, {\left (7 \, b^{4} c^{2} d - 11 \, a b^{3} c d^{2} + 4 \, a^{2} b^{2} d^{3}\right )} x\right )} \sqrt {d x + c}}{48 \, {\left (a^{3} b^{5} c^{2} - 2 \, a^{4} b^{4} c d + a^{5} b^{3} d^{2} + {\left (b^{8} c^{2} - 2 \, a b^{7} c d + a^{2} b^{6} d^{2}\right )} x^{3} + 3 \, {\left (a b^{7} c^{2} - 2 \, a^{2} b^{6} c d + a^{3} b^{5} d^{2}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} c^{2} - 2 \, a^{3} b^{5} c d + a^{4} b^{4} d^{2}\right )} x\right )}}, -\frac {3 \, {\left (b^{3} d^{3} x^{3} + 3 \, a b^{2} d^{3} x^{2} + 3 \, a^{2} b d^{3} x + a^{3} d^{3}\right )} \sqrt {-b^{2} c + a b d} \arctan \left (\frac {\sqrt {-b^{2} c + a b d} \sqrt {d x + c}}{b d x + b c}\right ) + {\left (8 \, b^{4} c^{3} - 10 \, a b^{3} c^{2} d - a^{2} b^{2} c d^{2} + 3 \, a^{3} b d^{3} + 3 \, {\left (b^{4} c d^{2} - a b^{3} d^{3}\right )} x^{2} + 2 \, {\left (7 \, b^{4} c^{2} d - 11 \, a b^{3} c d^{2} + 4 \, a^{2} b^{2} d^{3}\right )} x\right )} \sqrt {d x + c}}{24 \, {\left (a^{3} b^{5} c^{2} - 2 \, a^{4} b^{4} c d + a^{5} b^{3} d^{2} + {\left (b^{8} c^{2} - 2 \, a b^{7} c d + a^{2} b^{6} d^{2}\right )} x^{3} + 3 \, {\left (a b^{7} c^{2} - 2 \, a^{2} b^{6} c d + a^{3} b^{5} d^{2}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} c^{2} - 2 \, a^{3} b^{5} c d + a^{4} b^{4} d^{2}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(3/2)/(b*x+a)^4,x, algorithm="fricas")

[Out]

[-1/48*(3*(b^3*d^3*x^3 + 3*a*b^2*d^3*x^2 + 3*a^2*b*d^3*x + a^3*d^3)*sqrt(b^2*c - a*b*d)*log((b*d*x + 2*b*c - a
*d - 2*sqrt(b^2*c - a*b*d)*sqrt(d*x + c))/(b*x + a)) + 2*(8*b^4*c^3 - 10*a*b^3*c^2*d - a^2*b^2*c*d^2 + 3*a^3*b
*d^3 + 3*(b^4*c*d^2 - a*b^3*d^3)*x^2 + 2*(7*b^4*c^2*d - 11*a*b^3*c*d^2 + 4*a^2*b^2*d^3)*x)*sqrt(d*x + c))/(a^3
*b^5*c^2 - 2*a^4*b^4*c*d + a^5*b^3*d^2 + (b^8*c^2 - 2*a*b^7*c*d + a^2*b^6*d^2)*x^3 + 3*(a*b^7*c^2 - 2*a^2*b^6*
c*d + a^3*b^5*d^2)*x^2 + 3*(a^2*b^6*c^2 - 2*a^3*b^5*c*d + a^4*b^4*d^2)*x), -1/24*(3*(b^3*d^3*x^3 + 3*a*b^2*d^3
*x^2 + 3*a^2*b*d^3*x + a^3*d^3)*sqrt(-b^2*c + a*b*d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c))
+ (8*b^4*c^3 - 10*a*b^3*c^2*d - a^2*b^2*c*d^2 + 3*a^3*b*d^3 + 3*(b^4*c*d^2 - a*b^3*d^3)*x^2 + 2*(7*b^4*c^2*d -
11*a*b^3*c*d^2 + 4*a^2*b^2*d^3)*x)*sqrt(d*x + c))/(a^3*b^5*c^2 - 2*a^4*b^4*c*d + a^5*b^3*d^2 + (b^8*c^2 - 2*a
*b^7*c*d + a^2*b^6*d^2)*x^3 + 3*(a*b^7*c^2 - 2*a^2*b^6*c*d + a^3*b^5*d^2)*x^2 + 3*(a^2*b^6*c^2 - 2*a^3*b^5*c*d
+ a^4*b^4*d^2)*x)]

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giac [A]  time = 1.40, size = 185, normalized size = 1.36 \[ -\frac {d^{3} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{8 \, {\left (b^{3} c - a b^{2} d\right )} \sqrt {-b^{2} c + a b d}} - \frac {3 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d^{3} + 8 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} c d^{3} - 3 \, \sqrt {d x + c} b^{2} c^{2} d^{3} - 8 \, {\left (d x + c\right )}^{\frac {3}{2}} a b d^{4} + 6 \, \sqrt {d x + c} a b c d^{4} - 3 \, \sqrt {d x + c} a^{2} d^{5}}{24 \, {\left (b^{3} c - a b^{2} d\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(3/2)/(b*x+a)^4,x, algorithm="giac")

[Out]

-1/8*d^3*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^3*c - a*b^2*d)*sqrt(-b^2*c + a*b*d)) - 1/24*(3*(d*x
+ c)^(5/2)*b^2*d^3 + 8*(d*x + c)^(3/2)*b^2*c*d^3 - 3*sqrt(d*x + c)*b^2*c^2*d^3 - 8*(d*x + c)^(3/2)*a*b*d^4 + 6
*sqrt(d*x + c)*a*b*c*d^4 - 3*sqrt(d*x + c)*a^2*d^5)/((b^3*c - a*b^2*d)*((d*x + c)*b - b*c + a*d)^3)

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maple [A]  time = 0.02, size = 163, normalized size = 1.20 \[ -\frac {\sqrt {d x +c}\, a \,d^{4}}{8 \left (b d x +a d \right )^{3} b^{2}}+\frac {\sqrt {d x +c}\, c \,d^{3}}{8 \left (b d x +a d \right )^{3} b}+\frac {\left (d x +c \right )^{\frac {5}{2}} d^{3}}{8 \left (b d x +a d \right )^{3} \left (a d -b c \right )}-\frac {\left (d x +c \right )^{\frac {3}{2}} d^{3}}{3 \left (b d x +a d \right )^{3} b}+\frac {d^{3} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \left (a d -b c \right ) \sqrt {\left (a d -b c \right ) b}\, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(3/2)/(b*x+a)^4,x)

[Out]

1/8*d^3/(b*d*x+a*d)^3/(a*d-b*c)*(d*x+c)^(5/2)-1/3*d^3/(b*d*x+a*d)^3/b*(d*x+c)^(3/2)-1/8*d^4/(b*d*x+a*d)^3/b^2*
(d*x+c)^(1/2)*a+1/8*d^3/(b*d*x+a*d)^3/b*(d*x+c)^(1/2)*c+1/8*d^3/(a*d-b*c)/b^2/((a*d-b*c)*b)^(1/2)*arctan((d*x+
c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)

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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)^(3/2)/(b*x+a)^4,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 = 0.34, size = 209, normalized size = 1.54 \[ \frac {d^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{8\,b^{5/2}\,{\left (a\,d-b\,c\right )}^{3/2}}-\frac {\frac {d^3\,{\left (c+d\,x\right )}^{3/2}}{3\,b}-\frac {d^3\,{\left (c+d\,x\right )}^{5/2}}{8\,\left (a\,d-b\,c\right )}+\frac {d^3\,\left (a\,d-b\,c\right )\,\sqrt {c+d\,x}}{8\,b^2}}{\left (c+d\,x\right )\,\left (3\,a^2\,b\,d^2-6\,a\,b^2\,c\,d+3\,b^3\,c^2\right )+b^3\,{\left (c+d\,x\right )}^3-\left (3\,b^3\,c-3\,a\,b^2\,d\right )\,{\left (c+d\,x\right )}^2+a^3\,d^3-b^3\,c^3+3\,a\,b^2\,c^2\,d-3\,a^2\,b\,c\,d^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(3/2)/(a + b*x)^4,x)

[Out]

(d^3*atan((b^(1/2)*(c + d*x)^(1/2))/(a*d - b*c)^(1/2)))/(8*b^(5/2)*(a*d - b*c)^(3/2)) - ((d^3*(c + d*x)^(3/2))
/(3*b) - (d^3*(c + d*x)^(5/2))/(8*(a*d - b*c)) + (d^3*(a*d - b*c)*(c + d*x)^(1/2))/(8*b^2))/((c + d*x)*(3*b^3*
c^2 + 3*a^2*b*d^2 - 6*a*b^2*c*d) + b^3*(c + d*x)^3 - (3*b^3*c - 3*a*b^2*d)*(c + d*x)^2 + a^3*d^3 - b^3*c^3 + 3
*a*b^2*c^2*d - 3*a^2*b*c*d^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(3/2)/(b*x+a)**4,x)

[Out]

Timed out

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