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

Optimal. Leaf size=128 \[ \frac {5 d^{3/2} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{7/2}}+\frac {5 d^2 \sqrt {a+b x} \sqrt {c+d x}}{b^3}-\frac {10 d (c+d x)^{3/2}}{3 b^2 \sqrt {a+b x}}-\frac {2 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}} \]

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Rubi [A]  time = 0.07, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {47, 50, 63, 217, 206} \begin {gather*} \frac {5 d^2 \sqrt {a+b x} \sqrt {c+d x}}{b^3}+\frac {5 d^{3/2} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{7/2}}-\frac {10 d (c+d x)^{3/2}}{3 b^2 \sqrt {a+b x}}-\frac {2 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*d^2*Sqrt[a + b*x]*Sqrt[c + d*x])/b^3 - (10*d*(c + d*x)^(3/2))/(3*b^2*Sqrt[a + b*x]) - (2*(c + d*x)^(5/2))/(
3*b*(a + b*x)^(3/2)) + (5*d^(3/2)*(b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/b^(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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

\begin {align*} \int \frac {(c+d x)^{5/2}}{(a+b x)^{5/2}} \, dx &=-\frac {2 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}}+\frac {(5 d) \int \frac {(c+d x)^{3/2}}{(a+b x)^{3/2}} \, dx}{3 b}\\ &=-\frac {10 d (c+d x)^{3/2}}{3 b^2 \sqrt {a+b x}}-\frac {2 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}}+\frac {\left (5 d^2\right ) \int \frac {\sqrt {c+d x}}{\sqrt {a+b x}} \, dx}{b^2}\\ &=\frac {5 d^2 \sqrt {a+b x} \sqrt {c+d x}}{b^3}-\frac {10 d (c+d x)^{3/2}}{3 b^2 \sqrt {a+b x}}-\frac {2 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}}+\frac {\left (5 d^2 (b c-a d)\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{2 b^3}\\ &=\frac {5 d^2 \sqrt {a+b x} \sqrt {c+d x}}{b^3}-\frac {10 d (c+d x)^{3/2}}{3 b^2 \sqrt {a+b x}}-\frac {2 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}}+\frac {\left (5 d^2 (b c-a d)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x}\right )}{b^4}\\ &=\frac {5 d^2 \sqrt {a+b x} \sqrt {c+d x}}{b^3}-\frac {10 d (c+d x)^{3/2}}{3 b^2 \sqrt {a+b x}}-\frac {2 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}}+\frac {\left (5 d^2 (b c-a d)\right ) \operatorname {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{b^4}\\ &=\frac {5 d^2 \sqrt {a+b x} \sqrt {c+d x}}{b^3}-\frac {10 d (c+d x)^{3/2}}{3 b^2 \sqrt {a+b x}}-\frac {2 (c+d x)^{5/2}}{3 b (a+b x)^{3/2}}+\frac {5 d^{3/2} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.07, size = 73, normalized size = 0.57 \begin {gather*} -\frac {2 (c+d x)^{5/2} \, _2F_1\left (-\frac {5}{2},-\frac {3}{2};-\frac {1}{2};\frac {d (a+b x)}{a d-b c}\right )}{3 b (a+b x)^{3/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(c + d*x)^(5/2)*Hypergeometric2F1[-5/2, -3/2, -1/2, (d*(a + b*x))/(-(b*c) + a*d)])/(3*b*(a + b*x)^(3/2)*((
b*(c + d*x))/(b*c - a*d))^(5/2))

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IntegrateAlgebraic [A]  time = 0.00, size = 227, normalized size = 1.77 \begin {gather*} \frac {\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}} \left (15 a^2 d^4 \sqrt {c+d x}+20 a b d^3 (c+d x)^{3/2}-30 a b c d^3 \sqrt {c+d x}+15 b^2 c^2 d^2 \sqrt {c+d x}+3 b^2 d^2 (c+d x)^{5/2}-20 b^2 c d^2 (c+d x)^{3/2}\right )}{3 b^3 (-a d-b (c+d x)+b c)^2}-\frac {5 \sqrt {\frac {b}{d}} \left (b c d^2-a d^3\right ) \log \left (\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}-\sqrt {\frac {b}{d}} \sqrt {c+d x}\right )}{b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a - (b*c)/d + (b*(c + d*x))/d]*(15*b^2*c^2*d^2*Sqrt[c + d*x] - 30*a*b*c*d^3*Sqrt[c + d*x] + 15*a^2*d^4*S
qrt[c + d*x] - 20*b^2*c*d^2*(c + d*x)^(3/2) + 20*a*b*d^3*(c + d*x)^(3/2) + 3*b^2*d^2*(c + d*x)^(5/2)))/(3*b^3*
(b*c - a*d - b*(c + d*x))^2) - (5*Sqrt[b/d]*(b*c*d^2 - a*d^3)*Log[-(Sqrt[b/d]*Sqrt[c + d*x]) + Sqrt[a - (b*c)/
d + (b*(c + d*x))/d]])/b^4

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fricas [B]  time = 2.00, size = 475, normalized size = 3.71 \begin {gather*} \left [-\frac {15 \, {\left (a^{2} b c d - a^{3} d^{2} + {\left (b^{3} c d - a b^{2} d^{2}\right )} x^{2} + 2 \, {\left (a b^{2} c d - a^{2} b d^{2}\right )} x\right )} \sqrt {\frac {d}{b}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b^{2} d x + b^{2} c + a b d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {d}{b}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (3 \, b^{2} d^{2} x^{2} - 2 \, b^{2} c^{2} - 10 \, a b c d + 15 \, a^{2} d^{2} - 2 \, {\left (7 \, b^{2} c d - 10 \, a b d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{12 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac {15 \, {\left (a^{2} b c d - a^{3} d^{2} + {\left (b^{3} c d - a b^{2} d^{2}\right )} x^{2} + 2 \, {\left (a b^{2} c d - a^{2} b d^{2}\right )} x\right )} \sqrt {-\frac {d}{b}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {d}{b}}}{2 \, {\left (b d^{2} x^{2} + a c d + {\left (b c d + a d^{2}\right )} x\right )}}\right ) - 2 \, {\left (3 \, b^{2} d^{2} x^{2} - 2 \, b^{2} c^{2} - 10 \, a b c d + 15 \, a^{2} d^{2} - 2 \, {\left (7 \, b^{2} c d - 10 \, a b d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{6 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*(15*(a^2*b*c*d - a^3*d^2 + (b^3*c*d - a*b^2*d^2)*x^2 + 2*(a*b^2*c*d - a^2*b*d^2)*x)*sqrt(d/b)*log(8*b^2
*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 - 4*(2*b^2*d*x + b^2*c + a*b*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(d/b)
+ 8*(b^2*c*d + a*b*d^2)*x) - 4*(3*b^2*d^2*x^2 - 2*b^2*c^2 - 10*a*b*c*d + 15*a^2*d^2 - 2*(7*b^2*c*d - 10*a*b*d
^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3), -1/6*(15*(a^2*b*c*d - a^3*d^2 + (b^3*c*d
- a*b^2*d^2)*x^2 + 2*(a*b^2*c*d - a^2*b*d^2)*x)*sqrt(-d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt
(d*x + c)*sqrt(-d/b)/(b*d^2*x^2 + a*c*d + (b*c*d + a*d^2)*x)) - 2*(3*b^2*d^2*x^2 - 2*b^2*c^2 - 10*a*b*c*d + 15
*a^2*d^2 - 2*(7*b^2*c*d - 10*a*b*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)]

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giac [B]  time = 2.54, size = 650, normalized size = 5.08 \begin {gather*} \frac {\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} d^{2} {\left | b \right |}}{b^{5}} - \frac {5 \, {\left (\sqrt {b d} b c d {\left | b \right |} - \sqrt {b d} a d^{2} {\left | b \right |}\right )} \log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{2 \, b^{5}} - \frac {4 \, {\left (7 \, \sqrt {b d} b^{6} c^{4} d {\left | b \right |} - 28 \, \sqrt {b d} a b^{5} c^{3} d^{2} {\left | b \right |} + 42 \, \sqrt {b d} a^{2} b^{4} c^{2} d^{3} {\left | b \right |} - 28 \, \sqrt {b d} a^{3} b^{3} c d^{4} {\left | b \right |} + 7 \, \sqrt {b d} a^{4} b^{2} d^{5} {\left | b \right |} - 12 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{4} c^{3} d {\left | b \right |} + 36 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b^{3} c^{2} d^{2} {\left | b \right |} - 36 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{2} b^{2} c d^{3} {\left | b \right |} + 12 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{3} b d^{4} {\left | b \right |} + 9 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} b^{2} c^{2} d {\left | b \right |} - 18 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} a b c d^{2} {\left | b \right |} + 9 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} a^{2} d^{3} {\left | b \right |}\right )}}{3 \, {\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}^{3} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*d^2*abs(b)/b^5 - 5/2*(sqrt(b*d)*b*c*d*abs(b) - sqrt(b*d)*a*d
^2*abs(b))*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/b^5 - 4/3*(7*sqrt(b*d)*b^6*c
^4*d*abs(b) - 28*sqrt(b*d)*a*b^5*c^3*d^2*abs(b) + 42*sqrt(b*d)*a^2*b^4*c^2*d^3*abs(b) - 28*sqrt(b*d)*a^3*b^3*c
*d^4*abs(b) + 7*sqrt(b*d)*a^4*b^2*d^5*abs(b) - 12*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*
b*d - a*b*d))^2*b^4*c^3*d*abs(b) + 36*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d)
)^2*a*b^3*c^2*d^2*abs(b) - 36*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*
b^2*c*d^3*abs(b) + 12*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^3*b*d^4*ab
s(b) + 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^2*c^2*d*abs(b) - 18*sqr
t(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a*b*c*d^2*abs(b) + 9*sqrt(b*d)*(sqrt(
b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^2*d^3*abs(b))/((b^2*c - a*b*d - (sqrt(b*d)*sqrt(
b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)^3*b^4)

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maple [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d x +c \right )^{\frac {5}{2}}}{\left (b x +a \right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(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 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c+d\,x\right )}^{5/2}}{{\left (a+b\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)**(5/2),x)

[Out]

Timed out

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