3.13.99 \(\int \frac {(b d+2 c d x)^{7/2}}{(a+b x+c x^2)^2} \, dx\) [1299]

Optimal. Leaf size=150 \[ 20 c d^3 \sqrt {b d+2 c d x}-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}-10 c \sqrt [4]{b^2-4 a c} d^{7/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-10 c \sqrt [4]{b^2-4 a c} d^{7/2} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ) \]

[Out]

-d*(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)-10*c*(-4*a*c+b^2)^(1/4)*d^(7/2)*arctan((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(
1/4)/d^(1/2))-10*c*(-4*a*c+b^2)^(1/4)*d^(7/2)*arctanh((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))+20*c*d^3
*(2*c*d*x+b*d)^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {700, 706, 708, 335, 218, 212, 209} \begin {gather*} -10 c d^{7/2} \sqrt [4]{b^2-4 a c} \text {ArcTan}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-10 c d^{7/2} \sqrt [4]{b^2-4 a c} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}+20 c d^3 \sqrt {b d+2 c d x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

20*c*d^3*Sqrt[b*d + 2*c*d*x] - (d*(b*d + 2*c*d*x)^(5/2))/(a + b*x + c*x^2) - 10*c*(b^2 - 4*a*c)^(1/4)*d^(7/2)*
ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])] - 10*c*(b^2 - 4*a*c)^(1/4)*d^(7/2)*ArcTanh[Sqrt[d*(b
+ 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 700

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 706

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])

Rule 708

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

Rubi steps

\begin {align*} \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^2} \, dx &=-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}+\left (5 c d^2\right ) \int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx\\ &=20 c d^3 \sqrt {b d+2 c d x}-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}+\left (5 c \left (b^2-4 a c\right ) d^4\right ) \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )} \, dx\\ &=20 c d^3 \sqrt {b d+2 c d x}-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}+\frac {1}{2} \left (5 \left (b^2-4 a c\right ) d^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}\right )} \, dx,x,b d+2 c d x\right )\\ &=20 c d^3 \sqrt {b d+2 c d x}-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}+\left (5 \left (b^2-4 a c\right ) d^3\right ) \text {Subst}\left (\int \frac {1}{a-\frac {b^2}{4 c}+\frac {x^4}{4 c d^2}} \, dx,x,\sqrt {d (b+2 c x)}\right )\\ &=20 c d^3 \sqrt {b d+2 c d x}-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}-\left (10 c \sqrt {b^2-4 a c} d^4\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b^2-4 a c} d-x^2} \, dx,x,\sqrt {d (b+2 c x)}\right )-\left (10 c \sqrt {b^2-4 a c} d^4\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b^2-4 a c} d+x^2} \, dx,x,\sqrt {d (b+2 c x)}\right )\\ &=20 c d^3 \sqrt {b d+2 c d x}-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}-10 c \sqrt [4]{b^2-4 a c} d^{7/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-10 c \sqrt [4]{b^2-4 a c} d^{7/2} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.77, size = 256, normalized size = 1.71 \begin {gather*} (1+i) c (d (b+2 c x))^{7/2} \left (\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (-5 b^2+20 a c+4 (b+2 c x)^2\right )}{c (b+2 c x)^3 (a+x (b+c x))}-\frac {5 i \sqrt [4]{b^2-4 a c} \tan ^{-1}\left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{(b+2 c x)^{7/2}}+\frac {5 i \sqrt [4]{b^2-4 a c} \tan ^{-1}\left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{(b+2 c x)^{7/2}}+\frac {5 i \sqrt [4]{b^2-4 a c} \tanh ^{-1}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )}{(b+2 c x)^{7/2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + I)*c*(d*(b + 2*c*x))^(7/2)*(((1/2 - I/2)*(-5*b^2 + 20*a*c + 4*(b + 2*c*x)^2))/(c*(b + 2*c*x)^3*(a + x*(b
+ c*x))) - ((5*I)*(b^2 - 4*a*c)^(1/4)*ArcTan[1 - ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/(b + 2*c*x)^(
7/2) + ((5*I)*(b^2 - 4*a*c)^(1/4)*ArcTan[1 + ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/(b + 2*c*x)^(7/2)
+ ((5*I)*(b^2 - 4*a*c)^(1/4)*ArcTanh[((1 + I)*(b^2 - 4*a*c)^(1/4)*Sqrt[b + 2*c*x])/(Sqrt[b^2 - 4*a*c] + I*(b
+ 2*c*x))])/(b + 2*c*x)^(7/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(311\) vs. \(2(126)=252\).
time = 0.73, size = 312, normalized size = 2.08

method	result	size
derivativedivides	\(16 c \,d^{3} \left (\sqrt {2 c d x +b d}-d^{2} \left (\frac {\left (-\frac {a c}{4}+\frac {b^{2}}{16}\right ) \sqrt {2 c d x +b d}}{a c \,d^{2}-\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}}+\frac {5 \left (a c -\frac {b^{2}}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\right )\)	\(312\)
default	\(16 c \,d^{3} \left (\sqrt {2 c d x +b d}-d^{2} \left (\frac {\left (-\frac {a c}{4}+\frac {b^{2}}{16}\right ) \sqrt {2 c d x +b d}}{a c \,d^{2}-\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}}+\frac {5 \left (a c -\frac {b^{2}}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\right )\)	\(312\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

16*c*d^3*((2*c*d*x+b*d)^(1/2)-d^2*((-1/4*a*c+1/16*b^2)*(2*c*d*x+b*d)^(1/2)/(a*c*d^2-1/4*b^2*d^2+1/4*(2*c*d*x+b
*d)^2)+5/8*(a*c-1/4*b^2)/(4*a*c*d^2-b^2*d^2)^(3/4)*2^(1/2)*(ln((2*c*d*x+b*d+(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x
+b*d)^(1/2)*2^(1/2)+(4*a*c*d^2-b^2*d^2)^(1/2))/(2*c*d*x+b*d-(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)*2^(1
/2)+(4*a*c*d^2-b^2*d^2)^(1/2)))+2*arctan(2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)-2*arctan(-2^
(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1))))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^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(4*a*c-b^2>0)', see `assume?` f
or more deta

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 359 vs. \(2 (126) = 252\).
time = 3.05, size = 359, normalized size = 2.39 \begin {gather*} -\frac {20 \, \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \arctan \left (\frac {\left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac {3}{4}} \sqrt {2 \, c d x + b d} c d^{3} - \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac {3}{4}} \sqrt {2 \, c^{3} d^{7} x + b c^{2} d^{7} + \sqrt {{\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}}}}{{\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}}\right ) + 5 \, \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \log \left (5 \, \sqrt {2 \, c d x + b d} c d^{3} + 5 \, \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac {1}{4}}\right ) - 5 \, \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \log \left (5 \, \sqrt {2 \, c d x + b d} c d^{3} - 5 \, \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac {1}{4}}\right ) - {\left (16 \, c^{2} d^{3} x^{2} + 16 \, b c d^{3} x - {\left (b^{2} - 20 \, a c\right )} d^{3}\right )} \sqrt {2 \, c d x + b d}}{c x^{2} + b x + a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(20*((b^2*c^4 - 4*a*c^5)*d^14)^(1/4)*(c*x^2 + b*x + a)*arctan((((b^2*c^4 - 4*a*c^5)*d^14)^(3/4)*sqrt(2*c*d*x
+ b*d)*c*d^3 - ((b^2*c^4 - 4*a*c^5)*d^14)^(3/4)*sqrt(2*c^3*d^7*x + b*c^2*d^7 + sqrt((b^2*c^4 - 4*a*c^5)*d^14))
)/((b^2*c^4 - 4*a*c^5)*d^14)) + 5*((b^2*c^4 - 4*a*c^5)*d^14)^(1/4)*(c*x^2 + b*x + a)*log(5*sqrt(2*c*d*x + b*d)
*c*d^3 + 5*((b^2*c^4 - 4*a*c^5)*d^14)^(1/4)) - 5*((b^2*c^4 - 4*a*c^5)*d^14)^(1/4)*(c*x^2 + b*x + a)*log(5*sqrt
(2*c*d*x + b*d)*c*d^3 - 5*((b^2*c^4 - 4*a*c^5)*d^14)^(1/4)) - (16*c^2*d^3*x^2 + 16*b*c*d^3*x - (b^2 - 20*a*c)*
d^3)*sqrt(2*c*d*x + b*d))/(c*x^2 + b*x + a)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 441 vs. \(2 (126) = 252\).
time = 2.66, size = 441, normalized size = 2.94 \begin {gather*} -5 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c d^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) - 5 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c d^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) - \frac {5}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c d^{3} \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) + \frac {5}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c d^{3} \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) + 16 \, \sqrt {2 \, c d x + b d} c d^{3} + \frac {4 \, {\left (\sqrt {2 \, c d x + b d} b^{2} c d^{5} - 4 \, \sqrt {2 \, c d x + b d} a c^{2} d^{5}\right )}}{b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c*d^3*arctan(1/2*sqrt(2)*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4) + 2*sqr
t(2*c*d*x + b*d))/(-b^2*d^2 + 4*a*c*d^2)^(1/4)) - 5*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c*d^3*arctan(-1/2*sqr
t(2)*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4) - 2*sqrt(2*c*d*x + b*d))/(-b^2*d^2 + 4*a*c*d^2)^(1/4)) - 5/2*sqrt(2
)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c*d^3*log(2*c*d*x + b*d + sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*sqrt(2*c*d*x + b
*d) + sqrt(-b^2*d^2 + 4*a*c*d^2)) + 5/2*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c*d^3*log(2*c*d*x + b*d - sqrt(2)
*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*sqrt(2*c*d*x + b*d) + sqrt(-b^2*d^2 + 4*a*c*d^2)) + 16*sqrt(2*c*d*x + b*d)*c*d^3
+ 4*(sqrt(2*c*d*x + b*d)*b^2*c*d^5 - 4*sqrt(2*c*d*x + b*d)*a*c^2*d^5)/(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^
2)

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Mupad [B]
time = 0.62, size = 597, normalized size = 3.98 \begin {gather*} 16\,c\,d^3\,\sqrt {b\,d+2\,c\,d\,x}+\frac {\sqrt {b\,d+2\,c\,d\,x}\,\left (16\,a\,c^2\,d^5-4\,b^2\,c\,d^5\right )}{{\left (b\,d+2\,c\,d\,x\right )}^2-b^2\,d^2+4\,a\,c\,d^2}-10\,c\,d^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}-c\,d^{7/2}\,\mathrm {atan}\left (\frac {c\,d^{7/2}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (25600\,a^2\,c^4\,d^{10}-12800\,a\,b^2\,c^3\,d^{10}+1600\,b^4\,c^2\,d^{10}\right )-5\,c\,d^{7/2}\,{\left (b^2-4\,a\,c\right )}^{1/4}\,\left (5120\,a^2\,c^3\,d^7-2560\,a\,b^2\,c^2\,d^7+320\,b^4\,c\,d^7\right )\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}\,5{}\mathrm {i}+c\,d^{7/2}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (25600\,a^2\,c^4\,d^{10}-12800\,a\,b^2\,c^3\,d^{10}+1600\,b^4\,c^2\,d^{10}\right )+5\,c\,d^{7/2}\,{\left (b^2-4\,a\,c\right )}^{1/4}\,\left (5120\,a^2\,c^3\,d^7-2560\,a\,b^2\,c^2\,d^7+320\,b^4\,c\,d^7\right )\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}\,5{}\mathrm {i}}{5\,c\,d^{7/2}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (25600\,a^2\,c^4\,d^{10}-12800\,a\,b^2\,c^3\,d^{10}+1600\,b^4\,c^2\,d^{10}\right )-5\,c\,d^{7/2}\,{\left (b^2-4\,a\,c\right )}^{1/4}\,\left (5120\,a^2\,c^3\,d^7-2560\,a\,b^2\,c^2\,d^7+320\,b^4\,c\,d^7\right )\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}-5\,c\,d^{7/2}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (25600\,a^2\,c^4\,d^{10}-12800\,a\,b^2\,c^3\,d^{10}+1600\,b^4\,c^2\,d^{10}\right )+5\,c\,d^{7/2}\,{\left (b^2-4\,a\,c\right )}^{1/4}\,\left (5120\,a^2\,c^3\,d^7-2560\,a\,b^2\,c^2\,d^7+320\,b^4\,c\,d^7\right )\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}\,10{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16*c*d^3*(b*d + 2*c*d*x)^(1/2) + ((b*d + 2*c*d*x)^(1/2)*(16*a*c^2*d^5 - 4*b^2*c*d^5))/((b*d + 2*c*d*x)^2 - b^2
*d^2 + 4*a*c*d^2) - c*d^(7/2)*atan((c*d^(7/2)*((b*d + 2*c*d*x)^(1/2)*(25600*a^2*c^4*d^10 + 1600*b^4*c^2*d^10 -
12800*a*b^2*c^3*d^10) - 5*c*d^(7/2)*(b^2 - 4*a*c)^(1/4)*(320*b^4*c*d^7 + 5120*a^2*c^3*d^7 - 2560*a*b^2*c^2*d^
7))*(b^2 - 4*a*c)^(1/4)*5i + c*d^(7/2)*((b*d + 2*c*d*x)^(1/2)*(25600*a^2*c^4*d^10 + 1600*b^4*c^2*d^10 - 12800*
a*b^2*c^3*d^10) + 5*c*d^(7/2)*(b^2 - 4*a*c)^(1/4)*(320*b^4*c*d^7 + 5120*a^2*c^3*d^7 - 2560*a*b^2*c^2*d^7))*(b^
2 - 4*a*c)^(1/4)*5i)/(5*c*d^(7/2)*((b*d + 2*c*d*x)^(1/2)*(25600*a^2*c^4*d^10 + 1600*b^4*c^2*d^10 - 12800*a*b^2
*c^3*d^10) - 5*c*d^(7/2)*(b^2 - 4*a*c)^(1/4)*(320*b^4*c*d^7 + 5120*a^2*c^3*d^7 - 2560*a*b^2*c^2*d^7))*(b^2 - 4
*a*c)^(1/4) - 5*c*d^(7/2)*((b*d + 2*c*d*x)^(1/2)*(25600*a^2*c^4*d^10 + 1600*b^4*c^2*d^10 - 12800*a*b^2*c^3*d^1
0) + 5*c*d^(7/2)*(b^2 - 4*a*c)^(1/4)*(320*b^4*c*d^7 + 5120*a^2*c^3*d^7 - 2560*a*b^2*c^2*d^7))*(b^2 - 4*a*c)^(1
/4)))*(b^2 - 4*a*c)^(1/4)*10i - 10*c*d^(7/2)*atan((b*d + 2*c*d*x)^(1/2)/(d^(1/2)*(b^2 - 4*a*c)^(1/4)))*(b^2 -
4*a*c)^(1/4)

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