3.14.36 \(\int \frac {1}{(a+b x)^3 (c+d x)^{5/2}} \, dx\)

Optimal. Leaf size=167 \[ -\frac {35 b^{3/2} d^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 (b c-a d)^{9/2}}+\frac {35 b d^2}{4 \sqrt {c+d x} (b c-a d)^4}+\frac {35 d^2}{12 (c+d x)^{3/2} (b c-a d)^3}+\frac {7 d}{4 (a+b x) (c+d x)^{3/2} (b c-a d)^2}-\frac {1}{2 (a+b x)^2 (c+d x)^{3/2} (b c-a d)} \]

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Rubi [A]  time = 0.06, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {51, 63, 208} \begin {gather*} -\frac {35 b^{3/2} d^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 (b c-a d)^{9/2}}+\frac {35 b d^2}{4 \sqrt {c+d x} (b c-a d)^4}+\frac {35 d^2}{12 (c+d x)^{3/2} (b c-a d)^3}+\frac {7 d}{4 (a+b x) (c+d x)^{3/2} (b c-a d)^2}-\frac {1}{2 (a+b x)^2 (c+d x)^{3/2} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^3*(c + d*x)^(5/2)),x]

[Out]

(35*d^2)/(12*(b*c - a*d)^3*(c + d*x)^(3/2)) - 1/(2*(b*c - a*d)*(a + b*x)^2*(c + d*x)^(3/2)) + (7*d)/(4*(b*c -
a*d)^2*(a + b*x)*(c + d*x)^(3/2)) + (35*b*d^2)/(4*(b*c - a*d)^4*Sqrt[c + d*x]) - (35*b^(3/2)*d^2*ArcTanh[(Sqrt
[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(4*(b*c - a*d)^(9/2))

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 {1}{(a+b x)^3 (c+d x)^{5/2}} \, dx &=-\frac {1}{2 (b c-a d) (a+b x)^2 (c+d x)^{3/2}}-\frac {(7 d) \int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx}{4 (b c-a d)}\\ &=-\frac {1}{2 (b c-a d) (a+b x)^2 (c+d x)^{3/2}}+\frac {7 d}{4 (b c-a d)^2 (a+b x) (c+d x)^{3/2}}+\frac {\left (35 d^2\right ) \int \frac {1}{(a+b x) (c+d x)^{5/2}} \, dx}{8 (b c-a d)^2}\\ &=\frac {35 d^2}{12 (b c-a d)^3 (c+d x)^{3/2}}-\frac {1}{2 (b c-a d) (a+b x)^2 (c+d x)^{3/2}}+\frac {7 d}{4 (b c-a d)^2 (a+b x) (c+d x)^{3/2}}+\frac {\left (35 b d^2\right ) \int \frac {1}{(a+b x) (c+d x)^{3/2}} \, dx}{8 (b c-a d)^3}\\ &=\frac {35 d^2}{12 (b c-a d)^3 (c+d x)^{3/2}}-\frac {1}{2 (b c-a d) (a+b x)^2 (c+d x)^{3/2}}+\frac {7 d}{4 (b c-a d)^2 (a+b x) (c+d x)^{3/2}}+\frac {35 b d^2}{4 (b c-a d)^4 \sqrt {c+d x}}+\frac {\left (35 b^2 d^2\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{8 (b c-a d)^4}\\ &=\frac {35 d^2}{12 (b c-a d)^3 (c+d x)^{3/2}}-\frac {1}{2 (b c-a d) (a+b x)^2 (c+d x)^{3/2}}+\frac {7 d}{4 (b c-a d)^2 (a+b x) (c+d x)^{3/2}}+\frac {35 b d^2}{4 (b c-a d)^4 \sqrt {c+d x}}+\frac {\left (35 b^2 d\right ) \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 c-a d)^4}\\ &=\frac {35 d^2}{12 (b c-a d)^3 (c+d x)^{3/2}}-\frac {1}{2 (b c-a d) (a+b x)^2 (c+d x)^{3/2}}+\frac {7 d}{4 (b c-a d)^2 (a+b x) (c+d x)^{3/2}}+\frac {35 b d^2}{4 (b c-a d)^4 \sqrt {c+d x}}-\frac {35 b^{3/2} d^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 (b c-a d)^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 52, normalized size = 0.31 \begin {gather*} -\frac {2 d^2 \, _2F_1\left (-\frac {3}{2},3;-\frac {1}{2};-\frac {b (c+d x)}{a d-b c}\right )}{3 (c+d x)^{3/2} (a d-b c)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^3*(c + d*x)^(5/2)),x]

[Out]

(-2*d^2*Hypergeometric2F1[-3/2, 3, -1/2, -((b*(c + d*x))/(-(b*c) + a*d))])/(3*(-(b*c) + a*d)^3*(c + d*x)^(3/2)
)

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IntegrateAlgebraic [A]  time = 0.60, size = 223, normalized size = 1.34 \begin {gather*} -\frac {d^2 \left (8 a^3 d^3-56 a^2 b d^2 (c+d x)-24 a^2 b c d^2+24 a b^2 c^2 d-175 a b^2 d (c+d x)^2+112 a b^2 c d (c+d x)-8 b^3 c^3-56 b^3 c^2 (c+d x)-105 b^3 (c+d x)^3+175 b^3 c (c+d x)^2\right )}{12 (c+d x)^{3/2} (b c-a d)^4 (-a d-b (c+d x)+b c)^2}-\frac {35 b^{3/2} d^2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x} \sqrt {a d-b c}}{b c-a d}\right )}{4 (a d-b c)^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((a + b*x)^3*(c + d*x)^(5/2)),x]

[Out]

-1/12*(d^2*(-8*b^3*c^3 + 24*a*b^2*c^2*d - 24*a^2*b*c*d^2 + 8*a^3*d^3 - 56*b^3*c^2*(c + d*x) + 112*a*b^2*c*d*(c
+ d*x) - 56*a^2*b*d^2*(c + d*x) + 175*b^3*c*(c + d*x)^2 - 175*a*b^2*d*(c + d*x)^2 - 105*b^3*(c + d*x)^3))/((b
*c - a*d)^4*(c + d*x)^(3/2)*(b*c - a*d - b*(c + d*x))^2) - (35*b^(3/2)*d^2*ArcTan[(Sqrt[b]*Sqrt[-(b*c) + a*d]*
Sqrt[c + d*x])/(b*c - a*d)])/(4*(-(b*c) + a*d)^(9/2))

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fricas [B]  time = 1.36, size = 1226, normalized size = 7.34

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^3/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

[1/24*(105*(b^3*d^4*x^4 + a^2*b*c^2*d^2 + 2*(b^3*c*d^3 + a*b^2*d^4)*x^3 + (b^3*c^2*d^2 + 4*a*b^2*c*d^3 + a^2*b
*d^4)*x^2 + 2*(a*b^2*c^2*d^2 + a^2*b*c*d^3)*x)*sqrt(b/(b*c - a*d))*log((b*d*x + 2*b*c - a*d - 2*(b*c - a*d)*sq
rt(d*x + c)*sqrt(b/(b*c - a*d)))/(b*x + a)) + 2*(105*b^3*d^3*x^3 - 6*b^3*c^3 + 39*a*b^2*c^2*d + 80*a^2*b*c*d^2
- 8*a^3*d^3 + 35*(4*b^3*c*d^2 + 5*a*b^2*d^3)*x^2 + 7*(3*b^3*c^2*d + 34*a*b^2*c*d^2 + 8*a^2*b*d^3)*x)*sqrt(d*x
+ c))/(a^2*b^4*c^6 - 4*a^3*b^3*c^5*d + 6*a^4*b^2*c^4*d^2 - 4*a^5*b*c^3*d^3 + a^6*c^2*d^4 + (b^6*c^4*d^2 - 4*a
*b^5*c^3*d^3 + 6*a^2*b^4*c^2*d^4 - 4*a^3*b^3*c*d^5 + a^4*b^2*d^6)*x^4 + 2*(b^6*c^5*d - 3*a*b^5*c^4*d^2 + 2*a^2
*b^4*c^3*d^3 + 2*a^3*b^3*c^2*d^4 - 3*a^4*b^2*c*d^5 + a^5*b*d^6)*x^3 + (b^6*c^6 - 9*a^2*b^4*c^4*d^2 + 16*a^3*b^
3*c^3*d^3 - 9*a^4*b^2*c^2*d^4 + a^6*d^6)*x^2 + 2*(a*b^5*c^6 - 3*a^2*b^4*c^5*d + 2*a^3*b^3*c^4*d^2 + 2*a^4*b^2*
c^3*d^3 - 3*a^5*b*c^2*d^4 + a^6*c*d^5)*x), -1/12*(105*(b^3*d^4*x^4 + a^2*b*c^2*d^2 + 2*(b^3*c*d^3 + a*b^2*d^4)
*x^3 + (b^3*c^2*d^2 + 4*a*b^2*c*d^3 + a^2*b*d^4)*x^2 + 2*(a*b^2*c^2*d^2 + a^2*b*c*d^3)*x)*sqrt(-b/(b*c - a*d))
*arctan(-(b*c - a*d)*sqrt(d*x + c)*sqrt(-b/(b*c - a*d))/(b*d*x + b*c)) - (105*b^3*d^3*x^3 - 6*b^3*c^3 + 39*a*b
^2*c^2*d + 80*a^2*b*c*d^2 - 8*a^3*d^3 + 35*(4*b^3*c*d^2 + 5*a*b^2*d^3)*x^2 + 7*(3*b^3*c^2*d + 34*a*b^2*c*d^2 +
8*a^2*b*d^3)*x)*sqrt(d*x + c))/(a^2*b^4*c^6 - 4*a^3*b^3*c^5*d + 6*a^4*b^2*c^4*d^2 - 4*a^5*b*c^3*d^3 + a^6*c^2
*d^4 + (b^6*c^4*d^2 - 4*a*b^5*c^3*d^3 + 6*a^2*b^4*c^2*d^4 - 4*a^3*b^3*c*d^5 + a^4*b^2*d^6)*x^4 + 2*(b^6*c^5*d
- 3*a*b^5*c^4*d^2 + 2*a^2*b^4*c^3*d^3 + 2*a^3*b^3*c^2*d^4 - 3*a^4*b^2*c*d^5 + a^5*b*d^6)*x^3 + (b^6*c^6 - 9*a^
2*b^4*c^4*d^2 + 16*a^3*b^3*c^3*d^3 - 9*a^4*b^2*c^2*d^4 + a^6*d^6)*x^2 + 2*(a*b^5*c^6 - 3*a^2*b^4*c^5*d + 2*a^3
*b^3*c^4*d^2 + 2*a^4*b^2*c^3*d^3 - 3*a^5*b*c^2*d^4 + a^6*c*d^5)*x)]

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giac [B]  time = 1.20, size = 298, normalized size = 1.78 \begin {gather*} \frac {35 \, b^{2} d^{2} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{4 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt {-b^{2} c + a b d}} + \frac {2 \, {\left (9 \, {\left (d x + c\right )} b d^{2} + b c d^{2} - a d^{3}\right )}}{3 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} {\left (d x + c\right )}^{\frac {3}{2}}} + \frac {11 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{3} d^{2} - 13 \, \sqrt {d x + c} b^{3} c d^{2} + 13 \, \sqrt {d x + c} a b^{2} d^{3}}{4 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^3/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

35/4*b^2*d^2*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^
3*b*c*d^3 + a^4*d^4)*sqrt(-b^2*c + a*b*d)) + 2/3*(9*(d*x + c)*b*d^2 + b*c*d^2 - a*d^3)/((b^4*c^4 - 4*a*b^3*c^3
*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)*(d*x + c)^(3/2)) + 1/4*(11*(d*x + c)^(3/2)*b^3*d^2 - 13*sqrt
(d*x + c)*b^3*c*d^2 + 13*sqrt(d*x + c)*a*b^2*d^3)/((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^
3 + a^4*d^4)*((d*x + c)*b - b*c + a*d)^2)

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maple [A]  time = 0.02, size = 206, normalized size = 1.23 \begin {gather*} \frac {13 \sqrt {d x +c}\, a \,b^{2} d^{3}}{4 \left (a d -b c \right )^{4} \left (b d x +a d \right )^{2}}-\frac {13 \sqrt {d x +c}\, b^{3} c \,d^{2}}{4 \left (a d -b c \right )^{4} \left (b d x +a d \right )^{2}}+\frac {11 \left (d x +c \right )^{\frac {3}{2}} b^{3} d^{2}}{4 \left (a d -b c \right )^{4} \left (b d x +a d \right )^{2}}+\frac {35 b^{2} d^{2} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{4 \left (a d -b c \right )^{4} \sqrt {\left (a d -b c \right ) b}}+\frac {6 b \,d^{2}}{\left (a d -b c \right )^{4} \sqrt {d x +c}}-\frac {2 d^{2}}{3 \left (a d -b c \right )^{3} \left (d x +c \right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^3/(d*x+c)^(5/2),x)

[Out]

-2/3*d^2/(a*d-b*c)^3/(d*x+c)^(3/2)+6*d^2/(a*d-b*c)^4*b/(d*x+c)^(1/2)+11/4*d^2/(a*d-b*c)^4*b^3/(b*d*x+a*d)^2*(d
*x+c)^(3/2)+13/4*d^3/(a*d-b*c)^4*b^2/(b*d*x+a*d)^2*(d*x+c)^(1/2)*a-13/4*d^2/(a*d-b*c)^4*b^3/(b*d*x+a*d)^2*(d*x
+c)^(1/2)*c+35/4*d^2/(a*d-b*c)^4*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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^3/(d*x+c)^(5/2),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 = 0.28, size = 243, normalized size = 1.46 \begin {gather*} \frac {\frac {175\,b^2\,d^2\,{\left (c+d\,x\right )}^2}{12\,{\left (a\,d-b\,c\right )}^3}-\frac {2\,d^2}{3\,\left (a\,d-b\,c\right )}+\frac {35\,b^3\,d^2\,{\left (c+d\,x\right )}^3}{4\,{\left (a\,d-b\,c\right )}^4}+\frac {14\,b\,d^2\,\left (c+d\,x\right )}{3\,{\left (a\,d-b\,c\right )}^2}}{b^2\,{\left (c+d\,x\right )}^{7/2}-\left (2\,b^2\,c-2\,a\,b\,d\right )\,{\left (c+d\,x\right )}^{5/2}+{\left (c+d\,x\right )}^{3/2}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}+\frac {35\,b^{3/2}\,d^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}\,\left (a^4\,d^4-4\,a^3\,b\,c\,d^3+6\,a^2\,b^2\,c^2\,d^2-4\,a\,b^3\,c^3\,d+b^4\,c^4\right )}{{\left (a\,d-b\,c\right )}^{9/2}}\right )}{4\,{\left (a\,d-b\,c\right )}^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)^3*(c + d*x)^(5/2)),x)

[Out]

((175*b^2*d^2*(c + d*x)^2)/(12*(a*d - b*c)^3) - (2*d^2)/(3*(a*d - b*c)) + (35*b^3*d^2*(c + d*x)^3)/(4*(a*d - b
*c)^4) + (14*b*d^2*(c + d*x))/(3*(a*d - b*c)^2))/(b^2*(c + d*x)^(7/2) - (2*b^2*c - 2*a*b*d)*(c + d*x)^(5/2) +
(c + d*x)^(3/2)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)) + (35*b^(3/2)*d^2*atan((b^(1/2)*(c + d*x)^(1/2)*(a^4*d^4 + b^
4*c^4 + 6*a^2*b^2*c^2*d^2 - 4*a*b^3*c^3*d - 4*a^3*b*c*d^3))/(a*d - b*c)^(9/2)))/(4*(a*d - b*c)^(9/2))

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

[Out]

Timed out

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