3.15.87 \(\int \frac {(c+d x)^{5/2}}{(a+b x)^{7/2}} \, dx\) [1487]

Optimal. Leaf size=120 \[ -\frac {2 d^2 \sqrt {c+d x}}{b^3 \sqrt {a+b x}}-\frac {2 d (c+d x)^{3/2}}{3 b^2 (a+b x)^{3/2}}-\frac {2 (c+d x)^{5/2}}{5 b (a+b x)^{5/2}}+\frac {2 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{7/2}} \]

[Out]

-2/3*d*(d*x+c)^(3/2)/b^2/(b*x+a)^(3/2)-2/5*(d*x+c)^(5/2)/b/(b*x+a)^(5/2)+2*d^(5/2)*arctanh(d^(1/2)*(b*x+a)^(1/
2)/b^(1/2)/(d*x+c)^(1/2))/b^(7/2)-2*d^2*(d*x+c)^(1/2)/b^3/(b*x+a)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {49, 65, 223, 212} \begin {gather*} \frac {2 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{7/2}}-\frac {2 d^2 \sqrt {c+d x}}{b^3 \sqrt {a+b x}}-\frac {2 d (c+d x)^{3/2}}{3 b^2 (a+b x)^{3/2}}-\frac {2 (c+d x)^{5/2}}{5 b (a+b x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d^2*Sqrt[c + d*x])/(b^3*Sqrt[a + b*x]) - (2*d*(c + d*x)^(3/2))/(3*b^2*(a + b*x)^(3/2)) - (2*(c + d*x)^(5/2
))/(5*b*(a + b*x)^(5/2)) + (2*d^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/b^(7/2)

Rule 49

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 65

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 212

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

Rule 223

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)^{7/2}} \, dx &=-\frac {2 (c+d x)^{5/2}}{5 b (a+b x)^{5/2}}+\frac {d \int \frac {(c+d x)^{3/2}}{(a+b x)^{5/2}} \, dx}{b}\\ &=-\frac {2 d (c+d x)^{3/2}}{3 b^2 (a+b x)^{3/2}}-\frac {2 (c+d x)^{5/2}}{5 b (a+b x)^{5/2}}+\frac {d^2 \int \frac {\sqrt {c+d x}}{(a+b x)^{3/2}} \, dx}{b^2}\\ &=-\frac {2 d^2 \sqrt {c+d x}}{b^3 \sqrt {a+b x}}-\frac {2 d (c+d x)^{3/2}}{3 b^2 (a+b x)^{3/2}}-\frac {2 (c+d x)^{5/2}}{5 b (a+b x)^{5/2}}+\frac {d^3 \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{b^3}\\ &=-\frac {2 d^2 \sqrt {c+d x}}{b^3 \sqrt {a+b x}}-\frac {2 d (c+d x)^{3/2}}{3 b^2 (a+b x)^{3/2}}-\frac {2 (c+d x)^{5/2}}{5 b (a+b x)^{5/2}}+\frac {\left (2 d^3\right ) \text {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 {2 d^2 \sqrt {c+d x}}{b^3 \sqrt {a+b x}}-\frac {2 d (c+d x)^{3/2}}{3 b^2 (a+b x)^{3/2}}-\frac {2 (c+d x)^{5/2}}{5 b (a+b x)^{5/2}}+\frac {\left (2 d^3\right ) \text {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 {2 d^2 \sqrt {c+d x}}{b^3 \sqrt {a+b x}}-\frac {2 d (c+d x)^{3/2}}{3 b^2 (a+b x)^{3/2}}-\frac {2 (c+d x)^{5/2}}{5 b (a+b x)^{5/2}}+\frac {2 d^{5/2} \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 [A]
time = 0.13, size = 111, normalized size = 0.92 \begin {gather*} -\frac {2 \sqrt {c+d x} \left (15 a^2 d^2+5 a b d (c+7 d x)+b^2 \left (3 c^2+11 c d x+23 d^2 x^2\right )\right )}{15 b^3 (a+b x)^{5/2}}+\frac {2 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c + d*x]*(15*a^2*d^2 + 5*a*b*d*(c + 7*d*x) + b^2*(3*c^2 + 11*c*d*x + 23*d^2*x^2)))/(15*b^3*(a + b*x)^
(5/2)) + (2*d^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/b^(7/2)

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Mathics [F(-1)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[(c + d*x)^(5/2)/(a + b*x)^(7/2),x]')

[Out]

Timed out

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (d x +c \right )^{\frac {5}{2}}}{\left (b x +a \right )^{\frac {7}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(5/2)/(b*x+a)^(7/2),x)

[Out]

int((d*x+c)^(5/2)/(b*x+a)^(7/2),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)^(7/2),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 detail

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (92) = 184\).
time = 0.59, size = 463, normalized size = 3.86 \begin {gather*} \left [\frac {15 \, {\left (b^{3} d^{2} x^{3} + 3 \, a b^{2} d^{2} x^{2} + 3 \, a^{2} b d^{2} x + a^{3} d^{2}\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 (23 \, b^{2} d^{2} x^{2} + 3 \, b^{2} c^{2} + 5 \, a b c d + 15 \, a^{2} d^{2} + {\left (11 \, b^{2} c d + 35 \, a b d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{30 \, {\left (b^{6} x^{3} + 3 \, a b^{5} x^{2} + 3 \, a^{2} b^{4} x + a^{3} b^{3}\right )}}, -\frac {15 \, {\left (b^{3} d^{2} x^{3} + 3 \, a b^{2} d^{2} x^{2} + 3 \, a^{2} b d^{2} x + a^{3} d^{2}\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 (23 \, b^{2} d^{2} x^{2} + 3 \, b^{2} c^{2} + 5 \, a b c d + 15 \, a^{2} d^{2} + {\left (11 \, b^{2} c d + 35 \, a b d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{15 \, {\left (b^{6} x^{3} + 3 \, a b^{5} x^{2} + 3 \, a^{2} b^{4} x + a^{3} 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)^(7/2),x, algorithm="fricas")

[Out]

[1/30*(15*(b^3*d^2*x^3 + 3*a*b^2*d^2*x^2 + 3*a^2*b*d^2*x + a^3*d^2)*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*(23*b^2*d^2*x^2 + 3*b^2*c^2 + 5*a*b*c*d + 15*a^2*d^2 + (11*b^2*c*d + 35*a*b*d^2)*x)*sqrt(b*x + a)*sqr
t(d*x + c))/(b^6*x^3 + 3*a*b^5*x^2 + 3*a^2*b^4*x + a^3*b^3), -1/15*(15*(b^3*d^2*x^3 + 3*a*b^2*d^2*x^2 + 3*a^2*
b*d^2*x + a^3*d^2)*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*(23*b^2*d^2*x^2 + 3*b^2*c^2 + 5*a*b*c*d + 15*a^2*d^2 + (11*b^2*c*d + 35*a
*b*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^6*x^3 + 3*a*b^5*x^2 + 3*a^2*b^4*x + a^3*b^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)**(5/2)/(b*x+a)**(7/2),x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 343 vs. \(2 (92) = 184\).
time = 0.10, size = 421, normalized size = 3.51 \begin {gather*} \frac {2 \left (\left (-\frac {\left (345 b^{6} d^{6} c^{2}-690 b^{5} d^{7} a c+345 b^{4} d^{8} a^{2}\right ) \sqrt {c+d x} \sqrt {c+d x}}{225 b^{7} \left |d\right | c^{2}-450 b^{6} d \left |d\right | a c+225 b^{5} d^{2} \left |d\right | a^{2}}-\frac {-525 b^{6} d^{6} c^{3}+1575 b^{5} d^{7} a c^{2}-1575 b^{4} d^{8} a^{2} c+525 b^{3} d^{9} a^{3}}{225 b^{7} \left |d\right | c^{2}-450 b^{6} d \left |d\right | a c+225 b^{5} d^{2} \left |d\right | a^{2}}\right ) \sqrt {c+d x} \sqrt {c+d x}-\frac {225 b^{6} d^{6} c^{4}-900 b^{5} d^{7} a c^{3}+1350 b^{4} d^{8} a^{2} c^{2}-900 b^{3} d^{9} a^{3} c+225 b^{2} d^{10} a^{4}}{225 b^{7} \left |d\right | c^{2}-450 b^{6} d \left |d\right | a c+225 b^{5} d^{2} \left |d\right | a^{2}}\right ) \sqrt {c+d x} \sqrt {a d^{2}-b c d+b d \left (c+d x\right )}}{\left (a d^{2}-b c d+b d \left (c+d x\right )\right )^{3}}-\frac {2 d^{4} \ln \left |\sqrt {a d^{2}-b c d+b d \left (c+d x\right )}-\sqrt {b d} \sqrt {c+d x}\right |}{b^{3} \sqrt {b d} \left |d\right |} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)/(b*x+a)^(7/2),x)

[Out]

-2*d^4*log(abs(-sqrt(b*d)*sqrt(d*x + c) + sqrt((d*x + c)*b*d - b*c*d + a*d^2)))/(sqrt(b*d)*b^3*abs(d)) - 2/15*
((d*x + c)*(23*(b^6*c^2*d^6 - 2*a*b^5*c*d^7 + a^2*b^4*d^8)*(d*x + c)/(b^7*c^2*abs(d) - 2*a*b^6*c*d*abs(d) + a^
2*b^5*d^2*abs(d)) - 35*(b^6*c^3*d^6 - 3*a*b^5*c^2*d^7 + 3*a^2*b^4*c*d^8 - a^3*b^3*d^9)/(b^7*c^2*abs(d) - 2*a*b
^6*c*d*abs(d) + a^2*b^5*d^2*abs(d))) + 15*(b^6*c^4*d^6 - 4*a*b^5*c^3*d^7 + 6*a^2*b^4*c^2*d^8 - 4*a^3*b^3*c*d^9
+ a^4*b^2*d^10)/(b^7*c^2*abs(d) - 2*a*b^6*c*d*abs(d) + a^2*b^5*d^2*abs(d)))*sqrt(d*x + c)/((d*x + c)*b*d - b*
c*d + a*d^2)^(5/2)

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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 )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(5/2)/(a + b*x)^(7/2),x)

[Out]

int((c + d*x)^(5/2)/(a + b*x)^(7/2), x)

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