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

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

[Out]

5/3*d*(d*x+c)^(3/2)/b^2-(d*x+c)^(5/2)/b/(b*x+a)-5*d*(-a*d+b*c)^(3/2)*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^
(1/2))/b^(7/2)+5*d*(-a*d+b*c)*(d*x+c)^(1/2)/b^3

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Rubi [A]  time = 0.06, antiderivative size = 110, 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, 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*}

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Mathematica [C]  time = 0.02, 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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fricas [A]  time = 0.79, size = 330, normalized size = 3.00 \[ \left [-\frac {15 \, {\left (a b c d - a^{2} d^{2} + {\left (b^{2} c d - a b d^{2}\right )} x\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d + 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) - 2 \, {\left (2 \, b^{2} d^{2} x^{2} - 3 \, b^{2} c^{2} + 20 \, a b c d - 15 \, a^{2} d^{2} + 2 \, {\left (7 \, b^{2} c d - 5 \, a b d^{2}\right )} x\right )} \sqrt {d x + c}}{6 \, {\left (b^{4} x + a b^{3}\right )}}, -\frac {15 \, {\left (a b c d - a^{2} d^{2} + {\left (b^{2} c d - a b d^{2}\right )} x\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) - {\left (2 \, b^{2} d^{2} x^{2} - 3 \, b^{2} c^{2} + 20 \, a b c d - 15 \, a^{2} d^{2} + 2 \, {\left (7 \, b^{2} c d - 5 \, a b d^{2}\right )} x\right )} \sqrt {d x + c}}{3 \, {\left (b^{4} x + a b^{3}\right )}}\right ] \]

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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giac [A]  time = 1.28, size = 181, normalized size = 1.65 \[ \frac {5 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{3}} - \frac {\sqrt {d x + c} b^{2} c^{2} d - 2 \, \sqrt {d x + c} a b c d^{2} + \sqrt {d x + c} a^{2} d^{3}}{{\left ({\left (d x + c\right )} b - b c + a d\right )} b^{3}} + \frac {2 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} b^{4} d + 6 \, \sqrt {d x + c} b^{4} c d - 6 \, \sqrt {d x + c} a b^{3} d^{2}\right )}}{3 \, b^{6}} \]

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

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maple [B]  time = 0.02, size = 258, normalized size = 2.35 \[ \frac {5 a^{2} d^{3} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, b^{3}}-\frac {10 a c \,d^{2} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, b^{2}}+\frac {5 c^{2} d \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, b}-\frac {\sqrt {d x +c}\, a^{2} d^{3}}{\left (b d x +a d \right ) b^{3}}+\frac {2 \sqrt {d x +c}\, a c \,d^{2}}{\left (b d x +a d \right ) b^{2}}-\frac {\sqrt {d x +c}\, c^{2} d}{\left (b d x +a d \right ) b}-\frac {4 \sqrt {d x +c}\, a \,d^{2}}{b^{3}}+\frac {4 \sqrt {d x +c}\, c d}{b^{2}}+\frac {2 \left (d x +c \right )^{\frac {3}{2}} d}{3 b^{2}} \]

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

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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)^(5/2)/(b*x+a)^2,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 = 0.12, size = 161, normalized size = 1.46 \[ \frac {2\,d\,{\left (c+d\,x\right )}^{3/2}}{3\,b^2}-\frac {\sqrt {c+d\,x}\,\left (a^2\,d^3-2\,a\,b\,c\,d^2+b^2\,c^2\,d\right )}{b^4\,\left (c+d\,x\right )-b^4\,c+a\,b^3\,d}+\frac {5\,d\,\mathrm {atan}\left (\frac {\sqrt {b}\,d\,{\left (a\,d-b\,c\right )}^{3/2}\,\sqrt {c+d\,x}}{a^2\,d^3-2\,a\,b\,c\,d^2+b^2\,c^2\,d}\right )\,{\left (a\,d-b\,c\right )}^{3/2}}{b^{7/2}}+\frac {2\,d\,\left (2\,b^2\,c-2\,a\,b\,d\right )\,\sqrt {c+d\,x}}{b^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(5/2)/(a + b*x)^2,x)

[Out]

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

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

[Out]

Timed out

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