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

Optimal. Leaf size=148 \[ -\frac {5 (b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 \sqrt {b} d^{7/2}}+\frac {5 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^2}{8 d^3}-\frac {5 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)}{12 d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d} \]

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Rubi [A]  time = 0.07, antiderivative size = 148, 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 = {50, 63, 217, 206} \begin {gather*} \frac {5 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^2}{8 d^3}-\frac {5 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)}{12 d^2}-\frac {5 (b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 \sqrt {b} d^{7/2}}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(5/2)/Sqrt[c + d*x],x]

[Out]

(5*(b*c - a*d)^2*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*d^3) - (5*(b*c - a*d)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(12*d^2)
+ ((a + b*x)^(5/2)*Sqrt[c + d*x])/(3*d) - (5*(b*c - a*d)^3*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c +
d*x])])/(8*Sqrt[b]*d^(7/2))

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 {(a+b x)^{5/2}}{\sqrt {c+d x}} \, dx &=\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {(5 (b c-a d)) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx}{6 d}\\ &=-\frac {5 (b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}+\frac {\left (5 (b c-a d)^2\right ) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx}{8 d^2}\\ &=\frac {5 (b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 d^3}-\frac {5 (b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {\left (5 (b c-a d)^3\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{16 d^3}\\ &=\frac {5 (b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 d^3}-\frac {5 (b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {\left (5 (b c-a d)^3\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 )}{8 b d^3}\\ &=\frac {5 (b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 d^3}-\frac {5 (b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {\left (5 (b c-a d)^3\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 )}{8 b d^3}\\ &=\frac {5 (b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 d^3}-\frac {5 (b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 \sqrt {b} d^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.52, size = 150, normalized size = 1.01 \begin {gather*} \frac {\sqrt {d} \sqrt {a+b x} (c+d x) \left (33 a^2 d^2+2 a b d (13 d x-20 c)+b^2 \left (15 c^2-10 c d x+8 d^2 x^2\right )\right )-\frac {15 (b c-a d)^{7/2} \sqrt {\frac {b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{b}}{24 d^{7/2} \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d]*Sqrt[a + b*x]*(c + d*x)*(33*a^2*d^2 + 2*a*b*d*(-20*c + 13*d*x) + b^2*(15*c^2 - 10*c*d*x + 8*d^2*x^2))
- (15*(b*c - a*d)^(7/2)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/b)/
(24*d^(7/2)*Sqrt[c + d*x])

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IntegrateAlgebraic [A]  time = 0.00, size = 160, normalized size = 1.08 \begin {gather*} \frac {(a d-b c)^3 \left (\frac {15 b^2 (c+d x)^{5/2}}{(a+b x)^{5/2}}+\frac {33 d^2 \sqrt {c+d x}}{\sqrt {a+b x}}-\frac {40 b d (c+d x)^{3/2}}{(a+b x)^{3/2}}\right )}{24 d^3 \left (d-\frac {b (c+d x)}{a+b x}\right )^3}-\frac {5 (b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{8 \sqrt {b} d^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-(b*c) + a*d)^3*((33*d^2*Sqrt[c + d*x])/Sqrt[a + b*x] - (40*b*d*(c + d*x)^(3/2))/(a + b*x)^(3/2) + (15*b^2*(
c + d*x)^(5/2))/(a + b*x)^(5/2)))/(24*d^3*(d - (b*(c + d*x))/(a + b*x))^3) - (5*(b*c - a*d)^3*ArcTanh[(Sqrt[b]
*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(8*Sqrt[b]*d^(7/2))

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fricas [A]  time = 1.39, size = 412, normalized size = 2.78 \begin {gather*} \left [-\frac {15 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {b d} \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 d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (8 \, b^{3} d^{3} x^{2} + 15 \, b^{3} c^{2} d - 40 \, a b^{2} c d^{2} + 33 \, a^{2} b d^{3} - 2 \, {\left (5 \, b^{3} c d^{2} - 13 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, b d^{4}}, \frac {15 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (8 \, b^{3} d^{3} x^{2} + 15 \, b^{3} c^{2} d - 40 \, a b^{2} c d^{2} + 33 \, a^{2} b d^{3} - 2 \, {\left (5 \, b^{3} c d^{2} - 13 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, b d^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/96*(15*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(b*d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c
*d + a^2*d^2 + 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(8
*b^3*d^3*x^2 + 15*b^3*c^2*d - 40*a*b^2*c*d^2 + 33*a^2*b*d^3 - 2*(5*b^3*c*d^2 - 13*a*b^2*d^3)*x)*sqrt(b*x + a)*
sqrt(d*x + c))/(b*d^4), 1/48*(15*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(-b*d)*arctan(1/2*(2*
b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) + 2
*(8*b^3*d^3*x^2 + 15*b^3*c^2*d - 40*a*b^2*c*d^2 + 33*a^2*b*d^3 - 2*(5*b^3*c*d^2 - 13*a*b^2*d^3)*x)*sqrt(b*x +
a)*sqrt(d*x + c))/(b*d^4)]

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giac [A]  time = 1.06, size = 198, normalized size = 1.34 \begin {gather*} \frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )}}{b d} - \frac {5 \, {\left (b c d^{3} - a d^{4}\right )}}{b d^{5}}\right )} + \frac {15 \, {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )}}{b d^{5}}\right )} + \frac {15 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\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 |}\right )}{\sqrt {b d} d^{3}}\right )} b}{24 \, {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)/(b*d) - 5*(b*c*d^3 - a*d^4)/
(b*d^5)) + 15*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)/(b*d^5)) + 15*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a
^3*d^3)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*d^3))*b/abs(b)

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maple [B]  time = 0.00, size = 465, normalized size = 3.14 \begin {gather*} \frac {5 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{3} \ln \left (\frac {b d x +\frac {1}{2} a d +\frac {1}{2} b c}{\sqrt {b d}}+\sqrt {b d \,x^{2}+a c +\left (a d +b c \right ) x}\right )}{16 \sqrt {b x +a}\, \sqrt {d x +c}\, \sqrt {b d}}-\frac {15 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} b c \ln \left (\frac {b d x +\frac {1}{2} a d +\frac {1}{2} b c}{\sqrt {b d}}+\sqrt {b d \,x^{2}+a c +\left (a d +b c \right ) x}\right )}{16 \sqrt {b x +a}\, \sqrt {d x +c}\, \sqrt {b d}\, d}+\frac {15 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a \,b^{2} c^{2} \ln \left (\frac {b d x +\frac {1}{2} a d +\frac {1}{2} b c}{\sqrt {b d}}+\sqrt {b d \,x^{2}+a c +\left (a d +b c \right ) x}\right )}{16 \sqrt {b x +a}\, \sqrt {d x +c}\, \sqrt {b d}\, d^{2}}-\frac {5 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{3} c^{3} \ln \left (\frac {b d x +\frac {1}{2} a d +\frac {1}{2} b c}{\sqrt {b d}}+\sqrt {b d \,x^{2}+a c +\left (a d +b c \right ) x}\right )}{16 \sqrt {b x +a}\, \sqrt {d x +c}\, \sqrt {b d}\, d^{3}}+\frac {5 \sqrt {b x +a}\, \sqrt {d x +c}\, a^{2}}{8 d}-\frac {5 \sqrt {b x +a}\, \sqrt {d x +c}\, a b c}{4 d^{2}}+\frac {5 \sqrt {b x +a}\, \sqrt {d x +c}\, b^{2} c^{2}}{8 d^{3}}+\frac {5 \left (b x +a \right )^{\frac {3}{2}} \sqrt {d x +c}\, a}{12 d}-\frac {5 \left (b x +a \right )^{\frac {3}{2}} \sqrt {d x +c}\, b c}{12 d^{2}}+\frac {\left (b x +a \right )^{\frac {5}{2}} \sqrt {d x +c}}{3 d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/16*((b*x+a)*(d*x+c))^(1/2)/(b*x+a)^(1/2)/(d*x+c)^(1/2)/(b*d)^(1/2)*a^3*ln((b*d*x+1/2*a*d+1/2*b*c)/(b*d)^(1/2
)+(b*d*x^2+a*c+(a*d+b*c)*x)^(1/2))-15/16*((b*x+a)*(d*x+c))^(1/2)/(b*x+a)^(1/2)/(d*x+c)^(1/2)/(b*d)^(1/2)*a^2*b
*c/d*ln((b*d*x+1/2*a*d+1/2*b*c)/(b*d)^(1/2)+(b*d*x^2+a*c+(a*d+b*c)*x)^(1/2))+15/16*((b*x+a)*(d*x+c))^(1/2)/(b*
x+a)^(1/2)/(d*x+c)^(1/2)/(b*d)^(1/2)*a*b^2*c^2/d^2*ln((b*d*x+1/2*a*d+1/2*b*c)/(b*d)^(1/2)+(b*d*x^2+a*c+(a*d+b*
c)*x)^(1/2))-5/16*((b*x+a)*(d*x+c))^(1/2)/(b*x+a)^(1/2)/(d*x+c)^(1/2)/(b*d)^(1/2)*b^3*c^3/d^3*ln((b*d*x+1/2*a*
d+1/2*b*c)/(b*d)^(1/2)+(b*d*x^2+a*c+(a*d+b*c)*x)^(1/2))+5/8*(b*x+a)^(1/2)*(d*x+c)^(1/2)*a^2/d-5/4*(b*x+a)^(1/2
)*(d*x+c)^(1/2)*a*b*c/d^2+5/8*(b*x+a)^(1/2)*(d*x+c)^(1/2)*b^2*c^2/d^3+5/12*(b*x+a)^(3/2)*(d*x+c)^(1/2)*a/d-5/1
2*(b*x+a)^(3/2)*(d*x+c)^(1/2)*b*c/d^2+1/3*(b*x+a)^(5/2)*(d*x+c)^(1/2)/d

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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((b*x+a)^(5/2)/(d*x+c)^(1/2),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 (a+b\,x\right )}^{5/2}}{\sqrt {c+d\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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