3.13.79 \(\int \frac {\sqrt {c+d x}}{(a+b x)^5} \, dx\)

Optimal. Leaf size=182 \[ \frac {5 d^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{64 b^{3/2} (b c-a d)^{7/2}}-\frac {5 d^3 \sqrt {c+d x}}{64 b (a+b x) (b c-a d)^3}+\frac {5 d^2 \sqrt {c+d x}}{96 b (a+b x)^2 (b c-a d)^2}-\frac {d \sqrt {c+d x}}{24 b (a+b x)^3 (b c-a d)}-\frac {\sqrt {c+d x}}{4 b (a+b x)^4} \]

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Rubi [A]  time = 0.12, antiderivative size = 182, 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 {5 d^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{64 b^{3/2} (b c-a d)^{7/2}}-\frac {5 d^3 \sqrt {c+d x}}{64 b (a+b x) (b c-a d)^3}+\frac {5 d^2 \sqrt {c+d x}}{96 b (a+b x)^2 (b c-a d)^2}-\frac {d \sqrt {c+d x}}{24 b (a+b x)^3 (b c-a d)}-\frac {\sqrt {c+d x}}{4 b (a+b x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x]/(a + b*x)^5,x]

[Out]

-Sqrt[c + d*x]/(4*b*(a + b*x)^4) - (d*Sqrt[c + d*x])/(24*b*(b*c - a*d)*(a + b*x)^3) + (5*d^2*Sqrt[c + d*x])/(9
6*b*(b*c - a*d)^2*(a + b*x)^2) - (5*d^3*Sqrt[c + d*x])/(64*b*(b*c - a*d)^3*(a + b*x)) + (5*d^4*ArcTanh[(Sqrt[b
]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(64*b^(3/2)*(b*c - a*d)^(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 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 {\sqrt {c+d x}}{(a+b x)^5} \, dx &=-\frac {\sqrt {c+d x}}{4 b (a+b x)^4}+\frac {d \int \frac {1}{(a+b x)^4 \sqrt {c+d x}} \, dx}{8 b}\\ &=-\frac {\sqrt {c+d x}}{4 b (a+b x)^4}-\frac {d \sqrt {c+d x}}{24 b (b c-a d) (a+b x)^3}-\frac {\left (5 d^2\right ) \int \frac {1}{(a+b x)^3 \sqrt {c+d x}} \, dx}{48 b (b c-a d)}\\ &=-\frac {\sqrt {c+d x}}{4 b (a+b x)^4}-\frac {d \sqrt {c+d x}}{24 b (b c-a d) (a+b x)^3}+\frac {5 d^2 \sqrt {c+d x}}{96 b (b c-a d)^2 (a+b x)^2}+\frac {\left (5 d^3\right ) \int \frac {1}{(a+b x)^2 \sqrt {c+d x}} \, dx}{64 b (b c-a d)^2}\\ &=-\frac {\sqrt {c+d x}}{4 b (a+b x)^4}-\frac {d \sqrt {c+d x}}{24 b (b c-a d) (a+b x)^3}+\frac {5 d^2 \sqrt {c+d x}}{96 b (b c-a d)^2 (a+b x)^2}-\frac {5 d^3 \sqrt {c+d x}}{64 b (b c-a d)^3 (a+b x)}-\frac {\left (5 d^4\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{128 b (b c-a d)^3}\\ &=-\frac {\sqrt {c+d x}}{4 b (a+b x)^4}-\frac {d \sqrt {c+d x}}{24 b (b c-a d) (a+b x)^3}+\frac {5 d^2 \sqrt {c+d x}}{96 b (b c-a d)^2 (a+b x)^2}-\frac {5 d^3 \sqrt {c+d x}}{64 b (b c-a d)^3 (a+b x)}-\frac {\left (5 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 (b c-a d)^3}\\ &=-\frac {\sqrt {c+d x}}{4 b (a+b x)^4}-\frac {d \sqrt {c+d x}}{24 b (b c-a d) (a+b x)^3}+\frac {5 d^2 \sqrt {c+d x}}{96 b (b c-a d)^2 (a+b x)^2}-\frac {5 d^3 \sqrt {c+d x}}{64 b (b c-a d)^3 (a+b x)}+\frac {5 d^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{64 b^{3/2} (b c-a d)^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 52, normalized size = 0.29 \begin {gather*} \frac {2 d^4 (c+d x)^{3/2} \, _2F_1\left (\frac {3}{2},5;\frac {5}{2};-\frac {b (c+d x)}{a d-b c}\right )}{3 (a d-b c)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d^4*(c + d*x)^(3/2)*Hypergeometric2F1[3/2, 5, 5/2, -((b*(c + d*x))/(-(b*c) + a*d))])/(3*(-(b*c) + a*d)^5)

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IntegrateAlgebraic [A]  time = 1.07, size = 226, normalized size = 1.24 \begin {gather*} -\frac {d^4 \sqrt {c+d x} \left (-15 a^3 d^3+73 a^2 b d^2 (c+d x)+45 a^2 b c d^2-45 a b^2 c^2 d+55 a b^2 d (c+d x)^2-146 a b^2 c d (c+d x)+15 b^3 c^3+73 b^3 c^2 (c+d x)+15 b^3 (c+d x)^3-55 b^3 c (c+d x)^2\right )}{192 b (b c-a d)^3 (-a d-b (c+d x)+b c)^4}-\frac {5 d^4 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x} \sqrt {a d-b c}}{b c-a d}\right )}{64 b^{3/2} (a d-b c)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/192*(d^4*Sqrt[c + d*x]*(15*b^3*c^3 - 45*a*b^2*c^2*d + 45*a^2*b*c*d^2 - 15*a^3*d^3 + 73*b^3*c^2*(c + d*x) -
146*a*b^2*c*d*(c + d*x) + 73*a^2*b*d^2*(c + d*x) - 55*b^3*c*(c + d*x)^2 + 55*a*b^2*d*(c + d*x)^2 + 15*b^3*(c +
d*x)^3))/(b*(b*c - a*d)^3*(b*c - a*d - b*(c + d*x))^4) - (5*d^4*ArcTan[(Sqrt[b]*Sqrt[-(b*c) + a*d]*Sqrt[c + d
*x])/(b*c - a*d)])/(64*b^(3/2)*(-(b*c) + a*d)^(7/2))

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fricas [B]  time = 0.89, size = 1176, normalized size = 6.46 \begin {gather*} \left [-\frac {15 \, {\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 (48 \, b^{5} c^{4} - 184 \, a b^{4} c^{3} d + 254 \, a^{2} b^{3} c^{2} d^{2} - 133 \, a^{3} b^{2} c d^{3} + 15 \, a^{4} b d^{4} + 15 \, {\left (b^{5} c d^{3} - a b^{4} d^{4}\right )} x^{3} - 5 \, {\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 (8 \, b^{5} c^{3} d - 44 \, a b^{4} c^{2} d^{2} + 109 \, a^{2} b^{3} c d^{3} - 73 \, a^{3} b^{2} d^{4}\right )} x\right )} \sqrt {d x + c}}{384 \, {\left (a^{4} b^{6} c^{4} - 4 \, a^{5} b^{5} c^{3} d + 6 \, a^{6} b^{4} c^{2} d^{2} - 4 \, a^{7} b^{3} c d^{3} + a^{8} b^{2} d^{4} + {\left (b^{10} c^{4} - 4 \, a b^{9} c^{3} d + 6 \, a^{2} b^{8} c^{2} d^{2} - 4 \, a^{3} b^{7} c d^{3} + a^{4} b^{6} d^{4}\right )} x^{4} + 4 \, {\left (a b^{9} c^{4} - 4 \, a^{2} b^{8} c^{3} d + 6 \, a^{3} b^{7} c^{2} d^{2} - 4 \, a^{4} b^{6} c d^{3} + a^{5} b^{5} d^{4}\right )} x^{3} + 6 \, {\left (a^{2} b^{8} c^{4} - 4 \, a^{3} b^{7} c^{3} d + 6 \, a^{4} b^{6} c^{2} d^{2} - 4 \, a^{5} b^{5} c d^{3} + a^{6} b^{4} d^{4}\right )} x^{2} + 4 \, {\left (a^{3} b^{7} c^{4} - 4 \, a^{4} b^{6} c^{3} d + 6 \, a^{5} b^{5} c^{2} d^{2} - 4 \, a^{6} b^{4} c d^{3} + a^{7} b^{3} d^{4}\right )} x\right )}}, -\frac {15 \, {\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 (48 \, b^{5} c^{4} - 184 \, a b^{4} c^{3} d + 254 \, a^{2} b^{3} c^{2} d^{2} - 133 \, a^{3} b^{2} c d^{3} + 15 \, a^{4} b d^{4} + 15 \, {\left (b^{5} c d^{3} - a b^{4} d^{4}\right )} x^{3} - 5 \, {\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 (8 \, b^{5} c^{3} d - 44 \, a b^{4} c^{2} d^{2} + 109 \, a^{2} b^{3} c d^{3} - 73 \, a^{3} b^{2} d^{4}\right )} x\right )} \sqrt {d x + c}}{192 \, {\left (a^{4} b^{6} c^{4} - 4 \, a^{5} b^{5} c^{3} d + 6 \, a^{6} b^{4} c^{2} d^{2} - 4 \, a^{7} b^{3} c d^{3} + a^{8} b^{2} d^{4} + {\left (b^{10} c^{4} - 4 \, a b^{9} c^{3} d + 6 \, a^{2} b^{8} c^{2} d^{2} - 4 \, a^{3} b^{7} c d^{3} + a^{4} b^{6} d^{4}\right )} x^{4} + 4 \, {\left (a b^{9} c^{4} - 4 \, a^{2} b^{8} c^{3} d + 6 \, a^{3} b^{7} c^{2} d^{2} - 4 \, a^{4} b^{6} c d^{3} + a^{5} b^{5} d^{4}\right )} x^{3} + 6 \, {\left (a^{2} b^{8} c^{4} - 4 \, a^{3} b^{7} c^{3} d + 6 \, a^{4} b^{6} c^{2} d^{2} - 4 \, a^{5} b^{5} c d^{3} + a^{6} b^{4} d^{4}\right )} x^{2} + 4 \, {\left (a^{3} b^{7} c^{4} - 4 \, a^{4} b^{6} c^{3} d + 6 \, a^{5} b^{5} c^{2} d^{2} - 4 \, a^{6} b^{4} c d^{3} + a^{7} b^{3} d^{4}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/384*(15*(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)*
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*(48*b^5*c^4 - 184*a*b^4*c^3*d +
254*a^2*b^3*c^2*d^2 - 133*a^3*b^2*c*d^3 + 15*a^4*b*d^4 + 15*(b^5*c*d^3 - a*b^4*d^4)*x^3 - 5*(2*b^5*c^2*d^2 -
13*a*b^4*c*d^3 + 11*a^2*b^3*d^4)*x^2 + (8*b^5*c^3*d - 44*a*b^4*c^2*d^2 + 109*a^2*b^3*c*d^3 - 73*a^3*b^2*d^4)*x
)*sqrt(d*x + c))/(a^4*b^6*c^4 - 4*a^5*b^5*c^3*d + 6*a^6*b^4*c^2*d^2 - 4*a^7*b^3*c*d^3 + a^8*b^2*d^4 + (b^10*c^
4 - 4*a*b^9*c^3*d + 6*a^2*b^8*c^2*d^2 - 4*a^3*b^7*c*d^3 + a^4*b^6*d^4)*x^4 + 4*(a*b^9*c^4 - 4*a^2*b^8*c^3*d +
6*a^3*b^7*c^2*d^2 - 4*a^4*b^6*c*d^3 + a^5*b^5*d^4)*x^3 + 6*(a^2*b^8*c^4 - 4*a^3*b^7*c^3*d + 6*a^4*b^6*c^2*d^2
- 4*a^5*b^5*c*d^3 + a^6*b^4*d^4)*x^2 + 4*(a^3*b^7*c^4 - 4*a^4*b^6*c^3*d + 6*a^5*b^5*c^2*d^2 - 4*a^6*b^4*c*d^3
+ a^7*b^3*d^4)*x), -1/192*(15*(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)*sq
rt(-b^2*c + a*b*d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) + (48*b^5*c^4 - 184*a*b^4*c^3*d +
254*a^2*b^3*c^2*d^2 - 133*a^3*b^2*c*d^3 + 15*a^4*b*d^4 + 15*(b^5*c*d^3 - a*b^4*d^4)*x^3 - 5*(2*b^5*c^2*d^2 - 1
3*a*b^4*c*d^3 + 11*a^2*b^3*d^4)*x^2 + (8*b^5*c^3*d - 44*a*b^4*c^2*d^2 + 109*a^2*b^3*c*d^3 - 73*a^3*b^2*d^4)*x)
*sqrt(d*x + c))/(a^4*b^6*c^4 - 4*a^5*b^5*c^3*d + 6*a^6*b^4*c^2*d^2 - 4*a^7*b^3*c*d^3 + a^8*b^2*d^4 + (b^10*c^4
- 4*a*b^9*c^3*d + 6*a^2*b^8*c^2*d^2 - 4*a^3*b^7*c*d^3 + a^4*b^6*d^4)*x^4 + 4*(a*b^9*c^4 - 4*a^2*b^8*c^3*d + 6
*a^3*b^7*c^2*d^2 - 4*a^4*b^6*c*d^3 + a^5*b^5*d^4)*x^3 + 6*(a^2*b^8*c^4 - 4*a^3*b^7*c^3*d + 6*a^4*b^6*c^2*d^2 -
4*a^5*b^5*c*d^3 + a^6*b^4*d^4)*x^2 + 4*(a^3*b^7*c^4 - 4*a^4*b^6*c^3*d + 6*a^5*b^5*c^2*d^2 - 4*a^6*b^4*c*d^3 +
a^7*b^3*d^4)*x)]

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giac [B]  time = 1.39, size = 311, normalized size = 1.71 \begin {gather*} -\frac {5 \, d^{4} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{64 \, {\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} \sqrt {-b^{2} c + a b d}} - \frac {15 \, {\left (d x + c\right )}^{\frac {7}{2}} b^{3} d^{4} - 55 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{3} c d^{4} + 73 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{3} c^{2} d^{4} + 15 \, \sqrt {d x + c} b^{3} c^{3} d^{4} + 55 \, {\left (d x + c\right )}^{\frac {5}{2}} a b^{2} d^{5} - 146 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{2} c d^{5} - 45 \, \sqrt {d x + c} a b^{2} c^{2} d^{5} + 73 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} b d^{6} + 45 \, \sqrt {d x + c} a^{2} b c d^{6} - 15 \, \sqrt {d x + c} a^{3} d^{7}}{192 \, {\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\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)^(1/2)/(b*x+a)^5,x, algorithm="giac")

[Out]

-5/64*d^4*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^4*c^3 - 3*a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3
)*sqrt(-b^2*c + a*b*d)) - 1/192*(15*(d*x + c)^(7/2)*b^3*d^4 - 55*(d*x + c)^(5/2)*b^3*c*d^4 + 73*(d*x + c)^(3/2
)*b^3*c^2*d^4 + 15*sqrt(d*x + c)*b^3*c^3*d^4 + 55*(d*x + c)^(5/2)*a*b^2*d^5 - 146*(d*x + c)^(3/2)*a*b^2*c*d^5
- 45*sqrt(d*x + c)*a*b^2*c^2*d^5 + 73*(d*x + c)^(3/2)*a^2*b*d^6 + 45*sqrt(d*x + c)*a^2*b*c*d^6 - 15*sqrt(d*x +
c)*a^3*d^7)/((b^4*c^3 - 3*a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3)*((d*x + c)*b - b*c + a*d)^4)

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maple [A]  time = 0.02, size = 248, normalized size = 1.36 \begin {gather*} \frac {5 \left (d x +c \right )^{\frac {7}{2}} b^{2} d^{4}}{64 \left (b d x +a d \right )^{4} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}+\frac {55 \left (d x +c \right )^{\frac {5}{2}} b \,d^{4}}{192 \left (b d x +a d \right )^{4} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {5 d^{4} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{64 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \sqrt {\left (a d -b c \right ) b}\, b}+\frac {73 \left (d x +c \right )^{\frac {3}{2}} d^{4}}{192 \left (b d x +a d \right )^{4} \left (a d -b c \right )}-\frac {5 \sqrt {d x +c}\, d^{4}}{64 \left (b d x +a d \right )^{4} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/64*d^4/(b*d*x+a*d)^4*b^2/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)*(d*x+c)^(7/2)+55/192*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)^(5/2)+73/192*d^4/(b*d*x+a*d)^4/(a*d-b*c)*(d*x+c)^(3/2)-5/64*d^4/(b*d*x
+a*d)^4/b*(d*x+c)^(1/2)+5/64*d^4/b/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/((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)^(1/2)/(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 = 0.22, size = 297, normalized size = 1.63 \begin {gather*} \frac {\frac {73\,d^4\,{\left (c+d\,x\right )}^{3/2}}{192\,\left (a\,d-b\,c\right )}-\frac {5\,d^4\,\sqrt {c+d\,x}}{64\,b}+\frac {5\,b^2\,d^4\,{\left (c+d\,x\right )}^{7/2}}{64\,{\left (a\,d-b\,c\right )}^3}+\frac {55\,b\,d^4\,{\left (c+d\,x\right )}^{5/2}}{192\,{\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}+\frac {5\,d^4\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{64\,b^{3/2}\,{\left (a\,d-b\,c\right )}^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(1/2)/(a + b*x)^5,x)

[Out]

((73*d^4*(c + d*x)^(3/2))/(192*(a*d - b*c)) - (5*d^4*(c + d*x)^(1/2))/(64*b) + (5*b^2*d^4*(c + d*x)^(7/2))/(64
*(a*d - b*c)^3) + (55*b*d^4*(c + d*x)^(5/2))/(192*(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) + (5*d^
4*atan((b^(1/2)*(c + d*x)^(1/2))/(a*d - b*c)^(1/2)))/(64*b^(3/2)*(a*d - b*c)^(7/2))

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

[Out]

Timed out

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