3.1644 \(\int \frac {(c+d x)^{5/4}}{(a+b x)^{3/2}} \, dx\)

Optimal. Leaf size=132 \[ \frac {10 (b c-a d)^{5/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{3 b^{9/4} \sqrt {a+b x}}+\frac {10 d \sqrt {a+b x} \sqrt [4]{c+d x}}{3 b^2}-\frac {2 (c+d x)^{5/4}}{b \sqrt {a+b x}} \]

[Out]

-2*(d*x+c)^(5/4)/b/(b*x+a)^(1/2)+10/3*d*(d*x+c)^(1/4)*(b*x+a)^(1/2)/b^2+10/3*(-a*d+b*c)^(5/4)*EllipticF(b^(1/4
)*(d*x+c)^(1/4)/(-a*d+b*c)^(1/4),I)*(-d*(b*x+a)/(-a*d+b*c))^(1/2)/b^(9/4)/(b*x+a)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {47, 50, 63, 224, 221} \[ \frac {10 d \sqrt {a+b x} \sqrt [4]{c+d x}}{3 b^2}+\frac {10 (b c-a d)^{5/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{3 b^{9/4} \sqrt {a+b x}}-\frac {2 (c+d x)^{5/4}}{b \sqrt {a+b x}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/4)/(a + b*x)^(3/2),x]

[Out]

(10*d*Sqrt[a + b*x]*(c + d*x)^(1/4))/(3*b^2) - (2*(c + d*x)^(5/4))/(b*Sqrt[a + b*x]) + (10*(b*c - a*d)^(5/4)*S
qrt[-((d*(a + b*x))/(b*c - a*d))]*EllipticF[ArcSin[(b^(1/4)*(c + d*x)^(1/4))/(b*c - a*d)^(1/4)], -1])/(3*b^(9/
4)*Sqrt[a + b*x])

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 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + (b*x^4)/a]/Sqrt[a + b*x^4], Int[1/Sqrt[1 + (b*x^4)
/a], x], x] /; FreeQ[{a, b}, x] && NegQ[b/a] &&  !GtQ[a, 0]

Rubi steps

\begin {align*} \int \frac {(c+d x)^{5/4}}{(a+b x)^{3/2}} \, dx &=-\frac {2 (c+d x)^{5/4}}{b \sqrt {a+b x}}+\frac {(5 d) \int \frac {\sqrt [4]{c+d x}}{\sqrt {a+b x}} \, dx}{2 b}\\ &=\frac {10 d \sqrt {a+b x} \sqrt [4]{c+d x}}{3 b^2}-\frac {2 (c+d x)^{5/4}}{b \sqrt {a+b x}}+\frac {(5 d (b c-a d)) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}} \, dx}{6 b^2}\\ &=\frac {10 d \sqrt {a+b x} \sqrt [4]{c+d x}}{3 b^2}-\frac {2 (c+d x)^{5/4}}{b \sqrt {a+b x}}+\frac {(10 (b c-a d)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a-\frac {b c}{d}+\frac {b x^4}{d}}} \, dx,x,\sqrt [4]{c+d x}\right )}{3 b^2}\\ &=\frac {10 d \sqrt {a+b x} \sqrt [4]{c+d x}}{3 b^2}-\frac {2 (c+d x)^{5/4}}{b \sqrt {a+b x}}+\frac {\left (10 (b c-a d) \sqrt {\frac {d (a+b x)}{-b c+a d}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {b x^4}{\left (a-\frac {b c}{d}\right ) d}}} \, dx,x,\sqrt [4]{c+d x}\right )}{3 b^2 \sqrt {a+b x}}\\ &=\frac {10 d \sqrt {a+b x} \sqrt [4]{c+d x}}{3 b^2}-\frac {2 (c+d x)^{5/4}}{b \sqrt {a+b x}}+\frac {10 (b c-a d)^{5/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{3 b^{9/4} \sqrt {a+b x}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 71, normalized size = 0.54 \[ -\frac {2 (c+d x)^{5/4} \, _2F_1\left (-\frac {5}{4},-\frac {1}{2};\frac {1}{2};\frac {d (a+b x)}{a d-b c}\right )}{b \sqrt {a+b x} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(5/4)/(a + b*x)^(3/2),x]

[Out]

(-2*(c + d*x)^(5/4)*Hypergeometric2F1[-5/4, -1/2, 1/2, (d*(a + b*x))/(-(b*c) + a*d)])/(b*Sqrt[a + b*x]*((b*(c
+ d*x))/(b*c - a*d))^(5/4))

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x + a} {\left (d x + c\right )}^{\frac {5}{4}}}{b^{2} x^{2} + 2 \, a b x + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/4)/(b*x+a)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*x + a)*(d*x + c)^(5/4)/(b^2*x^2 + 2*a*b*x + a^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/4)/(b*x+a)^(3/2),x, algorithm="giac")

[Out]

integrate((d*x + c)^(5/4)/(b*x + a)^(3/2), x)

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x +c \right )^{\frac {5}{4}}}{\left (b x +a \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(5/4)/(b*x+a)^(3/2),x)

[Out]

int((d*x+c)^(5/4)/(b*x+a)^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/4)/(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

integrate((d*x + c)^(5/4)/(b*x + a)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c+d\,x\right )}^{5/4}}{{\left (a+b\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(5/4)/(a + b*x)^(3/2),x)

[Out]

int((c + d*x)^(5/4)/(a + b*x)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c + d x\right )^{\frac {5}{4}}}{\left (a + b x\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(5/4)/(b*x+a)**(3/2),x)

[Out]

Integral((c + d*x)**(5/4)/(a + b*x)**(3/2), x)

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