3.1343 \(\int \frac {(a+b x+c x^2)^{3/2}}{(b d+2 c d x)^{17/2}} \, dx\)

Optimal. Leaf size=268 \[ \frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{462 c^3 d^{17/2} \left (b^2-4 a c\right )^{7/4} \sqrt {a+b x+c x^2}}+\frac {\sqrt {a+b x+c x^2}}{231 c^2 d^7 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2}}+\frac {\sqrt {a+b x+c x^2}}{385 c^2 d^5 \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}-\frac {\sqrt {a+b x+c x^2}}{110 c^2 d^3 (b d+2 c d x)^{11/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{15 c d (b d+2 c d x)^{15/2}} \]

[Out]

-1/15*(c*x^2+b*x+a)^(3/2)/c/d/(2*c*d*x+b*d)^(15/2)-1/110*(c*x^2+b*x+a)^(1/2)/c^2/d^3/(2*c*d*x+b*d)^(11/2)+1/38
5*(c*x^2+b*x+a)^(1/2)/c^2/(-4*a*c+b^2)/d^5/(2*c*d*x+b*d)^(7/2)+1/231*(c*x^2+b*x+a)^(1/2)/c^2/(-4*a*c+b^2)^2/d^
7/(2*c*d*x+b*d)^(3/2)+1/462*EllipticF((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2),I)*(-c*(c*x^2+b*x+a)/(-4*
a*c+b^2))^(1/2)/c^3/(-4*a*c+b^2)^(7/4)/d^(17/2)/(c*x^2+b*x+a)^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 268, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {684, 693, 691, 689, 221} \[ \frac {\sqrt {a+b x+c x^2}}{231 c^2 d^7 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2}}+\frac {\sqrt {a+b x+c x^2}}{385 c^2 d^5 \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}+\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{462 c^3 d^{17/2} \left (b^2-4 a c\right )^{7/4} \sqrt {a+b x+c x^2}}-\frac {\sqrt {a+b x+c x^2}}{110 c^2 d^3 (b d+2 c d x)^{11/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{15 c d (b d+2 c d x)^{15/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^(17/2),x]

[Out]

-Sqrt[a + b*x + c*x^2]/(110*c^2*d^3*(b*d + 2*c*d*x)^(11/2)) + Sqrt[a + b*x + c*x^2]/(385*c^2*(b^2 - 4*a*c)*d^5
*(b*d + 2*c*d*x)^(7/2)) + Sqrt[a + b*x + c*x^2]/(231*c^2*(b^2 - 4*a*c)^2*d^7*(b*d + 2*c*d*x)^(3/2)) - (a + b*x
+ c*x^2)^(3/2)/(15*c*d*(b*d + 2*c*d*x)^(15/2)) + (Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[ArcS
in[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(462*c^3*(b^2 - 4*a*c)^(7/4)*d^(17/2)*Sqrt[a + b*x
+ c*x^2])

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 684

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 1)), x] - Dist[(b*p)/(d*e*(m + 1)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1
), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] &&
GtQ[p, 0] && LtQ[m, -1] &&  !(IntegerQ[m/2] && LtQ[m + 2*p + 3, 0]) && IntegerQ[2*p]

Rule 689

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[(4*Sqrt[-(c/(b^2 -
4*a*c))])/e, Subst[Int[1/Sqrt[Simp[1 - (b^2*x^4)/(d^2*(b^2 - 4*a*c)), x]], x], x, Sqrt[d + e*x]], x] /; FreeQ[
{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] && LtQ[c/(b^2 - 4*a*c), 0]

Rule 691

Int[((d_) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[-((c*(a + b*x + c
*x^2))/(b^2 - 4*a*c))]/Sqrt[a + b*x + c*x^2], Int[(d + e*x)^m/Sqrt[-((a*c)/(b^2 - 4*a*c)) - (b*c*x)/(b^2 - 4*a
*c) - (c^2*x^2)/(b^2 - 4*a*c)], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && EqQ[m^2, 1/4]

Rule 693

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-2*b*d*(d + e*x)^(m
+ 1)*(a + b*x + c*x^2)^(p + 1))/(d^2*(m + 1)*(b^2 - 4*a*c)), x] + Dist[(b^2*(m + 2*p + 3))/(d^2*(m + 1)*(b^2
- 4*a*c)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*
c, 0] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[m, -1] && (IntegerQ[2*p] || (IntegerQ[m] && Rationa
lQ[p]) || IntegerQ[(m + 2*p + 3)/2])

Rubi steps

\begin {align*} \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{17/2}} \, dx &=-\frac {\left (a+b x+c x^2\right )^{3/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac {\int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{13/2}} \, dx}{10 c d^2}\\ &=-\frac {\sqrt {a+b x+c x^2}}{110 c^2 d^3 (b d+2 c d x)^{11/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac {\int \frac {1}{(b d+2 c d x)^{9/2} \sqrt {a+b x+c x^2}} \, dx}{220 c^2 d^4}\\ &=-\frac {\sqrt {a+b x+c x^2}}{110 c^2 d^3 (b d+2 c d x)^{11/2}}+\frac {\sqrt {a+b x+c x^2}}{385 c^2 \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{7/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac {\int \frac {1}{(b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}} \, dx}{308 c^2 \left (b^2-4 a c\right ) d^6}\\ &=-\frac {\sqrt {a+b x+c x^2}}{110 c^2 d^3 (b d+2 c d x)^{11/2}}+\frac {\sqrt {a+b x+c x^2}}{385 c^2 \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{7/2}}+\frac {\sqrt {a+b x+c x^2}}{231 c^2 \left (b^2-4 a c\right )^2 d^7 (b d+2 c d x)^{3/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac {\int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx}{924 c^2 \left (b^2-4 a c\right )^2 d^8}\\ &=-\frac {\sqrt {a+b x+c x^2}}{110 c^2 d^3 (b d+2 c d x)^{11/2}}+\frac {\sqrt {a+b x+c x^2}}{385 c^2 \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{7/2}}+\frac {\sqrt {a+b x+c x^2}}{231 c^2 \left (b^2-4 a c\right )^2 d^7 (b d+2 c d x)^{3/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {-\frac {a c}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {c^2 x^2}{b^2-4 a c}}} \, dx}{924 c^2 \left (b^2-4 a c\right )^2 d^8 \sqrt {a+b x+c x^2}}\\ &=-\frac {\sqrt {a+b x+c x^2}}{110 c^2 d^3 (b d+2 c d x)^{11/2}}+\frac {\sqrt {a+b x+c x^2}}{385 c^2 \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{7/2}}+\frac {\sqrt {a+b x+c x^2}}{231 c^2 \left (b^2-4 a c\right )^2 d^7 (b d+2 c d x)^{3/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^4}{\left (b^2-4 a c\right ) d^2}}} \, dx,x,\sqrt {b d+2 c d x}\right )}{462 c^3 \left (b^2-4 a c\right )^2 d^9 \sqrt {a+b x+c x^2}}\\ &=-\frac {\sqrt {a+b x+c x^2}}{110 c^2 d^3 (b d+2 c d x)^{11/2}}+\frac {\sqrt {a+b x+c x^2}}{385 c^2 \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{7/2}}+\frac {\sqrt {a+b x+c x^2}}{231 c^2 \left (b^2-4 a c\right )^2 d^7 (b d+2 c d x)^{3/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{15 c d (b d+2 c d x)^{15/2}}+\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{462 c^3 \left (b^2-4 a c\right )^{7/4} d^{17/2} \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 107, normalized size = 0.40 \[ \frac {\left (b^2-4 a c\right ) \sqrt {a+x (b+c x)} \sqrt {d (b+2 c x)} \, _2F_1\left (-\frac {15}{4},-\frac {3}{2};-\frac {11}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{120 c^2 d^9 (b+2 c x)^8 \sqrt {\frac {c (a+x (b+c x))}{4 a c-b^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^(17/2),x]

[Out]

((b^2 - 4*a*c)*Sqrt[d*(b + 2*c*x)]*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-15/4, -3/2, -11/4, (b + 2*c*x)^2/(
b^2 - 4*a*c)])/(120*c^2*d^9*(b + 2*c*x)^8*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])

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fricas [F]  time = 0.81, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {2 \, c d x + b d} {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}{512 \, c^{9} d^{9} x^{9} + 2304 \, b c^{8} d^{9} x^{8} + 4608 \, b^{2} c^{7} d^{9} x^{7} + 5376 \, b^{3} c^{6} d^{9} x^{6} + 4032 \, b^{4} c^{5} d^{9} x^{5} + 2016 \, b^{5} c^{4} d^{9} x^{4} + 672 \, b^{6} c^{3} d^{9} x^{3} + 144 \, b^{7} c^{2} d^{9} x^{2} + 18 \, b^{8} c d^{9} x + b^{9} d^{9}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^(17/2),x, algorithm="fricas")

[Out]

integral(sqrt(2*c*d*x + b*d)*(c*x^2 + b*x + a)^(3/2)/(512*c^9*d^9*x^9 + 2304*b*c^8*d^9*x^8 + 4608*b^2*c^7*d^9*
x^7 + 5376*b^3*c^6*d^9*x^6 + 4032*b^4*c^5*d^9*x^5 + 2016*b^5*c^4*d^9*x^4 + 672*b^6*c^3*d^9*x^3 + 144*b^7*c^2*d
^9*x^2 + 18*b^8*c*d^9*x + b^9*d^9), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac {17}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^(17/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(3/2)/(2*c*d*x + b*d)^(17/2), x)

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maple [B]  time = 0.12, size = 1431, normalized size = 5.34 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^(17/2),x)

[Out]

1/4620*(c*x^2+b*x+a)^(1/2)*((2*c*x+b)*d)^(1/2)*(5*(-4*a*c+b^2)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2
)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*
EllipticF(1/2*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*b^7+1792*a^3*b^2*c^3-20
*a^2*b^4*c^2-10*a*b^6*c-4928*a^4*c^4+70*(-4*a*c+b^2)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(
1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(
1/2*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*x*b^6*c+1400*(-4*a*c+b^2)^(1/2)*(
(2*c*x+b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-2*c*x-b+(-4*a*
c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(1/2*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*
2^(1/2),2^(1/2))*x^3*b^4*c^3+420*(-4*a*c+b^2)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-
(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(1/2*((2
*c*x+b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*x^2*b^5*c^2+2240*(-4*a*c+b^2)^(1/2)*((2*
c*x+b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-2*c*x-b+(-4*a*c+b
^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(1/2*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(
1/2),2^(1/2))*x^6*b*c^6+3360*(-4*a*c+b^2)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c
*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(1/2*((2*c*x
+b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*x^5*b^2*c^5+2800*(-4*a*c+b^2)^(1/2)*((2*c*x+
b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^
(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(1/2*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2)
,2^(1/2))*x^4*b^3*c^4+8576*x^5*b^3*c^5-150*x^2*b^6*c^2-10*x*b^7*c+8832*x^6*b^2*c^6+5120*x^7*b*c^7+872*x^3*b^5*
c^3+4596*x^4*b^4*c^4+512*x^6*a*c^7-8384*x^4*a^2*c^6-12544*x^2*a^3*c^5+1280*x^8*c^8-160*x*a*b^5*c^2+1536*x^5*a*
b*c^6+6112*x^4*a*b^2*c^5-16768*x^3*a^2*b*c^5+9664*x^3*a*b^3*c^4-3168*x^2*a^2*b^2*c^4+4416*x^2*a*b^4*c^3-12544*
x*a^3*b*c^4+5216*x*a^2*b^3*c^3+640*(-4*a*c+b^2)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*
(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(1/2*(
(2*c*x+b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*x^7*c^7)/d^9/(2*c^2*x^3+3*b*c*x^2+2*a*
c*x+b^2*x+a*b)/(2*c*x+b)^7/c^3/(4*a*c-b^2)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac {17}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^(17/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(3/2)/(2*c*d*x + b*d)^(17/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,x^2+b\,x+a\right )}^{3/2}}{{\left (b\,d+2\,c\,d\,x\right )}^{17/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^(17/2),x)

[Out]

int((a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^(17/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**(3/2)/(2*c*d*x+b*d)**(17/2),x)

[Out]

Timed out

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