3.14.27 \(\int \frac {1}{(a+b x)^4 (c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=173 \[ -\frac {35 d^3}{8 \sqrt {c+d x} (b c-a d)^4}+\frac {35 \sqrt {b} d^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{8 (b c-a d)^{9/2}}-\frac {35 d^2}{24 (a+b x) \sqrt {c+d x} (b c-a d)^3}+\frac {7 d}{12 (a+b x)^2 \sqrt {c+d x} (b c-a d)^2}-\frac {1}{3 (a+b x)^3 \sqrt {c+d x} (b c-a d)} \]

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Rubi [A]  time = 0.07, antiderivative size = 173, 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 d^3}{8 \sqrt {c+d x} (b c-a d)^4}-\frac {35 d^2}{24 (a+b x) \sqrt {c+d x} (b c-a d)^3}+\frac {35 \sqrt {b} d^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{8 (b c-a d)^{9/2}}+\frac {7 d}{12 (a+b x)^2 \sqrt {c+d x} (b c-a d)^2}-\frac {1}{3 (a+b x)^3 \sqrt {c+d x} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^4*(c + d*x)^(3/2)),x]

[Out]

(-35*d^3)/(8*(b*c - a*d)^4*Sqrt[c + d*x]) - 1/(3*(b*c - a*d)*(a + b*x)^3*Sqrt[c + d*x]) + (7*d)/(12*(b*c - a*d
)^2*(a + b*x)^2*Sqrt[c + d*x]) - (35*d^2)/(24*(b*c - a*d)^3*(a + b*x)*Sqrt[c + d*x]) + (35*Sqrt[b]*d^3*ArcTanh
[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(8*(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)^4 (c+d x)^{3/2}} \, dx &=-\frac {1}{3 (b c-a d) (a+b x)^3 \sqrt {c+d x}}-\frac {(7 d) \int \frac {1}{(a+b x)^3 (c+d x)^{3/2}} \, dx}{6 (b c-a d)}\\ &=-\frac {1}{3 (b c-a d) (a+b x)^3 \sqrt {c+d x}}+\frac {7 d}{12 (b c-a d)^2 (a+b x)^2 \sqrt {c+d x}}+\frac {\left (35 d^2\right ) \int \frac {1}{(a+b x)^2 (c+d x)^{3/2}} \, dx}{24 (b c-a d)^2}\\ &=-\frac {1}{3 (b c-a d) (a+b x)^3 \sqrt {c+d x}}+\frac {7 d}{12 (b c-a d)^2 (a+b x)^2 \sqrt {c+d x}}-\frac {35 d^2}{24 (b c-a d)^3 (a+b x) \sqrt {c+d x}}-\frac {\left (35 d^3\right ) \int \frac {1}{(a+b x) (c+d x)^{3/2}} \, dx}{16 (b c-a d)^3}\\ &=-\frac {35 d^3}{8 (b c-a d)^4 \sqrt {c+d x}}-\frac {1}{3 (b c-a d) (a+b x)^3 \sqrt {c+d x}}+\frac {7 d}{12 (b c-a d)^2 (a+b x)^2 \sqrt {c+d x}}-\frac {35 d^2}{24 (b c-a d)^3 (a+b x) \sqrt {c+d x}}-\frac {\left (35 b d^3\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{16 (b c-a d)^4}\\ &=-\frac {35 d^3}{8 (b c-a d)^4 \sqrt {c+d x}}-\frac {1}{3 (b c-a d) (a+b x)^3 \sqrt {c+d x}}+\frac {7 d}{12 (b c-a d)^2 (a+b x)^2 \sqrt {c+d x}}-\frac {35 d^2}{24 (b c-a d)^3 (a+b x) \sqrt {c+d x}}-\frac {\left (35 b 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 c-a d)^4}\\ &=-\frac {35 d^3}{8 (b c-a d)^4 \sqrt {c+d x}}-\frac {1}{3 (b c-a d) (a+b x)^3 \sqrt {c+d x}}+\frac {7 d}{12 (b c-a d)^2 (a+b x)^2 \sqrt {c+d x}}-\frac {35 d^2}{24 (b c-a d)^3 (a+b x) \sqrt {c+d x}}+\frac {35 \sqrt {b} d^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{8 (b c-a d)^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 50, normalized size = 0.29 \begin {gather*} -\frac {2 d^3 \, _2F_1\left (-\frac {1}{2},4;\frac {1}{2};-\frac {b (c+d x)}{a d-b c}\right )}{\sqrt {c+d x} (a d-b c)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^4*(c + d*x)^(3/2)),x]

[Out]

(-2*d^3*Hypergeometric2F1[-1/2, 4, 1/2, -((b*(c + d*x))/(-(b*c) + a*d))])/((-(b*c) + a*d)^4*Sqrt[c + d*x])

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IntegrateAlgebraic [A]  time = 0.68, size = 223, normalized size = 1.29 \begin {gather*} \frac {d^3 \left (48 a^3 d^3+231 a^2 b d^2 (c+d x)-144 a^2 b c d^2+144 a b^2 c^2 d+280 a b^2 d (c+d x)^2-462 a b^2 c d (c+d x)-48 b^3 c^3+231 b^3 c^2 (c+d x)+105 b^3 (c+d x)^3-280 b^3 c (c+d x)^2\right )}{24 \sqrt {c+d x} (b c-a d)^4 (-a d-b (c+d x)+b c)^3}+\frac {35 \sqrt {b} d^3 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x} \sqrt {a d-b c}}{b c-a d}\right )}{8 (a d-b c)^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((a + b*x)^4*(c + d*x)^(3/2)),x]

[Out]

(d^3*(-48*b^3*c^3 + 144*a*b^2*c^2*d - 144*a^2*b*c*d^2 + 48*a^3*d^3 + 231*b^3*c^2*(c + d*x) - 462*a*b^2*c*d*(c
+ d*x) + 231*a^2*b*d^2*(c + d*x) - 280*b^3*c*(c + d*x)^2 + 280*a*b^2*d*(c + d*x)^2 + 105*b^3*(c + d*x)^3))/(24
*(b*c - a*d)^4*Sqrt[c + d*x]*(b*c - a*d - b*(c + d*x))^3) + (35*Sqrt[b]*d^3*ArcTan[(Sqrt[b]*Sqrt[-(b*c) + a*d]
*Sqrt[c + d*x])/(b*c - a*d)])/(8*(-(b*c) + a*d)^(9/2))

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fricas [B]  time = 1.53, size = 1204, normalized size = 6.96 \begin {gather*} \left [\frac {105 \, {\left (b^{3} d^{4} x^{4} + a^{3} c d^{3} + {\left (b^{3} c d^{3} + 3 \, a b^{2} d^{4}\right )} x^{3} + 3 \, {\left (a b^{2} c d^{3} + a^{2} b d^{4}\right )} x^{2} + {\left (3 \, a^{2} b c d^{3} + a^{3} d^{4}\right )} x\right )} \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b d x + 2 \, b c - a d + 2 \, {\left (b c - a d\right )} \sqrt {d x + c} \sqrt {\frac {b}{b c - a d}}}{b x + a}\right ) - 2 \, {\left (105 \, b^{3} d^{3} x^{3} + 8 \, b^{3} c^{3} - 38 \, a b^{2} c^{2} d + 87 \, a^{2} b c d^{2} + 48 \, a^{3} d^{3} + 35 \, {\left (b^{3} c d^{2} + 8 \, a b^{2} d^{3}\right )} x^{2} - 7 \, {\left (2 \, b^{3} c^{2} d - 14 \, a b^{2} c d^{2} - 33 \, a^{2} b d^{3}\right )} x\right )} \sqrt {d x + c}}{48 \, {\left (a^{3} b^{4} c^{5} - 4 \, a^{4} b^{3} c^{4} d + 6 \, a^{5} b^{2} c^{3} d^{2} - 4 \, a^{6} b c^{2} d^{3} + a^{7} c d^{4} + {\left (b^{7} c^{4} d - 4 \, a b^{6} c^{3} d^{2} + 6 \, a^{2} b^{5} c^{2} d^{3} - 4 \, a^{3} b^{4} c d^{4} + a^{4} b^{3} d^{5}\right )} x^{4} + {\left (b^{7} c^{5} - a b^{6} c^{4} d - 6 \, a^{2} b^{5} c^{3} d^{2} + 14 \, a^{3} b^{4} c^{2} d^{3} - 11 \, a^{4} b^{3} c d^{4} + 3 \, a^{5} b^{2} d^{5}\right )} x^{3} + 3 \, {\left (a b^{6} c^{5} - 3 \, a^{2} b^{5} c^{4} d + 2 \, a^{3} b^{4} c^{3} d^{2} + 2 \, a^{4} b^{3} c^{2} d^{3} - 3 \, a^{5} b^{2} c d^{4} + a^{6} b d^{5}\right )} x^{2} + {\left (3 \, a^{2} b^{5} c^{5} - 11 \, a^{3} b^{4} c^{4} d + 14 \, a^{4} b^{3} c^{3} d^{2} - 6 \, a^{5} b^{2} c^{2} d^{3} - a^{6} b c d^{4} + a^{7} d^{5}\right )} x\right )}}, \frac {105 \, {\left (b^{3} d^{4} x^{4} + a^{3} c d^{3} + {\left (b^{3} c d^{3} + 3 \, a b^{2} d^{4}\right )} x^{3} + 3 \, {\left (a b^{2} c d^{3} + a^{2} b d^{4}\right )} x^{2} + {\left (3 \, a^{2} b c d^{3} + a^{3} d^{4}\right )} x\right )} \sqrt {-\frac {b}{b c - a d}} \arctan \left (-\frac {{\left (b c - a d\right )} \sqrt {d x + c} \sqrt {-\frac {b}{b c - a d}}}{b d x + b c}\right ) - {\left (105 \, b^{3} d^{3} x^{3} + 8 \, b^{3} c^{3} - 38 \, a b^{2} c^{2} d + 87 \, a^{2} b c d^{2} + 48 \, a^{3} d^{3} + 35 \, {\left (b^{3} c d^{2} + 8 \, a b^{2} d^{3}\right )} x^{2} - 7 \, {\left (2 \, b^{3} c^{2} d - 14 \, a b^{2} c d^{2} - 33 \, a^{2} b d^{3}\right )} x\right )} \sqrt {d x + c}}{24 \, {\left (a^{3} b^{4} c^{5} - 4 \, a^{4} b^{3} c^{4} d + 6 \, a^{5} b^{2} c^{3} d^{2} - 4 \, a^{6} b c^{2} d^{3} + a^{7} c d^{4} + {\left (b^{7} c^{4} d - 4 \, a b^{6} c^{3} d^{2} + 6 \, a^{2} b^{5} c^{2} d^{3} - 4 \, a^{3} b^{4} c d^{4} + a^{4} b^{3} d^{5}\right )} x^{4} + {\left (b^{7} c^{5} - a b^{6} c^{4} d - 6 \, a^{2} b^{5} c^{3} d^{2} + 14 \, a^{3} b^{4} c^{2} d^{3} - 11 \, a^{4} b^{3} c d^{4} + 3 \, a^{5} b^{2} d^{5}\right )} x^{3} + 3 \, {\left (a b^{6} c^{5} - 3 \, a^{2} b^{5} c^{4} d + 2 \, a^{3} b^{4} c^{3} d^{2} + 2 \, a^{4} b^{3} c^{2} d^{3} - 3 \, a^{5} b^{2} c d^{4} + a^{6} b d^{5}\right )} x^{2} + {\left (3 \, a^{2} b^{5} c^{5} - 11 \, a^{3} b^{4} c^{4} d + 14 \, a^{4} b^{3} c^{3} d^{2} - 6 \, a^{5} b^{2} c^{2} d^{3} - a^{6} b c d^{4} + a^{7} d^{5}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^4/(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

[1/48*(105*(b^3*d^4*x^4 + a^3*c*d^3 + (b^3*c*d^3 + 3*a*b^2*d^4)*x^3 + 3*(a*b^2*c*d^3 + a^2*b*d^4)*x^2 + (3*a^2
*b*c*d^3 + a^3*d^4)*x)*sqrt(b/(b*c - a*d))*log((b*d*x + 2*b*c - a*d + 2*(b*c - a*d)*sqrt(d*x + c)*sqrt(b/(b*c
- a*d)))/(b*x + a)) - 2*(105*b^3*d^3*x^3 + 8*b^3*c^3 - 38*a*b^2*c^2*d + 87*a^2*b*c*d^2 + 48*a^3*d^3 + 35*(b^3*
c*d^2 + 8*a*b^2*d^3)*x^2 - 7*(2*b^3*c^2*d - 14*a*b^2*c*d^2 - 33*a^2*b*d^3)*x)*sqrt(d*x + c))/(a^3*b^4*c^5 - 4*
a^4*b^3*c^4*d + 6*a^5*b^2*c^3*d^2 - 4*a^6*b*c^2*d^3 + a^7*c*d^4 + (b^7*c^4*d - 4*a*b^6*c^3*d^2 + 6*a^2*b^5*c^2
*d^3 - 4*a^3*b^4*c*d^4 + a^4*b^3*d^5)*x^4 + (b^7*c^5 - a*b^6*c^4*d - 6*a^2*b^5*c^3*d^2 + 14*a^3*b^4*c^2*d^3 -
11*a^4*b^3*c*d^4 + 3*a^5*b^2*d^5)*x^3 + 3*(a*b^6*c^5 - 3*a^2*b^5*c^4*d + 2*a^3*b^4*c^3*d^2 + 2*a^4*b^3*c^2*d^3
- 3*a^5*b^2*c*d^4 + a^6*b*d^5)*x^2 + (3*a^2*b^5*c^5 - 11*a^3*b^4*c^4*d + 14*a^4*b^3*c^3*d^2 - 6*a^5*b^2*c^2*d
^3 - a^6*b*c*d^4 + a^7*d^5)*x), 1/24*(105*(b^3*d^4*x^4 + a^3*c*d^3 + (b^3*c*d^3 + 3*a*b^2*d^4)*x^3 + 3*(a*b^2*
c*d^3 + a^2*b*d^4)*x^2 + (3*a^2*b*c*d^3 + a^3*d^4)*x)*sqrt(-b/(b*c - a*d))*arctan(-(b*c - a*d)*sqrt(d*x + c)*s
qrt(-b/(b*c - a*d))/(b*d*x + b*c)) - (105*b^3*d^3*x^3 + 8*b^3*c^3 - 38*a*b^2*c^2*d + 87*a^2*b*c*d^2 + 48*a^3*d
^3 + 35*(b^3*c*d^2 + 8*a*b^2*d^3)*x^2 - 7*(2*b^3*c^2*d - 14*a*b^2*c*d^2 - 33*a^2*b*d^3)*x)*sqrt(d*x + c))/(a^3
*b^4*c^5 - 4*a^4*b^3*c^4*d + 6*a^5*b^2*c^3*d^2 - 4*a^6*b*c^2*d^3 + a^7*c*d^4 + (b^7*c^4*d - 4*a*b^6*c^3*d^2 +
6*a^2*b^5*c^2*d^3 - 4*a^3*b^4*c*d^4 + a^4*b^3*d^5)*x^4 + (b^7*c^5 - a*b^6*c^4*d - 6*a^2*b^5*c^3*d^2 + 14*a^3*b
^4*c^2*d^3 - 11*a^4*b^3*c*d^4 + 3*a^5*b^2*d^5)*x^3 + 3*(a*b^6*c^5 - 3*a^2*b^5*c^4*d + 2*a^3*b^4*c^3*d^2 + 2*a^
4*b^3*c^2*d^3 - 3*a^5*b^2*c*d^4 + a^6*b*d^5)*x^2 + (3*a^2*b^5*c^5 - 11*a^3*b^4*c^4*d + 14*a^4*b^3*c^3*d^2 - 6*
a^5*b^2*c^2*d^3 - a^6*b*c*d^4 + a^7*d^5)*x)]

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giac [B]  time = 1.31, size = 326, normalized size = 1.88 \begin {gather*} -\frac {35 \, b d^{3} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{8 \, {\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 \, d^{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 )} \sqrt {d x + c}} - \frac {57 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{3} d^{3} - 136 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{3} c d^{3} + 87 \, \sqrt {d x + c} b^{3} c^{2} d^{3} + 136 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{2} d^{4} - 174 \, \sqrt {d x + c} a b^{2} c d^{4} + 87 \, \sqrt {d x + c} a^{2} b d^{5}}{24 \, {\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 )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^4/(d*x+c)^(3/2),x, algorithm="giac")

[Out]

-35/8*b*d^3*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*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)*sqrt(d*x + c)) - 1/24*(57*(d*x + c)^(5/2)*b^3*d^3 - 136*(d*x + c)^(3/2)*b^3*c*d^3 + 87*sqrt(d*x +
c)*b^3*c^2*d^3 + 136*(d*x + c)^(3/2)*a*b^2*d^4 - 174*sqrt(d*x + c)*a*b^2*c*d^4 + 87*sqrt(d*x + c)*a^2*b*d^5)/
((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)^3)

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maple [B]  time = 0.02, size = 292, normalized size = 1.69 \begin {gather*} -\frac {29 \sqrt {d x +c}\, a^{2} b \,d^{5}}{8 \left (a d -b c \right )^{4} \left (b d x +a d \right )^{3}}+\frac {29 \sqrt {d x +c}\, a \,b^{2} c \,d^{4}}{4 \left (a d -b c \right )^{4} \left (b d x +a d \right )^{3}}-\frac {29 \sqrt {d x +c}\, b^{3} c^{2} d^{3}}{8 \left (a d -b c \right )^{4} \left (b d x +a d \right )^{3}}-\frac {17 \left (d x +c \right )^{\frac {3}{2}} a \,b^{2} d^{4}}{3 \left (a d -b c \right )^{4} \left (b d x +a d \right )^{3}}+\frac {17 \left (d x +c \right )^{\frac {3}{2}} b^{3} c \,d^{3}}{3 \left (a d -b c \right )^{4} \left (b d x +a d \right )^{3}}-\frac {19 \left (d x +c \right )^{\frac {5}{2}} b^{3} d^{3}}{8 \left (a d -b c \right )^{4} \left (b d x +a d \right )^{3}}-\frac {35 b \,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 )^{4} \sqrt {\left (a d -b c \right ) b}}-\frac {2 d^{3}}{\left (a d -b c \right )^{4} \sqrt {d x +c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^4/(d*x+c)^(3/2),x)

[Out]

-2*d^3/(a*d-b*c)^4/(d*x+c)^(1/2)-19/8*d^3/(a*d-b*c)^4*b^3/(b*d*x+a*d)^3*(d*x+c)^(5/2)-17/3*d^4/(a*d-b*c)^4*b^2
/(b*d*x+a*d)^3*(d*x+c)^(3/2)*a+17/3*d^3/(a*d-b*c)^4*b^3/(b*d*x+a*d)^3*(d*x+c)^(3/2)*c-29/8*d^5/(a*d-b*c)^4*b/(
b*d*x+a*d)^3*(d*x+c)^(1/2)*a^2+29/4*d^4/(a*d-b*c)^4*b^2/(b*d*x+a*d)^3*(d*x+c)^(1/2)*a*c-29/8*d^3/(a*d-b*c)^4*b
^3/(b*d*x+a*d)^3*(d*x+c)^(1/2)*c^2-35/8*d^3/(a*d-b*c)^4*b/((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)^4/(d*x+c)^(3/2),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 = 0.54, size = 294, normalized size = 1.70 \begin {gather*} -\frac {\frac {2\,d^3}{a\,d-b\,c}+\frac {35\,b^2\,d^3\,{\left (c+d\,x\right )}^2}{3\,{\left (a\,d-b\,c\right )}^3}+\frac {35\,b^3\,d^3\,{\left (c+d\,x\right )}^3}{8\,{\left (a\,d-b\,c\right )}^4}+\frac {77\,b\,d^3\,\left (c+d\,x\right )}{8\,{\left (a\,d-b\,c\right )}^2}}{\sqrt {c+d\,x}\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3\right )+b^3\,{\left (c+d\,x\right )}^{7/2}-\left (3\,b^3\,c-3\,a\,b^2\,d\right )\,{\left (c+d\,x\right )}^{5/2}+{\left (c+d\,x\right )}^{3/2}\,\left (3\,a^2\,b\,d^2-6\,a\,b^2\,c\,d+3\,b^3\,c^2\right )}-\frac {35\,\sqrt {b}\,d^3\,\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 )}{8\,{\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)^4*(c + d*x)^(3/2)),x)

[Out]

- ((2*d^3)/(a*d - b*c) + (35*b^2*d^3*(c + d*x)^2)/(3*(a*d - b*c)^3) + (35*b^3*d^3*(c + d*x)^3)/(8*(a*d - b*c)^
4) + (77*b*d^3*(c + d*x))/(8*(a*d - b*c)^2))/((c + d*x)^(1/2)*(a^3*d^3 - b^3*c^3 + 3*a*b^2*c^2*d - 3*a^2*b*c*d
^2) + b^3*(c + d*x)^(7/2) - (3*b^3*c - 3*a*b^2*d)*(c + d*x)^(5/2) + (c + d*x)^(3/2)*(3*b^3*c^2 + 3*a^2*b*d^2 -
6*a*b^2*c*d)) - (35*b^(1/2)*d^3*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)))/(8*(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)**4/(d*x+c)**(3/2),x)

[Out]

Timed out

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