3.1379 \(\int \frac {(b d+2 c d x)^{11/2}}{(a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=205 \[ \frac {60 d^{11/2} \left (b^2-4 a c\right )^{9/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{7 \sqrt {a+b x+c x^2}}+\frac {120}{7} c d^5 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}+\frac {72}{7} c d^3 \sqrt {a+b x+c x^2} (b d+2 c d x)^{5/2}-\frac {2 d (b d+2 c d x)^{9/2}}{\sqrt {a+b x+c x^2}} \]

[Out]

-2*d*(2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)^(1/2)+72/7*c*d^3*(2*c*d*x+b*d)^(5/2)*(c*x^2+b*x+a)^(1/2)+120/7*c*(-4*a*
c+b^2)*d^5*(2*c*d*x+b*d)^(1/2)*(c*x^2+b*x+a)^(1/2)+60/7*(-4*a*c+b^2)^(9/4)*d^(11/2)*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*x^2+b*x+a)^(1/2)

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Rubi [A]  time = 0.18, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {686, 692, 691, 689, 221} \[ \frac {120}{7} c d^5 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}+\frac {60 d^{11/2} \left (b^2-4 a c\right )^{9/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{7 \sqrt {a+b x+c x^2}}+\frac {72}{7} c d^3 \sqrt {a+b x+c x^2} (b d+2 c d x)^{5/2}-\frac {2 d (b d+2 c d x)^{9/2}}{\sqrt {a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^(3/2),x]

[Out]

(-2*d*(b*d + 2*c*d*x)^(9/2))/Sqrt[a + b*x + c*x^2] + (120*c*(b^2 - 4*a*c)*d^5*Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x
+ c*x^2])/7 + (72*c*d^3*(b*d + 2*c*d*x)^(5/2)*Sqrt[a + b*x + c*x^2])/7 + (60*(b^2 - 4*a*c)^(9/4)*d^(11/2)*Sqr
t[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[ArcSin[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])],
-1])/(7*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 686

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*(d + e*x)^(m - 1)*
(a + b*x + c*x^2)^(p + 1))/(b*(p + 1)), x] - Dist[(d*e*(m - 1))/(b*(p + 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] && LtQ[p, -1] && GtQ[m, 1] && 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 692

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*d*(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1))/(b*(m + 2*p + 1)), x] + Dist[(d^2*(m - 1)*(b^2 - 4*a*c))/(b^2*(m + 2*p + 1)), In
t[(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] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] &
& RationalQ[p]) || OddQ[m])

Rubi steps

\begin {align*} \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx &=-\frac {2 d (b d+2 c d x)^{9/2}}{\sqrt {a+b x+c x^2}}+\left (18 c d^2\right ) \int \frac {(b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}} \, dx\\ &=-\frac {2 d (b d+2 c d x)^{9/2}}{\sqrt {a+b x+c x^2}}+\frac {72}{7} c d^3 (b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}+\frac {1}{7} \left (90 c \left (b^2-4 a c\right ) d^4\right ) \int \frac {(b d+2 c d x)^{3/2}}{\sqrt {a+b x+c x^2}} \, dx\\ &=-\frac {2 d (b d+2 c d x)^{9/2}}{\sqrt {a+b x+c x^2}}+\frac {120}{7} c \left (b^2-4 a c\right ) d^5 \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}+\frac {72}{7} c d^3 (b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}+\frac {1}{7} \left (30 c \left (b^2-4 a c\right )^2 d^6\right ) \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx\\ &=-\frac {2 d (b d+2 c d x)^{9/2}}{\sqrt {a+b x+c x^2}}+\frac {120}{7} c \left (b^2-4 a c\right ) d^5 \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}+\frac {72}{7} c d^3 (b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}+\frac {\left (30 c \left (b^2-4 a c\right )^2 d^6 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \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}{7 \sqrt {a+b x+c x^2}}\\ &=-\frac {2 d (b d+2 c d x)^{9/2}}{\sqrt {a+b x+c x^2}}+\frac {120}{7} c \left (b^2-4 a c\right ) d^5 \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}+\frac {72}{7} c d^3 (b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}+\frac {\left (60 \left (b^2-4 a c\right )^2 d^5 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \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 )}{7 \sqrt {a+b x+c x^2}}\\ &=-\frac {2 d (b d+2 c d x)^{9/2}}{\sqrt {a+b x+c x^2}}+\frac {120}{7} c \left (b^2-4 a c\right ) d^5 \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}+\frac {72}{7} c d^3 (b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}+\frac {60 \left (b^2-4 a c\right )^{9/4} d^{11/2} \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 )}{7 \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.20, size = 172, normalized size = 0.84 \[ \frac {2 d^5 \sqrt {d (b+2 c x)} \left (16 c^2 \left (-15 a^2-6 a c x^2+2 c^2 x^4\right )+30 \left (b^2-4 a c\right )^2 \sqrt {\frac {c (a+x (b+c x))}{4 a c-b^2}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )+24 b^2 c \left (4 a+3 c x^2\right )+32 b c^2 x \left (2 c x^2-3 a\right )-7 b^4+40 b^3 c x\right )}{7 \sqrt {a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^(3/2),x]

[Out]

(2*d^5*Sqrt[d*(b + 2*c*x)]*(-7*b^4 + 40*b^3*c*x + 32*b*c^2*x*(-3*a + 2*c*x^2) + 24*b^2*c*(4*a + 3*c*x^2) + 16*
c^2*(-15*a^2 - 6*a*c*x^2 + 2*c^2*x^4) + 30*(b^2 - 4*a*c)^2*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]*Hypergeo
metric2F1[1/4, 1/2, 5/4, (b + 2*c*x)^2/(b^2 - 4*a*c)]))/(7*Sqrt[a + x*(b + c*x)])

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fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (32 \, c^{5} d^{5} x^{5} + 80 \, b c^{4} d^{5} x^{4} + 80 \, b^{2} c^{3} d^{5} x^{3} + 40 \, b^{3} c^{2} d^{5} x^{2} + 10 \, b^{4} c d^{5} x + b^{5} d^{5}\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

integral((32*c^5*d^5*x^5 + 80*b*c^4*d^5*x^4 + 80*b^2*c^3*d^5*x^3 + 40*b^3*c^2*d^5*x^2 + 10*b^4*c*d^5*x + b^5*d
^5)*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a)/(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (2 \, c d x + b d\right )}^{\frac {11}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

integrate((2*c*d*x + b*d)^(11/2)/(c*x^2 + b*x + a)^(3/2), x)

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maple [B]  time = 0.19, size = 569, normalized size = 2.78 \[ \frac {2 \sqrt {\left (2 c x +b \right ) d}\, \sqrt {c \,x^{2}+b x +a}\, \left (64 c^{5} x^{5}+160 b \,c^{4} x^{4}-192 a \,c^{4} x^{3}+208 b^{2} c^{3} x^{3}-288 a b \,c^{3} x^{2}+152 b^{3} c^{2} x^{2}-480 a^{2} c^{3} x +96 a \,b^{2} c^{2} x +26 b^{4} c x -240 a^{2} b \,c^{2}+240 \sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-4 a c +b^{2}}\, a^{2} c^{2} \EllipticF \left (\frac {\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right )+96 a \,b^{3} c -120 \sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-4 a c +b^{2}}\, a \,b^{2} c \EllipticF \left (\frac {\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right )-7 b^{5}+15 \sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-4 a c +b^{2}}\, b^{4} \EllipticF \left (\frac {\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right )\right ) d^{5}}{7 \left (2 c^{2} x^{3}+3 b c \,x^{2}+2 a c x +b^{2} x +a b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^(3/2),x)

[Out]

2/7*((2*c*x+b)*d)^(1/2)*(c*x^2+b*x+a)^(1/2)*d^5*(64*c^5*x^5+240*((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)*Ellip
ticF(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))*(-4*a*c+b^2)^(1/2)*a^2*c^2-1
20*((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))*(-4*a*c+b^2)^(1/2)*a*b^2*c+15*((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))*(-4*a*c+b^2)^(1/2)*b^4+160*x^4*b*c^4-192*
a*c^4*x^3+208*b^2*c^3*x^3-288*x^2*a*b*c^3+152*x^2*b^3*c^2-480*a^2*c^3*x+96*a*b^2*c^2*x+26*b^4*c*x-240*a^2*b*c^
2+96*a*b^3*c-7*b^5)/(2*c^2*x^3+3*b*c*x^2+2*a*c*x+b^2*x+a*b)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (2 \, c d x + b d\right )}^{\frac {11}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")

[Out]

integrate((2*c*d*x + b*d)^(11/2)/(c*x^2 + b*x + a)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (b\,d+2\,c\,d\,x\right )}^{11/2}}{{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^(3/2),x)

[Out]

int((b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^(3/2), 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((2*c*d*x+b*d)**(11/2)/(c*x**2+b*x+a)**(3/2),x)

[Out]

Timed out

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