3.1201 \(\int \frac{\sqrt{a+b x+c x^2}}{(b d+2 c d x)^7} \, dx\)

Optimal. Leaf size=175 \[ \frac{\tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{64 c^{3/2} d^7 \left (b^2-4 a c\right )^{5/2}}+\frac{\sqrt{a+b x+c x^2}}{32 c d^7 \left (b^2-4 a c\right )^2 (b+2 c x)^2}+\frac{\sqrt{a+b x+c x^2}}{48 c d^7 \left (b^2-4 a c\right ) (b+2 c x)^4}-\frac{\sqrt{a+b x+c x^2}}{12 c d^7 (b+2 c x)^6} \]

[Out]

-Sqrt[a + b*x + c*x^2]/(12*c*d^7*(b + 2*c*x)^6) + Sqrt[a + b*x + c*x^2]/(48*c*(b^2 - 4*a*c)*d^7*(b + 2*c*x)^4)
 + Sqrt[a + b*x + c*x^2]/(32*c*(b^2 - 4*a*c)^2*d^7*(b + 2*c*x)^2) + ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/S
qrt[b^2 - 4*a*c]]/(64*c^(3/2)*(b^2 - 4*a*c)^(5/2)*d^7)

________________________________________________________________________________________

Rubi [A]  time = 0.125092, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {684, 693, 688, 205} \[ \frac{\tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{64 c^{3/2} d^7 \left (b^2-4 a c\right )^{5/2}}+\frac{\sqrt{a+b x+c x^2}}{32 c d^7 \left (b^2-4 a c\right )^2 (b+2 c x)^2}+\frac{\sqrt{a+b x+c x^2}}{48 c d^7 \left (b^2-4 a c\right ) (b+2 c x)^4}-\frac{\sqrt{a+b x+c x^2}}{12 c d^7 (b+2 c x)^6} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x + c*x^2]/(b*d + 2*c*d*x)^7,x]

[Out]

-Sqrt[a + b*x + c*x^2]/(12*c*d^7*(b + 2*c*x)^6) + Sqrt[a + b*x + c*x^2]/(48*c*(b^2 - 4*a*c)*d^7*(b + 2*c*x)^4)
 + Sqrt[a + b*x + c*x^2]/(32*c*(b^2 - 4*a*c)^2*d^7*(b + 2*c*x)^2) + ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/S
qrt[b^2 - 4*a*c]]/(64*c^(3/2)*(b^2 - 4*a*c)^(5/2)*d^7)

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 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])

Rule 688

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\sqrt{a+b x+c x^2}}{(b d+2 c d x)^7} \, dx &=-\frac{\sqrt{a+b x+c x^2}}{12 c d^7 (b+2 c x)^6}+\frac{\int \frac{1}{(b d+2 c d x)^5 \sqrt{a+b x+c x^2}} \, dx}{24 c d^2}\\ &=-\frac{\sqrt{a+b x+c x^2}}{12 c d^7 (b+2 c x)^6}+\frac{\sqrt{a+b x+c x^2}}{48 c \left (b^2-4 a c\right ) d^7 (b+2 c x)^4}+\frac{\int \frac{1}{(b d+2 c d x)^3 \sqrt{a+b x+c x^2}} \, dx}{32 c \left (b^2-4 a c\right ) d^4}\\ &=-\frac{\sqrt{a+b x+c x^2}}{12 c d^7 (b+2 c x)^6}+\frac{\sqrt{a+b x+c x^2}}{48 c \left (b^2-4 a c\right ) d^7 (b+2 c x)^4}+\frac{\sqrt{a+b x+c x^2}}{32 c \left (b^2-4 a c\right )^2 d^7 (b+2 c x)^2}+\frac{\int \frac{1}{(b d+2 c d x) \sqrt{a+b x+c x^2}} \, dx}{64 c \left (b^2-4 a c\right )^2 d^6}\\ &=-\frac{\sqrt{a+b x+c x^2}}{12 c d^7 (b+2 c x)^6}+\frac{\sqrt{a+b x+c x^2}}{48 c \left (b^2-4 a c\right ) d^7 (b+2 c x)^4}+\frac{\sqrt{a+b x+c x^2}}{32 c \left (b^2-4 a c\right )^2 d^7 (b+2 c x)^2}+\frac{\operatorname{Subst}\left (\int \frac{1}{2 b^2 c d-8 a c^2 d+8 c^2 d x^2} \, dx,x,\sqrt{a+b x+c x^2}\right )}{16 \left (b^2-4 a c\right )^2 d^6}\\ &=-\frac{\sqrt{a+b x+c x^2}}{12 c d^7 (b+2 c x)^6}+\frac{\sqrt{a+b x+c x^2}}{48 c \left (b^2-4 a c\right ) d^7 (b+2 c x)^4}+\frac{\sqrt{a+b x+c x^2}}{32 c \left (b^2-4 a c\right )^2 d^7 (b+2 c x)^2}+\frac{\tan ^{-1}\left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}\right )}{64 c^{3/2} \left (b^2-4 a c\right )^{5/2} d^7}\\ \end{align*}

Mathematica [C]  time = 0.0246535, size = 62, normalized size = 0.35 \[ \frac{2 (a+x (b+c x))^{3/2} \, _2F_1\left (\frac{3}{2},4;\frac{5}{2};\frac{4 c (a+x (b+c x))}{4 a c-b^2}\right )}{3 d^7 \left (b^2-4 a c\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x + c*x^2]/(b*d + 2*c*d*x)^7,x]

[Out]

(2*(a + x*(b + c*x))^(3/2)*Hypergeometric2F1[3/2, 4, 5/2, (4*c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])/(3*(b^2 - 4
*a*c)^4*d^7)

________________________________________________________________________________________

Maple [B]  time = 0.202, size = 460, normalized size = 2.6 \begin{align*} -{\frac{1}{192\,{d}^{7}{c}^{6} \left ( 4\,ac-{b}^{2} \right ) } \left ( \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}} \right ) ^{{\frac{3}{2}}} \left ( x+{\frac{b}{2\,c}} \right ) ^{-6}}+{\frac{1}{64\,{c}^{4}{d}^{7} \left ( 4\,ac-{b}^{2} \right ) ^{2}} \left ( \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}} \right ) ^{{\frac{3}{2}}} \left ( x+{\frac{b}{2\,c}} \right ) ^{-4}}-{\frac{1}{32\,{d}^{7}{c}^{2} \left ( 4\,ac-{b}^{2} \right ) ^{3}} \left ( \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}} \right ) ^{{\frac{3}{2}}} \left ( x+{\frac{b}{2\,c}} \right ) ^{-2}}+{\frac{1}{64\,{d}^{7}c \left ( 4\,ac-{b}^{2} \right ) ^{3}}\sqrt{4\, \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{c}}}}-{\frac{a}{16\,{d}^{7}c \left ( 4\,ac-{b}^{2} \right ) ^{3}}\ln \left ({ \left ({\frac{4\,ac-{b}^{2}}{2\,c}}+{\frac{1}{2}\sqrt{{\frac{4\,ac-{b}^{2}}{c}}}\sqrt{4\, \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{c}}}} \right ) \left ( x+{\frac{b}{2\,c}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{4\,ac-{b}^{2}}{c}}}}}}+{\frac{{b}^{2}}{64\,{d}^{7}{c}^{2} \left ( 4\,ac-{b}^{2} \right ) ^{3}}\ln \left ({ \left ({\frac{4\,ac-{b}^{2}}{2\,c}}+{\frac{1}{2}\sqrt{{\frac{4\,ac-{b}^{2}}{c}}}\sqrt{4\, \left ( x+1/2\,{\frac{b}{c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{c}}}} \right ) \left ( x+{\frac{b}{2\,c}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{4\,ac-{b}^{2}}{c}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^7,x)

[Out]

-1/192/d^7/c^6/(4*a*c-b^2)/(x+1/2*b/c)^6*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)+1/64/d^7/c^4/(4*a*c-b^2)^2/
(x+1/2*b/c)^4*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)-1/32/d^7/c^2/(4*a*c-b^2)^3/(x+1/2*b/c)^2*((x+1/2*b/c)^
2*c+1/4*(4*a*c-b^2)/c)^(3/2)+1/64/d^7/c/(4*a*c-b^2)^3*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)-1/16/d^7/c/(4*a*
c-b^2)^3/((4*a*c-b^2)/c)^(1/2)*ln((1/2*(4*a*c-b^2)/c+1/2*((4*a*c-b^2)/c)^(1/2)*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/
c)^(1/2))/(x+1/2*b/c))*a+1/64/d^7/c^2/(4*a*c-b^2)^3/((4*a*c-b^2)/c)^(1/2)*ln((1/2*(4*a*c-b^2)/c+1/2*((4*a*c-b^
2)/c)^(1/2)*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2))/(x+1/2*b/c))*b^2

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^7,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 43.5559, size = 2624, normalized size = 14.99 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^7,x, algorithm="fricas")

[Out]

[-1/384*(3*(64*c^6*x^6 + 192*b*c^5*x^5 + 240*b^2*c^4*x^4 + 160*b^3*c^3*x^3 + 60*b^4*c^2*x^2 + 12*b^5*c*x + b^6
)*sqrt(-b^2*c + 4*a*c^2)*log(-(4*c^2*x^2 + 4*b*c*x - b^2 + 8*a*c - 4*sqrt(-b^2*c + 4*a*c^2)*sqrt(c*x^2 + b*x +
 a))/(4*c^2*x^2 + 4*b*c*x + b^2)) + 4*(3*b^6*c - 68*a*b^4*c^2 + 352*a^2*b^2*c^3 - 512*a^3*c^4 - 48*(b^2*c^5 -
4*a*c^6)*x^4 - 96*(b^3*c^4 - 4*a*b*c^5)*x^3 - 16*(5*b^4*c^3 - 22*a*b^2*c^4 + 8*a^2*c^5)*x^2 - 32*(b^5*c^2 - 5*
a*b^3*c^3 + 4*a^2*b*c^4)*x)*sqrt(c*x^2 + b*x + a))/(64*(b^6*c^8 - 12*a*b^4*c^9 + 48*a^2*b^2*c^10 - 64*a^3*c^11
)*d^7*x^6 + 192*(b^7*c^7 - 12*a*b^5*c^8 + 48*a^2*b^3*c^9 - 64*a^3*b*c^10)*d^7*x^5 + 240*(b^8*c^6 - 12*a*b^6*c^
7 + 48*a^2*b^4*c^8 - 64*a^3*b^2*c^9)*d^7*x^4 + 160*(b^9*c^5 - 12*a*b^7*c^6 + 48*a^2*b^5*c^7 - 64*a^3*b^3*c^8)*
d^7*x^3 + 60*(b^10*c^4 - 12*a*b^8*c^5 + 48*a^2*b^6*c^6 - 64*a^3*b^4*c^7)*d^7*x^2 + 12*(b^11*c^3 - 12*a*b^9*c^4
 + 48*a^2*b^7*c^5 - 64*a^3*b^5*c^6)*d^7*x + (b^12*c^2 - 12*a*b^10*c^3 + 48*a^2*b^8*c^4 - 64*a^3*b^6*c^5)*d^7),
 -1/192*(3*(64*c^6*x^6 + 192*b*c^5*x^5 + 240*b^2*c^4*x^4 + 160*b^3*c^3*x^3 + 60*b^4*c^2*x^2 + 12*b^5*c*x + b^6
)*sqrt(b^2*c - 4*a*c^2)*arctan(1/2*sqrt(b^2*c - 4*a*c^2)*sqrt(c*x^2 + b*x + a)/(c^2*x^2 + b*c*x + a*c)) + 2*(3
*b^6*c - 68*a*b^4*c^2 + 352*a^2*b^2*c^3 - 512*a^3*c^4 - 48*(b^2*c^5 - 4*a*c^6)*x^4 - 96*(b^3*c^4 - 4*a*b*c^5)*
x^3 - 16*(5*b^4*c^3 - 22*a*b^2*c^4 + 8*a^2*c^5)*x^2 - 32*(b^5*c^2 - 5*a*b^3*c^3 + 4*a^2*b*c^4)*x)*sqrt(c*x^2 +
 b*x + a))/(64*(b^6*c^8 - 12*a*b^4*c^9 + 48*a^2*b^2*c^10 - 64*a^3*c^11)*d^7*x^6 + 192*(b^7*c^7 - 12*a*b^5*c^8
+ 48*a^2*b^3*c^9 - 64*a^3*b*c^10)*d^7*x^5 + 240*(b^8*c^6 - 12*a*b^6*c^7 + 48*a^2*b^4*c^8 - 64*a^3*b^2*c^9)*d^7
*x^4 + 160*(b^9*c^5 - 12*a*b^7*c^6 + 48*a^2*b^5*c^7 - 64*a^3*b^3*c^8)*d^7*x^3 + 60*(b^10*c^4 - 12*a*b^8*c^5 +
48*a^2*b^6*c^6 - 64*a^3*b^4*c^7)*d^7*x^2 + 12*(b^11*c^3 - 12*a*b^9*c^4 + 48*a^2*b^7*c^5 - 64*a^3*b^5*c^6)*d^7*
x + (b^12*c^2 - 12*a*b^10*c^3 + 48*a^2*b^8*c^4 - 64*a^3*b^6*c^5)*d^7)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sqrt{a + b x + c x^{2}}}{b^{7} + 14 b^{6} c x + 84 b^{5} c^{2} x^{2} + 280 b^{4} c^{3} x^{3} + 560 b^{3} c^{4} x^{4} + 672 b^{2} c^{5} x^{5} + 448 b c^{6} x^{6} + 128 c^{7} x^{7}}\, dx}{d^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**(1/2)/(2*c*d*x+b*d)**7,x)

[Out]

Integral(sqrt(a + b*x + c*x**2)/(b**7 + 14*b**6*c*x + 84*b**5*c**2*x**2 + 280*b**4*c**3*x**3 + 560*b**3*c**4*x
**4 + 672*b**2*c**5*x**5 + 448*b*c**6*x**6 + 128*c**7*x**7), x)/d**7

________________________________________________________________________________________

Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^7,x, algorithm="giac")

[Out]

Exception raised: TypeError