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

Optimal. Leaf size=172 \[ -\frac {3 d^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{64 b^{5/2} (b c-a d)^{5/2}}+\frac {3 d^3 \sqrt {c+d x}}{64 b^2 (a+b x) (b c-a d)^2}-\frac {d^2 \sqrt {c+d x}}{32 b^2 (a+b x)^2 (b c-a d)}-\frac {d \sqrt {c+d x}}{8 b^2 (a+b x)^3}-\frac {(c+d x)^{3/2}}{4 b (a+b x)^4} \]

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Rubi [A]  time = 0.07, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {47, 51, 63, 208} \begin {gather*} \frac {3 d^3 \sqrt {c+d x}}{64 b^2 (a+b x) (b c-a d)^2}-\frac {d^2 \sqrt {c+d x}}{32 b^2 (a+b x)^2 (b c-a d)}-\frac {3 d^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{64 b^{5/2} (b c-a d)^{5/2}}-\frac {d \sqrt {c+d x}}{8 b^2 (a+b x)^3}-\frac {(c+d x)^{3/2}}{4 b (a+b x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(3/2)/(a + b*x)^5,x]

[Out]

-(d*Sqrt[c + d*x])/(8*b^2*(a + b*x)^3) - (d^2*Sqrt[c + d*x])/(32*b^2*(b*c - a*d)*(a + b*x)^2) + (3*d^3*Sqrt[c
+ d*x])/(64*b^2*(b*c - a*d)^2*(a + b*x)) - (c + d*x)^(3/2)/(4*b*(a + b*x)^4) - (3*d^4*ArcTanh[(Sqrt[b]*Sqrt[c
+ d*x])/Sqrt[b*c - a*d]])/(64*b^(5/2)*(b*c - a*d)^(5/2))

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)^5} \, dx &=-\frac {(c+d x)^{3/2}}{4 b (a+b x)^4}+\frac {(3 d) \int \frac {\sqrt {c+d x}}{(a+b x)^4} \, dx}{8 b}\\ &=-\frac {d \sqrt {c+d x}}{8 b^2 (a+b x)^3}-\frac {(c+d x)^{3/2}}{4 b (a+b x)^4}+\frac {d^2 \int \frac {1}{(a+b x)^3 \sqrt {c+d x}} \, dx}{16 b^2}\\ &=-\frac {d \sqrt {c+d x}}{8 b^2 (a+b x)^3}-\frac {d^2 \sqrt {c+d x}}{32 b^2 (b c-a d) (a+b x)^2}-\frac {(c+d x)^{3/2}}{4 b (a+b x)^4}-\frac {\left (3 d^3\right ) \int \frac {1}{(a+b x)^2 \sqrt {c+d x}} \, dx}{64 b^2 (b c-a d)}\\ &=-\frac {d \sqrt {c+d x}}{8 b^2 (a+b x)^3}-\frac {d^2 \sqrt {c+d x}}{32 b^2 (b c-a d) (a+b x)^2}+\frac {3 d^3 \sqrt {c+d x}}{64 b^2 (b c-a d)^2 (a+b x)}-\frac {(c+d x)^{3/2}}{4 b (a+b x)^4}+\frac {\left (3 d^4\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{128 b^2 (b c-a d)^2}\\ &=-\frac {d \sqrt {c+d x}}{8 b^2 (a+b x)^3}-\frac {d^2 \sqrt {c+d x}}{32 b^2 (b c-a d) (a+b x)^2}+\frac {3 d^3 \sqrt {c+d x}}{64 b^2 (b c-a d)^2 (a+b x)}-\frac {(c+d x)^{3/2}}{4 b (a+b x)^4}+\frac {\left (3 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{64 b^2 (b c-a d)^2}\\ &=-\frac {d \sqrt {c+d x}}{8 b^2 (a+b x)^3}-\frac {d^2 \sqrt {c+d x}}{32 b^2 (b c-a d) (a+b x)^2}+\frac {3 d^3 \sqrt {c+d x}}{64 b^2 (b c-a d)^2 (a+b x)}-\frac {(c+d x)^{3/2}}{4 b (a+b x)^4}-\frac {3 d^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{64 b^{5/2} (b c-a d)^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 52, normalized size = 0.30 \begin {gather*} \frac {2 d^4 (c+d x)^{5/2} \, _2F_1\left (\frac {5}{2},5;\frac {7}{2};-\frac {b (c+d x)}{a d-b c}\right )}{5 (a d-b c)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 1.01, size = 226, normalized size = 1.31 \begin {gather*} \frac {d^4 \sqrt {c+d x} \left (-3 a^3 d^3-11 a^2 b d^2 (c+d x)+9 a^2 b c d^2-9 a b^2 c^2 d+11 a b^2 d (c+d x)^2+22 a b^2 c d (c+d x)+3 b^3 c^3-11 b^3 c^2 (c+d x)+3 b^3 (c+d x)^3-11 b^3 c (c+d x)^2\right )}{64 b^2 (b c-a d)^2 (-a d-b (c+d x)+b c)^4}-\frac {3 d^4 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x} \sqrt {a d-b c}}{b c-a d}\right )}{64 b^{5/2} (a d-b c)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^4*Sqrt[c + d*x]*(3*b^3*c^3 - 9*a*b^2*c^2*d + 9*a^2*b*c*d^2 - 3*a^3*d^3 - 11*b^3*c^2*(c + d*x) + 22*a*b^2*c*
d*(c + d*x) - 11*a^2*b*d^2*(c + d*x) - 11*b^3*c*(c + d*x)^2 + 11*a*b^2*d*(c + d*x)^2 + 3*b^3*(c + d*x)^3))/(64
*b^2*(b*c - a*d)^2*(b*c - a*d - b*(c + d*x))^4) - (3*d^4*ArcTan[(Sqrt[b]*Sqrt[-(b*c) + a*d]*Sqrt[c + d*x])/(b*
c - a*d)])/(64*b^(5/2)*(-(b*c) + a*d)^(5/2))

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fricas [B]  time = 1.54, size = 1043, normalized size = 6.06 \begin {gather*} \left [\frac {3 \, {\left (b^{4} d^{4} x^{4} + 4 \, a b^{3} d^{4} x^{3} + 6 \, a^{2} b^{2} d^{4} x^{2} + 4 \, a^{3} b d^{4} x + a^{4} d^{4}\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 (16 \, b^{5} c^{4} - 40 \, a b^{4} c^{3} d + 26 \, a^{2} b^{3} c^{2} d^{2} + a^{3} b^{2} c d^{3} - 3 \, a^{4} b d^{4} - 3 \, {\left (b^{5} c d^{3} - a b^{4} d^{4}\right )} x^{3} + {\left (2 \, b^{5} c^{2} d^{2} - 13 \, a b^{4} c d^{3} + 11 \, a^{2} b^{3} d^{4}\right )} x^{2} + {\left (24 \, b^{5} c^{3} d - 68 \, a b^{4} c^{2} d^{2} + 55 \, a^{2} b^{3} c d^{3} - 11 \, a^{3} b^{2} d^{4}\right )} x\right )} \sqrt {d x + c}}{128 \, {\left (a^{4} b^{6} c^{3} - 3 \, a^{5} b^{5} c^{2} d + 3 \, a^{6} b^{4} c d^{2} - a^{7} b^{3} d^{3} + {\left (b^{10} c^{3} - 3 \, a b^{9} c^{2} d + 3 \, a^{2} b^{8} c d^{2} - a^{3} b^{7} d^{3}\right )} x^{4} + 4 \, {\left (a b^{9} c^{3} - 3 \, a^{2} b^{8} c^{2} d + 3 \, a^{3} b^{7} c d^{2} - a^{4} b^{6} d^{3}\right )} x^{3} + 6 \, {\left (a^{2} b^{8} c^{3} - 3 \, a^{3} b^{7} c^{2} d + 3 \, a^{4} b^{6} c d^{2} - a^{5} b^{5} d^{3}\right )} x^{2} + 4 \, {\left (a^{3} b^{7} c^{3} - 3 \, a^{4} b^{6} c^{2} d + 3 \, a^{5} b^{5} c d^{2} - a^{6} b^{4} d^{3}\right )} x\right )}}, \frac {3 \, {\left (b^{4} d^{4} x^{4} + 4 \, a b^{3} d^{4} x^{3} + 6 \, a^{2} b^{2} d^{4} x^{2} + 4 \, a^{3} b d^{4} x + a^{4} d^{4}\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 (16 \, b^{5} c^{4} - 40 \, a b^{4} c^{3} d + 26 \, a^{2} b^{3} c^{2} d^{2} + a^{3} b^{2} c d^{3} - 3 \, a^{4} b d^{4} - 3 \, {\left (b^{5} c d^{3} - a b^{4} d^{4}\right )} x^{3} + {\left (2 \, b^{5} c^{2} d^{2} - 13 \, a b^{4} c d^{3} + 11 \, a^{2} b^{3} d^{4}\right )} x^{2} + {\left (24 \, b^{5} c^{3} d - 68 \, a b^{4} c^{2} d^{2} + 55 \, a^{2} b^{3} c d^{3} - 11 \, a^{3} b^{2} d^{4}\right )} x\right )} \sqrt {d x + c}}{64 \, {\left (a^{4} b^{6} c^{3} - 3 \, a^{5} b^{5} c^{2} d + 3 \, a^{6} b^{4} c d^{2} - a^{7} b^{3} d^{3} + {\left (b^{10} c^{3} - 3 \, a b^{9} c^{2} d + 3 \, a^{2} b^{8} c d^{2} - a^{3} b^{7} d^{3}\right )} x^{4} + 4 \, {\left (a b^{9} c^{3} - 3 \, a^{2} b^{8} c^{2} d + 3 \, a^{3} b^{7} c d^{2} - a^{4} b^{6} d^{3}\right )} x^{3} + 6 \, {\left (a^{2} b^{8} c^{3} - 3 \, a^{3} b^{7} c^{2} d + 3 \, a^{4} b^{6} c d^{2} - a^{5} b^{5} d^{3}\right )} x^{2} + 4 \, {\left (a^{3} b^{7} c^{3} - 3 \, a^{4} b^{6} c^{2} d + 3 \, a^{5} b^{5} c d^{2} - a^{6} b^{4} d^{3}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/128*(3*(b^4*d^4*x^4 + 4*a*b^3*d^4*x^3 + 6*a^2*b^2*d^4*x^2 + 4*a^3*b*d^4*x + a^4*d^4)*sqrt(b^2*c - a*b*d)*lo
g((b*d*x + 2*b*c - a*d - 2*sqrt(b^2*c - a*b*d)*sqrt(d*x + c))/(b*x + a)) - 2*(16*b^5*c^4 - 40*a*b^4*c^3*d + 26
*a^2*b^3*c^2*d^2 + a^3*b^2*c*d^3 - 3*a^4*b*d^4 - 3*(b^5*c*d^3 - a*b^4*d^4)*x^3 + (2*b^5*c^2*d^2 - 13*a*b^4*c*d
^3 + 11*a^2*b^3*d^4)*x^2 + (24*b^5*c^3*d - 68*a*b^4*c^2*d^2 + 55*a^2*b^3*c*d^3 - 11*a^3*b^2*d^4)*x)*sqrt(d*x +
c))/(a^4*b^6*c^3 - 3*a^5*b^5*c^2*d + 3*a^6*b^4*c*d^2 - a^7*b^3*d^3 + (b^10*c^3 - 3*a*b^9*c^2*d + 3*a^2*b^8*c*
d^2 - a^3*b^7*d^3)*x^4 + 4*(a*b^9*c^3 - 3*a^2*b^8*c^2*d + 3*a^3*b^7*c*d^2 - a^4*b^6*d^3)*x^3 + 6*(a^2*b^8*c^3
- 3*a^3*b^7*c^2*d + 3*a^4*b^6*c*d^2 - a^5*b^5*d^3)*x^2 + 4*(a^3*b^7*c^3 - 3*a^4*b^6*c^2*d + 3*a^5*b^5*c*d^2 -
a^6*b^4*d^3)*x), 1/64*(3*(b^4*d^4*x^4 + 4*a*b^3*d^4*x^3 + 6*a^2*b^2*d^4*x^2 + 4*a^3*b*d^4*x + a^4*d^4)*sqrt(-b
^2*c + a*b*d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) - (16*b^5*c^4 - 40*a*b^4*c^3*d + 26*a^2
*b^3*c^2*d^2 + a^3*b^2*c*d^3 - 3*a^4*b*d^4 - 3*(b^5*c*d^3 - a*b^4*d^4)*x^3 + (2*b^5*c^2*d^2 - 13*a*b^4*c*d^3 +
11*a^2*b^3*d^4)*x^2 + (24*b^5*c^3*d - 68*a*b^4*c^2*d^2 + 55*a^2*b^3*c*d^3 - 11*a^3*b^2*d^4)*x)*sqrt(d*x + c))
/(a^4*b^6*c^3 - 3*a^5*b^5*c^2*d + 3*a^6*b^4*c*d^2 - a^7*b^3*d^3 + (b^10*c^3 - 3*a*b^9*c^2*d + 3*a^2*b^8*c*d^2
- a^3*b^7*d^3)*x^4 + 4*(a*b^9*c^3 - 3*a^2*b^8*c^2*d + 3*a^3*b^7*c*d^2 - a^4*b^6*d^3)*x^3 + 6*(a^2*b^8*c^3 - 3*
a^3*b^7*c^2*d + 3*a^4*b^6*c*d^2 - a^5*b^5*d^3)*x^2 + 4*(a^3*b^7*c^3 - 3*a^4*b^6*c^2*d + 3*a^5*b^5*c*d^2 - a^6*
b^4*d^3)*x)]

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giac [A]  time = 1.47, size = 285, normalized size = 1.66 \begin {gather*} \frac {3 \, d^{4} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{64 \, {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} \sqrt {-b^{2} c + a b d}} + \frac {3 \, {\left (d x + c\right )}^{\frac {7}{2}} b^{3} d^{4} - 11 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{3} c d^{4} - 11 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{3} c^{2} d^{4} + 3 \, \sqrt {d x + c} b^{3} c^{3} d^{4} + 11 \, {\left (d x + c\right )}^{\frac {5}{2}} a b^{2} d^{5} + 22 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{2} c d^{5} - 9 \, \sqrt {d x + c} a b^{2} c^{2} d^{5} - 11 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} b d^{6} + 9 \, \sqrt {d x + c} a^{2} b c d^{6} - 3 \, \sqrt {d x + c} a^{3} d^{7}}{64 \, {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/64*d^4*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2)*sqrt(-b^2*c + a*b
*d)) + 1/64*(3*(d*x + c)^(7/2)*b^3*d^4 - 11*(d*x + c)^(5/2)*b^3*c*d^4 - 11*(d*x + c)^(3/2)*b^3*c^2*d^4 + 3*sqr
t(d*x + c)*b^3*c^3*d^4 + 11*(d*x + c)^(5/2)*a*b^2*d^5 + 22*(d*x + c)^(3/2)*a*b^2*c*d^5 - 9*sqrt(d*x + c)*a*b^2
*c^2*d^5 - 11*(d*x + c)^(3/2)*a^2*b*d^6 + 9*sqrt(d*x + c)*a^2*b*c*d^6 - 3*sqrt(d*x + c)*a^3*d^7)/((b^4*c^2 - 2
*a*b^3*c*d + a^2*b^2*d^2)*((d*x + c)*b - b*c + a*d)^4)

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maple [A]  time = 0.02, size = 222, normalized size = 1.29 \begin {gather*} \frac {3 \left (d x +c \right )^{\frac {7}{2}} b \,d^{4}}{64 \left (b d x +a d \right )^{4} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}-\frac {3 \sqrt {d x +c}\, a \,d^{5}}{64 \left (b d x +a d \right )^{4} b^{2}}+\frac {3 \sqrt {d x +c}\, c \,d^{4}}{64 \left (b d x +a d \right )^{4} b}+\frac {11 \left (d x +c \right )^{\frac {5}{2}} d^{4}}{64 \left (b d x +a d \right )^{4} \left (a d -b c \right )}-\frac {11 \left (d x +c \right )^{\frac {3}{2}} d^{4}}{64 \left (b d x +a d \right )^{4} b}+\frac {3 d^{4} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{64 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {\left (a d -b c \right ) b}\, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/64*d^4/(b*d*x+a*d)^4*b/(a^2*d^2-2*a*b*c*d+b^2*c^2)*(d*x+c)^(7/2)+11/64*d^4/(b*d*x+a*d)^4/(a*d-b*c)*(d*x+c)^(
5/2)-11/64*d^4/(b*d*x+a*d)^4/b*(d*x+c)^(3/2)-3/64*d^5/(b*d*x+a*d)^4/b^2*(d*x+c)^(1/2)*a+3/64*d^4/(b*d*x+a*d)^4
/b*(d*x+c)^(1/2)*c+3/64*d^4/(a^2*d^2-2*a*b*c*d+b^2*c^2)/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((d*x+c)^(3/2)/(b*x+a)^5,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 = 0.37, size = 296, normalized size = 1.72 \begin {gather*} \frac {3\,d^4\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{64\,b^{5/2}\,{\left (a\,d-b\,c\right )}^{5/2}}-\frac {\frac {11\,d^4\,{\left (c+d\,x\right )}^{3/2}}{64\,b}-\frac {11\,d^4\,{\left (c+d\,x\right )}^{5/2}}{64\,\left (a\,d-b\,c\right )}+\frac {3\,d^4\,\left (a\,d-b\,c\right )\,\sqrt {c+d\,x}}{64\,b^2}-\frac {3\,b\,d^4\,{\left (c+d\,x\right )}^{7/2}}{64\,{\left (a\,d-b\,c\right )}^2}}{b^4\,{\left (c+d\,x\right )}^4-\left (4\,b^4\,c-4\,a\,b^3\,d\right )\,{\left (c+d\,x\right )}^3-\left (c+d\,x\right )\,\left (-4\,a^3\,b\,d^3+12\,a^2\,b^2\,c\,d^2-12\,a\,b^3\,c^2\,d+4\,b^4\,c^3\right )+a^4\,d^4+b^4\,c^4+{\left (c+d\,x\right )}^2\,\left (6\,a^2\,b^2\,d^2-12\,a\,b^3\,c\,d+6\,b^4\,c^2\right )+6\,a^2\,b^2\,c^2\,d^2-4\,a\,b^3\,c^3\,d-4\,a^3\,b\,c\,d^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(3/2)/(a + b*x)^5,x)

[Out]

(3*d^4*atan((b^(1/2)*(c + d*x)^(1/2))/(a*d - b*c)^(1/2)))/(64*b^(5/2)*(a*d - b*c)^(5/2)) - ((11*d^4*(c + d*x)^
(3/2))/(64*b) - (11*d^4*(c + d*x)^(5/2))/(64*(a*d - b*c)) + (3*d^4*(a*d - b*c)*(c + d*x)^(1/2))/(64*b^2) - (3*
b*d^4*(c + d*x)^(7/2))/(64*(a*d - b*c)^2))/(b^4*(c + d*x)^4 - (4*b^4*c - 4*a*b^3*d)*(c + d*x)^3 - (c + d*x)*(4
*b^4*c^3 - 4*a^3*b*d^3 + 12*a^2*b^2*c*d^2 - 12*a*b^3*c^2*d) + a^4*d^4 + b^4*c^4 + (c + d*x)^2*(6*b^4*c^2 + 6*a
^2*b^2*d^2 - 12*a*b^3*c*d) + 6*a^2*b^2*c^2*d^2 - 4*a*b^3*c^3*d - 4*a^3*b*c*d^3)

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

[Out]

Timed out

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