3.14.2 \(\int \frac {(c+d x)^{5/2}}{(a+b x)^4} \, dx\)

Optimal. Leaf size=126 \[ -\frac {5 d^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{8 b^{7/2} \sqrt {b c-a d}}-\frac {5 d^2 \sqrt {c+d x}}{8 b^3 (a+b x)}-\frac {5 d (c+d x)^{3/2}}{12 b^2 (a+b x)^2}-\frac {(c+d x)^{5/2}}{3 b (a+b x)^3} \]

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Rubi [A]  time = 0.05, antiderivative size = 126, normalized size of antiderivative = 1.00, 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 = {47, 63, 208} \begin {gather*} -\frac {5 d^2 \sqrt {c+d x}}{8 b^3 (a+b x)}-\frac {5 d^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{8 b^{7/2} \sqrt {b c-a d}}-\frac {5 d (c+d x)^{3/2}}{12 b^2 (a+b x)^2}-\frac {(c+d x)^{5/2}}{3 b (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/2)/(a + b*x)^4,x]

[Out]

(-5*d^2*Sqrt[c + d*x])/(8*b^3*(a + b*x)) - (5*d*(c + d*x)^(3/2))/(12*b^2*(a + b*x)^2) - (c + d*x)^(5/2)/(3*b*(
a + b*x)^3) - (5*d^3*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(8*b^(7/2)*Sqrt[b*c - a*d])

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 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)^4} \, dx &=-\frac {(c+d x)^{5/2}}{3 b (a+b x)^3}+\frac {(5 d) \int \frac {(c+d x)^{3/2}}{(a+b x)^3} \, dx}{6 b}\\ &=-\frac {5 d (c+d x)^{3/2}}{12 b^2 (a+b x)^2}-\frac {(c+d x)^{5/2}}{3 b (a+b x)^3}+\frac {\left (5 d^2\right ) \int \frac {\sqrt {c+d x}}{(a+b x)^2} \, dx}{8 b^2}\\ &=-\frac {5 d^2 \sqrt {c+d x}}{8 b^3 (a+b x)}-\frac {5 d (c+d x)^{3/2}}{12 b^2 (a+b x)^2}-\frac {(c+d x)^{5/2}}{3 b (a+b x)^3}+\frac {\left (5 d^3\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{16 b^3}\\ &=-\frac {5 d^2 \sqrt {c+d x}}{8 b^3 (a+b x)}-\frac {5 d (c+d x)^{3/2}}{12 b^2 (a+b x)^2}-\frac {(c+d x)^{5/2}}{3 b (a+b x)^3}+\frac {\left (5 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 )}{8 b^3}\\ &=-\frac {5 d^2 \sqrt {c+d x}}{8 b^3 (a+b x)}-\frac {5 d (c+d x)^{3/2}}{12 b^2 (a+b x)^2}-\frac {(c+d x)^{5/2}}{3 b (a+b x)^3}-\frac {5 d^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{8 b^{7/2} \sqrt {b c-a d}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 119, normalized size = 0.94 \begin {gather*} \frac {5 d^3 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {a d-b c}}\right )}{8 b^{7/2} \sqrt {a d-b c}}-\frac {\sqrt {c+d x} \left (15 a^2 d^2+10 a b d (c+4 d x)+b^2 \left (8 c^2+26 c d x+33 d^2 x^2\right )\right )}{24 b^3 (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*(Sqrt[c + d*x]*(15*a^2*d^2 + 10*a*b*d*(c + 4*d*x) + b^2*(8*c^2 + 26*c*d*x + 33*d^2*x^2)))/(b^3*(a + b*x)
^3) + (5*d^3*ArcTan[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/(8*b^(7/2)*Sqrt[-(b*c) + a*d])

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IntegrateAlgebraic [A]  time = 0.62, size = 155, normalized size = 1.23 \begin {gather*} -\frac {d^3 \sqrt {c+d x} \left (15 a^2 d^2+40 a b d (c+d x)-30 a b c d+15 b^2 c^2+33 b^2 (c+d x)^2-40 b^2 c (c+d x)\right )}{24 b^3 (a d+b (c+d x)-b c)^3}-\frac {5 d^3 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x} \sqrt {a d-b c}}{b c-a d}\right )}{8 b^{7/2} \sqrt {a d-b c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*(d^3*Sqrt[c + d*x]*(15*b^2*c^2 - 30*a*b*c*d + 15*a^2*d^2 - 40*b^2*c*(c + d*x) + 40*a*b*d*(c + d*x) + 33*
b^2*(c + d*x)^2))/(b^3*(-(b*c) + a*d + b*(c + d*x))^3) - (5*d^3*ArcTan[(Sqrt[b]*Sqrt[-(b*c) + a*d]*Sqrt[c + d*
x])/(b*c - a*d)])/(8*b^(7/2)*Sqrt[-(b*c) + a*d])

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fricas [B]  time = 1.20, size = 563, normalized size = 4.47 \begin {gather*} \left [\frac {15 \, {\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} + 2 \, a b^{3} c^{2} d + 5 \, a^{2} b^{2} c d^{2} - 15 \, a^{3} b d^{3} + 33 \, {\left (b^{4} c d^{2} - a b^{3} d^{3}\right )} x^{2} + 2 \, {\left (13 \, b^{4} c^{2} d + 7 \, a b^{3} c d^{2} - 20 \, a^{2} b^{2} d^{3}\right )} x\right )} \sqrt {d x + c}}{48 \, {\left (a^{3} b^{5} c - a^{4} b^{4} d + {\left (b^{8} c - a b^{7} d\right )} x^{3} + 3 \, {\left (a b^{7} c - a^{2} b^{6} d\right )} x^{2} + 3 \, {\left (a^{2} b^{6} c - a^{3} b^{5} d\right )} x\right )}}, \frac {15 \, {\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} + 2 \, a b^{3} c^{2} d + 5 \, a^{2} b^{2} c d^{2} - 15 \, a^{3} b d^{3} + 33 \, {\left (b^{4} c d^{2} - a b^{3} d^{3}\right )} x^{2} + 2 \, {\left (13 \, b^{4} c^{2} d + 7 \, a b^{3} c d^{2} - 20 \, a^{2} b^{2} d^{3}\right )} x\right )} \sqrt {d x + c}}{24 \, {\left (a^{3} b^{5} c - a^{4} b^{4} d + {\left (b^{8} c - a b^{7} d\right )} x^{3} + 3 \, {\left (a b^{7} c - a^{2} b^{6} d\right )} x^{2} + 3 \, {\left (a^{2} b^{6} c - a^{3} b^{5} d\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)/(b*x+a)^4,x, algorithm="fricas")

[Out]

[1/48*(15*(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 + 2*a*b^3*c^2*d + 5*a^2*b^2*c*d^2 - 15*a^3
*b*d^3 + 33*(b^4*c*d^2 - a*b^3*d^3)*x^2 + 2*(13*b^4*c^2*d + 7*a*b^3*c*d^2 - 20*a^2*b^2*d^3)*x)*sqrt(d*x + c))/
(a^3*b^5*c - a^4*b^4*d + (b^8*c - a*b^7*d)*x^3 + 3*(a*b^7*c - a^2*b^6*d)*x^2 + 3*(a^2*b^6*c - a^3*b^5*d)*x), 1
/24*(15*(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 + 2*a*b^3*c^2*d + 5*a^2*b^2*c*d^2 - 15*a^3*b*d^3 + 33*(b^4*c*d^
2 - a*b^3*d^3)*x^2 + 2*(13*b^4*c^2*d + 7*a*b^3*c*d^2 - 20*a^2*b^2*d^3)*x)*sqrt(d*x + c))/(a^3*b^5*c - a^4*b^4*
d + (b^8*c - a*b^7*d)*x^3 + 3*(a*b^7*c - a^2*b^6*d)*x^2 + 3*(a^2*b^6*c - a^3*b^5*d)*x)]

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giac [A]  time = 0.98, size = 161, normalized size = 1.28 \begin {gather*} \frac {5 \, d^{3} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{8 \, \sqrt {-b^{2} c + a b d} b^{3}} - \frac {33 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d^{3} - 40 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} c d^{3} + 15 \, \sqrt {d x + c} b^{2} c^{2} d^{3} + 40 \, {\left (d x + c\right )}^{\frac {3}{2}} a b d^{4} - 30 \, \sqrt {d x + c} a b c d^{4} + 15 \, \sqrt {d x + c} a^{2} d^{5}}{24 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}^{3} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)/(b*x+a)^4,x, algorithm="giac")

[Out]

5/8*d^3*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b^3) - 1/24*(33*(d*x + c)^(5/2)*b^2
*d^3 - 40*(d*x + c)^(3/2)*b^2*c*d^3 + 15*sqrt(d*x + c)*b^2*c^2*d^3 + 40*(d*x + c)^(3/2)*a*b*d^4 - 30*sqrt(d*x
+ c)*a*b*c*d^4 + 15*sqrt(d*x + c)*a^2*d^5)/(((d*x + c)*b - b*c + a*d)^3*b^3)

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maple [A]  time = 0.02, size = 204, normalized size = 1.62 \begin {gather*} -\frac {5 \sqrt {d x +c}\, a^{2} d^{5}}{8 \left (b d x +a d \right )^{3} b^{3}}+\frac {5 \sqrt {d x +c}\, a c \,d^{4}}{4 \left (b d x +a d \right )^{3} b^{2}}-\frac {5 \sqrt {d x +c}\, c^{2} d^{3}}{8 \left (b d x +a d \right )^{3} b}-\frac {5 \left (d x +c \right )^{\frac {3}{2}} a \,d^{4}}{3 \left (b d x +a d \right )^{3} b^{2}}+\frac {5 \left (d x +c \right )^{\frac {3}{2}} c \,d^{3}}{3 \left (b d x +a d \right )^{3} b}-\frac {11 \left (d x +c \right )^{\frac {5}{2}} d^{3}}{8 \left (b d x +a d \right )^{3} b}+\frac {5 d^{3} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \sqrt {\left (a d -b c \right ) b}\, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(5/2)/(b*x+a)^4,x)

[Out]

-11/8*d^3/(b*d*x+a*d)^3/b*(d*x+c)^(5/2)-5/3*d^4/(b*d*x+a*d)^3/b^2*(d*x+c)^(3/2)*a+5/3*d^3/(b*d*x+a*d)^3/b*(d*x
+c)^(3/2)*c-5/8*d^5/(b*d*x+a*d)^3/b^3*(d*x+c)^(1/2)*a^2+5/4*d^4/(b*d*x+a*d)^3/b^2*(d*x+c)^(1/2)*a*c-5/8*d^3/(b
*d*x+a*d)^3/b*(d*x+c)^(1/2)*c^2+5/8*d^3/b^3/((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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)/(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 = 0.36, size = 222, normalized size = 1.76 \begin {gather*} \frac {5\,d^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{8\,b^{7/2}\,\sqrt {a\,d-b\,c}}-\frac {\frac {11\,d^3\,{\left (c+d\,x\right )}^{5/2}}{8\,b}+\frac {5\,d^3\,\sqrt {c+d\,x}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{8\,b^3}+\frac {5\,d^3\,\left (a\,d-b\,c\right )\,{\left (c+d\,x\right )}^{3/2}}{3\,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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(5/2)/(a + b*x)^4,x)

[Out]

(5*d^3*atan((b^(1/2)*(c + d*x)^(1/2))/(a*d - b*c)^(1/2)))/(8*b^(7/2)*(a*d - b*c)^(1/2)) - ((11*d^3*(c + d*x)^(
5/2))/(8*b) + (5*d^3*(c + d*x)^(1/2)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d))/(8*b^3) + (5*d^3*(a*d - b*c)*(c + d*x)^(
3/2))/(3*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 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(5/2)/(b*x+a)**4,x)

[Out]

Timed out

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