3.14.4 \(\int \frac {(c+d x)^{5/2}}{(a+b x)^6} \, dx\)

Optimal. Leaf size=198 \[ -\frac {3 d^5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{128 b^{7/2} (b c-a d)^{5/2}}+\frac {3 d^4 \sqrt {c+d x}}{128 b^3 (a+b x) (b c-a d)^2}-\frac {d^3 \sqrt {c+d x}}{64 b^3 (a+b x)^2 (b c-a d)}-\frac {d^2 \sqrt {c+d x}}{16 b^3 (a+b x)^3}-\frac {d (c+d x)^{3/2}}{8 b^2 (a+b x)^4}-\frac {(c+d x)^{5/2}}{5 b (a+b x)^5} \]

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Rubi [A]  time = 0.09, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 7, 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^4 \sqrt {c+d x}}{128 b^3 (a+b x) (b c-a d)^2}-\frac {d^3 \sqrt {c+d x}}{64 b^3 (a+b x)^2 (b c-a d)}-\frac {d^2 \sqrt {c+d x}}{16 b^3 (a+b x)^3}-\frac {3 d^5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{128 b^{7/2} (b c-a d)^{5/2}}-\frac {d (c+d x)^{3/2}}{8 b^2 (a+b x)^4}-\frac {(c+d x)^{5/2}}{5 b (a+b x)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/2)/(a + b*x)^6,x]

[Out]

-(d^2*Sqrt[c + d*x])/(16*b^3*(a + b*x)^3) - (d^3*Sqrt[c + d*x])/(64*b^3*(b*c - a*d)*(a + b*x)^2) + (3*d^4*Sqrt
[c + d*x])/(128*b^3*(b*c - a*d)^2*(a + b*x)) - (d*(c + d*x)^(3/2))/(8*b^2*(a + b*x)^4) - (c + d*x)^(5/2)/(5*b*
(a + b*x)^5) - (3*d^5*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(128*b^(7/2)*(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)^{5/2}}{(a+b x)^6} \, dx &=-\frac {(c+d x)^{5/2}}{5 b (a+b x)^5}+\frac {d \int \frac {(c+d x)^{3/2}}{(a+b x)^5} \, dx}{2 b}\\ &=-\frac {d (c+d x)^{3/2}}{8 b^2 (a+b x)^4}-\frac {(c+d x)^{5/2}}{5 b (a+b x)^5}+\frac {\left (3 d^2\right ) \int \frac {\sqrt {c+d x}}{(a+b x)^4} \, dx}{16 b^2}\\ &=-\frac {d^2 \sqrt {c+d x}}{16 b^3 (a+b x)^3}-\frac {d (c+d x)^{3/2}}{8 b^2 (a+b x)^4}-\frac {(c+d x)^{5/2}}{5 b (a+b x)^5}+\frac {d^3 \int \frac {1}{(a+b x)^3 \sqrt {c+d x}} \, dx}{32 b^3}\\ &=-\frac {d^2 \sqrt {c+d x}}{16 b^3 (a+b x)^3}-\frac {d^3 \sqrt {c+d x}}{64 b^3 (b c-a d) (a+b x)^2}-\frac {d (c+d x)^{3/2}}{8 b^2 (a+b x)^4}-\frac {(c+d x)^{5/2}}{5 b (a+b x)^5}-\frac {\left (3 d^4\right ) \int \frac {1}{(a+b x)^2 \sqrt {c+d x}} \, dx}{128 b^3 (b c-a d)}\\ &=-\frac {d^2 \sqrt {c+d x}}{16 b^3 (a+b x)^3}-\frac {d^3 \sqrt {c+d x}}{64 b^3 (b c-a d) (a+b x)^2}+\frac {3 d^4 \sqrt {c+d x}}{128 b^3 (b c-a d)^2 (a+b x)}-\frac {d (c+d x)^{3/2}}{8 b^2 (a+b x)^4}-\frac {(c+d x)^{5/2}}{5 b (a+b x)^5}+\frac {\left (3 d^5\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{256 b^3 (b c-a d)^2}\\ &=-\frac {d^2 \sqrt {c+d x}}{16 b^3 (a+b x)^3}-\frac {d^3 \sqrt {c+d x}}{64 b^3 (b c-a d) (a+b x)^2}+\frac {3 d^4 \sqrt {c+d x}}{128 b^3 (b c-a d)^2 (a+b x)}-\frac {d (c+d x)^{3/2}}{8 b^2 (a+b x)^4}-\frac {(c+d x)^{5/2}}{5 b (a+b x)^5}+\frac {\left (3 d^4\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{128 b^3 (b c-a d)^2}\\ &=-\frac {d^2 \sqrt {c+d x}}{16 b^3 (a+b x)^3}-\frac {d^3 \sqrt {c+d x}}{64 b^3 (b c-a d) (a+b x)^2}+\frac {3 d^4 \sqrt {c+d x}}{128 b^3 (b c-a d)^2 (a+b x)}-\frac {d (c+d x)^{3/2}}{8 b^2 (a+b x)^4}-\frac {(c+d x)^{5/2}}{5 b (a+b x)^5}-\frac {3 d^5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{128 b^{7/2} (b c-a d)^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 52, normalized size = 0.26 \begin {gather*} \frac {2 d^5 (c+d x)^{7/2} \, _2F_1\left (\frac {7}{2},6;\frac {9}{2};-\frac {b (c+d x)}{a d-b c}\right )}{7 (a d-b c)^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(5/2)/(a + b*x)^6,x]

[Out]

(2*d^5*(c + d*x)^(7/2)*Hypergeometric2F1[7/2, 6, 9/2, -((b*(c + d*x))/(-(b*c) + a*d))])/(7*(-(b*c) + a*d)^6)

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IntegrateAlgebraic [A]  time = 1.43, size = 307, normalized size = 1.55 \begin {gather*} \frac {d^5 \sqrt {c+d x} \left (15 a^4 d^4+70 a^3 b d^3 (c+d x)-60 a^3 b c d^3+90 a^2 b^2 c^2 d^2+128 a^2 b^2 d^2 (c+d x)^2-210 a^2 b^2 c d^2 (c+d x)-60 a b^3 c^3 d+210 a b^3 c^2 d (c+d x)-70 a b^3 d (c+d x)^3-256 a b^3 c d (c+d x)^2+15 b^4 c^4-70 b^4 c^3 (c+d x)+128 b^4 c^2 (c+d x)^2-15 b^4 (c+d x)^4+70 b^4 c (c+d x)^3\right )}{640 b^3 (b c-a d)^2 (-a d-b (c+d x)+b c)^5}-\frac {3 d^5 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x} \sqrt {a d-b c}}{b c-a d}\right )}{128 b^{7/2} (a d-b c)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(c + d*x)^(5/2)/(a + b*x)^6,x]

[Out]

(d^5*Sqrt[c + d*x]*(15*b^4*c^4 - 60*a*b^3*c^3*d + 90*a^2*b^2*c^2*d^2 - 60*a^3*b*c*d^3 + 15*a^4*d^4 - 70*b^4*c^
3*(c + d*x) + 210*a*b^3*c^2*d*(c + d*x) - 210*a^2*b^2*c*d^2*(c + d*x) + 70*a^3*b*d^3*(c + d*x) + 128*b^4*c^2*(
c + d*x)^2 - 256*a*b^3*c*d*(c + d*x)^2 + 128*a^2*b^2*d^2*(c + d*x)^2 + 70*b^4*c*(c + d*x)^3 - 70*a*b^3*d*(c +
d*x)^3 - 15*b^4*(c + d*x)^4))/(640*b^3*(b*c - a*d)^2*(b*c - a*d - b*(c + d*x))^5) - (3*d^5*ArcTan[(Sqrt[b]*Sqr
t[-(b*c) + a*d]*Sqrt[c + d*x])/(b*c - a*d)])/(128*b^(7/2)*(-(b*c) + a*d)^(5/2))

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fricas [B]  time = 1.49, size = 1337, normalized size = 6.75

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)/(b*x+a)^6,x, algorithm="fricas")

[Out]

[1/1280*(15*(b^5*d^5*x^5 + 5*a*b^4*d^5*x^4 + 10*a^2*b^3*d^5*x^3 + 10*a^3*b^2*d^5*x^2 + 5*a^4*b*d^5*x + a^5*d^5
)*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*(128*b^6*
c^5 - 304*a*b^5*c^4*d + 184*a^2*b^4*c^3*d^2 + 2*a^3*b^3*c^2*d^3 + 5*a^4*b^2*c*d^4 - 15*a^5*b*d^5 - 15*(b^6*c*d
^4 - a*b^5*d^5)*x^4 + 10*(b^6*c^2*d^3 - 8*a*b^5*c*d^4 + 7*a^2*b^4*d^5)*x^3 + 2*(124*b^6*c^3*d^2 - 357*a*b^5*c^
2*d^3 + 297*a^2*b^4*c*d^4 - 64*a^3*b^3*d^5)*x^2 + 2*(168*b^6*c^4*d - 424*a*b^5*c^3*d^2 + 279*a^2*b^4*c^2*d^3 +
12*a^3*b^3*c*d^4 - 35*a^4*b^2*d^5)*x)*sqrt(d*x + c))/(a^5*b^7*c^3 - 3*a^6*b^6*c^2*d + 3*a^7*b^5*c*d^2 - a^8*b
^4*d^3 + (b^12*c^3 - 3*a*b^11*c^2*d + 3*a^2*b^10*c*d^2 - a^3*b^9*d^3)*x^5 + 5*(a*b^11*c^3 - 3*a^2*b^10*c^2*d +
3*a^3*b^9*c*d^2 - a^4*b^8*d^3)*x^4 + 10*(a^2*b^10*c^3 - 3*a^3*b^9*c^2*d + 3*a^4*b^8*c*d^2 - a^5*b^7*d^3)*x^3
+ 10*(a^3*b^9*c^3 - 3*a^4*b^8*c^2*d + 3*a^5*b^7*c*d^2 - a^6*b^6*d^3)*x^2 + 5*(a^4*b^8*c^3 - 3*a^5*b^7*c^2*d +
3*a^6*b^6*c*d^2 - a^7*b^5*d^3)*x), 1/640*(15*(b^5*d^5*x^5 + 5*a*b^4*d^5*x^4 + 10*a^2*b^3*d^5*x^3 + 10*a^3*b^2*
d^5*x^2 + 5*a^4*b*d^5*x + a^5*d^5)*sqrt(-b^2*c + a*b*d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c
)) - (128*b^6*c^5 - 304*a*b^5*c^4*d + 184*a^2*b^4*c^3*d^2 + 2*a^3*b^3*c^2*d^3 + 5*a^4*b^2*c*d^4 - 15*a^5*b*d^5
- 15*(b^6*c*d^4 - a*b^5*d^5)*x^4 + 10*(b^6*c^2*d^3 - 8*a*b^5*c*d^4 + 7*a^2*b^4*d^5)*x^3 + 2*(124*b^6*c^3*d^2
- 357*a*b^5*c^2*d^3 + 297*a^2*b^4*c*d^4 - 64*a^3*b^3*d^5)*x^2 + 2*(168*b^6*c^4*d - 424*a*b^5*c^3*d^2 + 279*a^2
*b^4*c^2*d^3 + 12*a^3*b^3*c*d^4 - 35*a^4*b^2*d^5)*x)*sqrt(d*x + c))/(a^5*b^7*c^3 - 3*a^6*b^6*c^2*d + 3*a^7*b^5
*c*d^2 - a^8*b^4*d^3 + (b^12*c^3 - 3*a*b^11*c^2*d + 3*a^2*b^10*c*d^2 - a^3*b^9*d^3)*x^5 + 5*(a*b^11*c^3 - 3*a^
2*b^10*c^2*d + 3*a^3*b^9*c*d^2 - a^4*b^8*d^3)*x^4 + 10*(a^2*b^10*c^3 - 3*a^3*b^9*c^2*d + 3*a^4*b^8*c*d^2 - a^5
*b^7*d^3)*x^3 + 10*(a^3*b^9*c^3 - 3*a^4*b^8*c^2*d + 3*a^5*b^7*c*d^2 - a^6*b^6*d^3)*x^2 + 5*(a^4*b^8*c^3 - 3*a^
5*b^7*c^2*d + 3*a^6*b^6*c*d^2 - a^7*b^5*d^3)*x)]

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giac [B]  time = 1.25, size = 380, normalized size = 1.92 \begin {gather*} \frac {3 \, d^{5} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{128 \, {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} \sqrt {-b^{2} c + a b d}} + \frac {15 \, {\left (d x + c\right )}^{\frac {9}{2}} b^{4} d^{5} - 70 \, {\left (d x + c\right )}^{\frac {7}{2}} b^{4} c d^{5} - 128 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{4} c^{2} d^{5} + 70 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{4} c^{3} d^{5} - 15 \, \sqrt {d x + c} b^{4} c^{4} d^{5} + 70 \, {\left (d x + c\right )}^{\frac {7}{2}} a b^{3} d^{6} + 256 \, {\left (d x + c\right )}^{\frac {5}{2}} a b^{3} c d^{6} - 210 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{3} c^{2} d^{6} + 60 \, \sqrt {d x + c} a b^{3} c^{3} d^{6} - 128 \, {\left (d x + c\right )}^{\frac {5}{2}} a^{2} b^{2} d^{7} + 210 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} b^{2} c d^{7} - 90 \, \sqrt {d x + c} a^{2} b^{2} c^{2} d^{7} - 70 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{3} b d^{8} + 60 \, \sqrt {d x + c} a^{3} b c d^{8} - 15 \, \sqrt {d x + c} a^{4} d^{9}}{640 \, {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)/(b*x+a)^6,x, algorithm="giac")

[Out]

3/128*d^5*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^5*c^2 - 2*a*b^4*c*d + a^2*b^3*d^2)*sqrt(-b^2*c + a*
b*d)) + 1/640*(15*(d*x + c)^(9/2)*b^4*d^5 - 70*(d*x + c)^(7/2)*b^4*c*d^5 - 128*(d*x + c)^(5/2)*b^4*c^2*d^5 + 7
0*(d*x + c)^(3/2)*b^4*c^3*d^5 - 15*sqrt(d*x + c)*b^4*c^4*d^5 + 70*(d*x + c)^(7/2)*a*b^3*d^6 + 256*(d*x + c)^(5
/2)*a*b^3*c*d^6 - 210*(d*x + c)^(3/2)*a*b^3*c^2*d^6 + 60*sqrt(d*x + c)*a*b^3*c^3*d^6 - 128*(d*x + c)^(5/2)*a^2
*b^2*d^7 + 210*(d*x + c)^(3/2)*a^2*b^2*c*d^7 - 90*sqrt(d*x + c)*a^2*b^2*c^2*d^7 - 70*(d*x + c)^(3/2)*a^3*b*d^8
+ 60*sqrt(d*x + c)*a^3*b*c*d^8 - 15*sqrt(d*x + c)*a^4*d^9)/((b^5*c^2 - 2*a*b^4*c*d + a^2*b^3*d^2)*((d*x + c)*
b - b*c + a*d)^5)

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maple [A]  time = 0.02, size = 305, normalized size = 1.54 \begin {gather*} -\frac {3 \sqrt {d x +c}\, a^{2} d^{7}}{128 \left (b d x +a d \right )^{5} b^{3}}+\frac {3 \sqrt {d x +c}\, a c \,d^{6}}{64 \left (b d x +a d \right )^{5} b^{2}}+\frac {3 \left (d x +c \right )^{\frac {9}{2}} b \,d^{5}}{128 \left (b d x +a d \right )^{5} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}-\frac {3 \sqrt {d x +c}\, c^{2} d^{5}}{128 \left (b d x +a d \right )^{5} b}-\frac {7 \left (d x +c \right )^{\frac {3}{2}} a \,d^{6}}{64 \left (b d x +a d \right )^{5} b^{2}}+\frac {7 \left (d x +c \right )^{\frac {3}{2}} c \,d^{5}}{64 \left (b d x +a d \right )^{5} b}+\frac {7 \left (d x +c \right )^{\frac {7}{2}} d^{5}}{64 \left (b d x +a d \right )^{5} \left (a d -b c \right )}-\frac {\left (d x +c \right )^{\frac {5}{2}} d^{5}}{5 \left (b d x +a d \right )^{5} b}+\frac {3 d^{5} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{128 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \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)^6,x)

[Out]

3/128*d^5/(b*d*x+a*d)^5*b/(a^2*d^2-2*a*b*c*d+b^2*c^2)*(d*x+c)^(9/2)+7/64*d^5/(b*d*x+a*d)^5/(a*d-b*c)*(d*x+c)^(
7/2)-1/5*d^5/(b*d*x+a*d)^5/b*(d*x+c)^(5/2)-7/64*d^6/(b*d*x+a*d)^5/b^2*(d*x+c)^(3/2)*a+7/64*d^5/(b*d*x+a*d)^5/b
*(d*x+c)^(3/2)*c-3/128*d^7/(b*d*x+a*d)^5/b^3*(d*x+c)^(1/2)*a^2+3/64*d^6/(b*d*x+a*d)^5/b^2*(d*x+c)^(1/2)*a*c-3/
128*d^5/(b*d*x+a*d)^5/b*(d*x+c)^(1/2)*c^2+3/128*d^5/b^3/(a^2*d^2-2*a*b*c*d+b^2*c^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)^(5/2)/(b*x+a)^6,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 = 0.50, size = 411, normalized size = 2.08 \begin {gather*} \frac {3\,d^5\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{128\,b^{7/2}\,{\left (a\,d-b\,c\right )}^{5/2}}-\frac {\frac {d^5\,{\left (c+d\,x\right )}^{5/2}}{5\,b}-\frac {7\,d^5\,{\left (c+d\,x\right )}^{7/2}}{64\,\left (a\,d-b\,c\right )}+\frac {3\,d^5\,\sqrt {c+d\,x}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{128\,b^3}+\frac {7\,d^5\,\left (a\,d-b\,c\right )\,{\left (c+d\,x\right )}^{3/2}}{64\,b^2}-\frac {3\,b\,d^5\,{\left (c+d\,x\right )}^{9/2}}{128\,{\left (a\,d-b\,c\right )}^2}}{b^5\,{\left (c+d\,x\right )}^5-{\left (c+d\,x\right )}^2\,\left (-10\,a^3\,b^2\,d^3+30\,a^2\,b^3\,c\,d^2-30\,a\,b^4\,c^2\,d+10\,b^5\,c^3\right )-\left (5\,b^5\,c-5\,a\,b^4\,d\right )\,{\left (c+d\,x\right )}^4+a^5\,d^5-b^5\,c^5+{\left (c+d\,x\right )}^3\,\left (10\,a^2\,b^3\,d^2-20\,a\,b^4\,c\,d+10\,b^5\,c^2\right )+\left (c+d\,x\right )\,\left (5\,a^4\,b\,d^4-20\,a^3\,b^2\,c\,d^3+30\,a^2\,b^3\,c^2\,d^2-20\,a\,b^4\,c^3\,d+5\,b^5\,c^4\right )-10\,a^2\,b^3\,c^3\,d^2+10\,a^3\,b^2\,c^2\,d^3+5\,a\,b^4\,c^4\,d-5\,a^4\,b\,c\,d^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(5/2)/(a + b*x)^6,x)

[Out]

(3*d^5*atan((b^(1/2)*(c + d*x)^(1/2))/(a*d - b*c)^(1/2)))/(128*b^(7/2)*(a*d - b*c)^(5/2)) - ((d^5*(c + d*x)^(5
/2))/(5*b) - (7*d^5*(c + d*x)^(7/2))/(64*(a*d - b*c)) + (3*d^5*(c + d*x)^(1/2)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)
)/(128*b^3) + (7*d^5*(a*d - b*c)*(c + d*x)^(3/2))/(64*b^2) - (3*b*d^5*(c + d*x)^(9/2))/(128*(a*d - b*c)^2))/(b
^5*(c + d*x)^5 - (c + d*x)^2*(10*b^5*c^3 - 10*a^3*b^2*d^3 + 30*a^2*b^3*c*d^2 - 30*a*b^4*c^2*d) - (5*b^5*c - 5*
a*b^4*d)*(c + d*x)^4 + a^5*d^5 - b^5*c^5 + (c + d*x)^3*(10*b^5*c^2 + 10*a^2*b^3*d^2 - 20*a*b^4*c*d) + (c + d*x
)*(5*b^5*c^4 + 5*a^4*b*d^4 - 20*a^3*b^2*c*d^3 + 30*a^2*b^3*c^2*d^2 - 20*a*b^4*c^3*d) - 10*a^2*b^3*c^3*d^2 + 10
*a^3*b^2*c^2*d^3 + 5*a*b^4*c^4*d - 5*a^4*b*c*d^4)

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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)**6,x)

[Out]

Timed out

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