3.7.76 \(\int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}} \, dx\) [676]

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

[Out]

-5/8*(-a*d+b*c)^3*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/d^(7/2)/b^(1/2)-5/12*(-a*d+b*c)*(b*x+a)
^(3/2)*(d*x+c)^(1/2)/d^2+1/3*(b*x+a)^(5/2)*(d*x+c)^(1/2)/d+5/8*(-a*d+b*c)^2*(b*x+a)^(1/2)*(d*x+c)^(1/2)/d^3

________________________________________________________________________________________

Rubi [A]
time = 0.05, 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 = {52, 65, 223, 212} \begin {gather*} -\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} \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 52

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 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 {(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 ) \text {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 ) \text {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*}

________________________________________________________________________________________

Mathematica [A]
time = 0.05, size = 125, normalized size = 0.84 \begin {gather*} \frac {\sqrt {a+b x} \sqrt {c+d x} \left (33 a^2 d^2+2 a b d (-20 c+13 d x)+b^2 \left (15 c^2-10 c d x+8 d^2 x^2\right )\right )}{24 d^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]

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

[Out]

(Sqrt[a + b*x]*Sqrt[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)))/(2
4*d^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))

________________________________________________________________________________________

Maple [A]
time = 0.07, size = 173, normalized size = 1.17

method	result	size
default	\(\frac {\left (b x +a \right )^{\frac {5}{2}} \sqrt {d x +c}}{3 d}-\frac {5 \left (-a d +b c \right ) \left (\frac {\left (b x +a \right )^{\frac {3}{2}} \sqrt {d x +c}}{2 d}-\frac {3 \left (-a d +b c \right ) \left (\frac {\sqrt {b x +a}\, \sqrt {d x +c}}{d}-\frac {\left (-a d +b c \right ) \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d x}{\sqrt {b d}}+\sqrt {b d \,x^{2}+\left (a d +b c \right ) x +a c}\right )}{2 d \sqrt {b x +a}\, \sqrt {d x +c}\, \sqrt {b d}}\right )}{4 d}\right )}{6 d}\)	\(173\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*x+a)^(5/2)*(d*x+c)^(1/2)/d-5/6*(-a*d+b*c)/d*(1/2*(b*x+a)^(3/2)*(d*x+c)^(1/2)/d-3/4*(-a*d+b*c)/d*((b*x+a
)^(1/2)*(d*x+c)^(1/2)/d-1/2*(-a*d+b*c)/d*((d*x+c)*(b*x+a))^(1/2)/(b*x+a)^(1/2)/(d*x+c)^(1/2)*ln((1/2*a*d+1/2*b
*c+b*d*x)/(b*d)^(1/2)+(b*d*x^2+(a*d+b*c)*x+a*c)^(1/2))/(b*d)^(1/2)))

________________________________________________________________________________________

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 detail

________________________________________________________________________________________

Fricas [A]
time = 1.24, 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)]

________________________________________________________________________________________

Sympy [F(-1)] Timed out
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

________________________________________________________________________________________

Giac [A]
time = 2.85, 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)

________________________________________________________________________________________

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)

________________________________________________________________________________________