3.684 \(\int \frac {(a+b x)^{5/2}}{(c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=138 \[ \frac {15 \sqrt {b} (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 d^{7/2}}-\frac {15 b \sqrt {a+b x} \sqrt {c+d x} (b c-a d)}{4 d^3}+\frac {5 b (a+b x)^{3/2} \sqrt {c+d x}}{2 d^2}-\frac {2 (a+b x)^{5/2}}{d \sqrt {c+d x}} \]

[Out]

15/4*(-a*d+b*c)^2*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))*b^(1/2)/d^(7/2)-2*(b*x+a)^(5/2)/d/(d*x+
c)^(1/2)+5/2*b*(b*x+a)^(3/2)*(d*x+c)^(1/2)/d^2-15/4*b*(-a*d+b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/d^3

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Rubi [A]  time = 0.07, antiderivative size = 138, 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} \[ \frac {5 b (a+b x)^{3/2} \sqrt {c+d x}}{2 d^2}-\frac {15 b \sqrt {a+b x} \sqrt {c+d x} (b c-a d)}{4 d^3}+\frac {15 \sqrt {b} (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 d^{7/2}}-\frac {2 (a+b x)^{5/2}}{d \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*x)^(5/2))/(d*Sqrt[c + d*x]) - (15*b*(b*c - a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*d^3) + (5*b*(a + b*
x)^(3/2)*Sqrt[c + d*x])/(2*d^2) + (15*Sqrt[b]*(b*c - a*d)^2*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c +
d*x])])/(4*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 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}}{(c+d x)^{3/2}} \, dx &=-\frac {2 (a+b x)^{5/2}}{d \sqrt {c+d x}}+\frac {(5 b) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx}{d}\\ &=-\frac {2 (a+b x)^{5/2}}{d \sqrt {c+d x}}+\frac {5 b (a+b x)^{3/2} \sqrt {c+d x}}{2 d^2}-\frac {(15 b (b c-a d)) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx}{4 d^2}\\ &=-\frac {2 (a+b x)^{5/2}}{d \sqrt {c+d x}}-\frac {15 b (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d^3}+\frac {5 b (a+b x)^{3/2} \sqrt {c+d x}}{2 d^2}+\frac {\left (15 b (b c-a d)^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{8 d^3}\\ &=-\frac {2 (a+b x)^{5/2}}{d \sqrt {c+d x}}-\frac {15 b (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d^3}+\frac {5 b (a+b x)^{3/2} \sqrt {c+d x}}{2 d^2}+\frac {\left (15 (b c-a d)^2\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 )}{4 d^3}\\ &=-\frac {2 (a+b x)^{5/2}}{d \sqrt {c+d x}}-\frac {15 b (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d^3}+\frac {5 b (a+b x)^{3/2} \sqrt {c+d x}}{2 d^2}+\frac {\left (15 (b c-a d)^2\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 )}{4 d^3}\\ &=-\frac {2 (a+b x)^{5/2}}{d \sqrt {c+d x}}-\frac {15 b (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d^3}+\frac {5 b (a+b x)^{3/2} \sqrt {c+d x}}{2 d^2}+\frac {15 \sqrt {b} (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 d^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 73, normalized size = 0.53 \[ \frac {2 (a+b x)^{7/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{3/2} \, _2F_1\left (\frac {3}{2},\frac {7}{2};\frac {9}{2};\frac {d (a+b x)}{a d-b c}\right )}{7 b (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)^(7/2)*((b*(c + d*x))/(b*c - a*d))^(3/2)*Hypergeometric2F1[3/2, 7/2, 9/2, (d*(a + b*x))/(-(b*c) +
a*d)])/(7*b*(c + d*x)^(3/2))

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fricas [A]  time = 1.44, size = 441, normalized size = 3.20 \[ \left [\frac {15 \, {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2} + {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \sqrt {\frac {b}{d}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d^{2} x + b c d + a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {b}{d}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (2 \, b^{2} d^{2} x^{2} - 15 \, b^{2} c^{2} + 25 \, a b c d - 8 \, a^{2} d^{2} - {\left (5 \, b^{2} c d - 9 \, a b d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, {\left (d^{4} x + c d^{3}\right )}}, -\frac {15 \, {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2} + {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} d x^{2} + a b c + {\left (b^{2} c + a b d\right )} x\right )}}\right ) - 2 \, {\left (2 \, b^{2} d^{2} x^{2} - 15 \, b^{2} c^{2} + 25 \, a b c d - 8 \, a^{2} d^{2} - {\left (5 \, b^{2} c d - 9 \, a b d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, {\left (d^{4} x + c d^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)/(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

[1/16*(15*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2 + (b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*x)*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^2*x + b*c*d + a*d^2)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(b/d) + 8*
(b^2*c*d + a*b*d^2)*x) + 4*(2*b^2*d^2*x^2 - 15*b^2*c^2 + 25*a*b*c*d - 8*a^2*d^2 - (5*b^2*c*d - 9*a*b*d^2)*x)*s
qrt(b*x + a)*sqrt(d*x + c))/(d^4*x + c*d^3), -1/8*(15*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2 + (b^2*c^2*d - 2*a*b*
c*d^2 + a^2*d^3)*x)*sqrt(-b/d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-b/d)/(b^2*d*
x^2 + a*b*c + (b^2*c + a*b*d)*x)) - 2*(2*b^2*d^2*x^2 - 15*b^2*c^2 + 25*a*b*c*d - 8*a^2*d^2 - (5*b^2*c*d - 9*a*
b*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(d^4*x + c*d^3)]

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giac [A]  time = 1.38, size = 201, normalized size = 1.46 \[ \frac {\sqrt {b x + a} {\left ({\left (b x + a\right )} {\left (\frac {2 \, {\left (b x + a\right )} b^{2}}{d {\left | b \right |}} - \frac {5 \, {\left (b^{3} c d^{3} - a b^{2} d^{4}\right )}}{d^{5} {\left | b \right |}}\right )} - \frac {15 \, {\left (b^{4} c^{2} d^{2} - 2 \, a b^{3} c d^{3} + a^{2} b^{2} d^{4}\right )}}{d^{5} {\left | b \right |}}\right )}}{4 \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} - \frac {15 \, {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{4 \, \sqrt {b d} d^{3} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)/(d*x+c)^(3/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x +a \right )^{\frac {5}{2}}}{\left (d x +c \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)/(d*x+c)^(3/2),x)

[Out]

int((b*x+a)^(5/2)/(d*x+c)^(3/2),x)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)/(d*x+c)^(3/2),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 \[ \int \frac {{\left (a+b\,x\right )}^{5/2}}{{\left (c+d\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(5/2)/(c + d*x)^(3/2),x)

[Out]

int((a + b*x)^(5/2)/(c + d*x)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x\right )^{\frac {5}{2}}}{\left (c + d x\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(5/2)/(d*x+c)**(3/2),x)

[Out]

Integral((a + b*x)**(5/2)/(c + d*x)**(3/2), x)

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