3.1774 \(\int \frac{(a+b x)^2 (e+f x)^{5/2}}{c+d x} \, dx\)

Optimal. Leaf size=208 \[ -\frac{2 b (e+f x)^{7/2} (-2 a d f+b c f+b d e)}{7 d^2 f^2}+\frac{2 (e+f x)^{5/2} (b c-a d)^2}{5 d^3}+\frac{2 (e+f x)^{3/2} (b c-a d)^2 (d e-c f)}{3 d^4}+\frac{2 \sqrt{e+f x} (b c-a d)^2 (d e-c f)^2}{d^5}-\frac{2 (b c-a d)^2 (d e-c f)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{d^{11/2}}+\frac{2 b^2 (e+f x)^{9/2}}{9 d f^2} \]

[Out]

(2*(b*c - a*d)^2*(d*e - c*f)^2*Sqrt[e + f*x])/d^5 + (2*(b*c - a*d)^2*(d*e - c*f)*(e + f*x)^(3/2))/(3*d^4) + (2
*(b*c - a*d)^2*(e + f*x)^(5/2))/(5*d^3) - (2*b*(b*d*e + b*c*f - 2*a*d*f)*(e + f*x)^(7/2))/(7*d^2*f^2) + (2*b^2
*(e + f*x)^(9/2))/(9*d*f^2) - (2*(b*c - a*d)^2*(d*e - c*f)^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*
f]])/d^(11/2)

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Rubi [A]  time = 0.215273, antiderivative size = 208, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {88, 50, 63, 208} \[ -\frac{2 b (e+f x)^{7/2} (-2 a d f+b c f+b d e)}{7 d^2 f^2}+\frac{2 (e+f x)^{5/2} (b c-a d)^2}{5 d^3}+\frac{2 (e+f x)^{3/2} (b c-a d)^2 (d e-c f)}{3 d^4}+\frac{2 \sqrt{e+f x} (b c-a d)^2 (d e-c f)^2}{d^5}-\frac{2 (b c-a d)^2 (d e-c f)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{d^{11/2}}+\frac{2 b^2 (e+f x)^{9/2}}{9 d f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*c - a*d)^2*(d*e - c*f)^2*Sqrt[e + f*x])/d^5 + (2*(b*c - a*d)^2*(d*e - c*f)*(e + f*x)^(3/2))/(3*d^4) + (2
*(b*c - a*d)^2*(e + f*x)^(5/2))/(5*d^3) - (2*b*(b*d*e + b*c*f - 2*a*d*f)*(e + f*x)^(7/2))/(7*d^2*f^2) + (2*b^2
*(e + f*x)^(9/2))/(9*d*f^2) - (2*(b*c - a*d)^2*(d*e - c*f)^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*
f]])/d^(11/2)

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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 208

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

Rubi steps

\begin{align*} \int \frac{(a+b x)^2 (e+f x)^{5/2}}{c+d x} \, dx &=\int \left (-\frac{b (b d e+b c f-2 a d f) (e+f x)^{5/2}}{d^2 f}+\frac{(-b c+a d)^2 (e+f x)^{5/2}}{d^2 (c+d x)}+\frac{b^2 (e+f x)^{7/2}}{d f}\right ) \, dx\\ &=-\frac{2 b (b d e+b c f-2 a d f) (e+f x)^{7/2}}{7 d^2 f^2}+\frac{2 b^2 (e+f x)^{9/2}}{9 d f^2}+\frac{(b c-a d)^2 \int \frac{(e+f x)^{5/2}}{c+d x} \, dx}{d^2}\\ &=\frac{2 (b c-a d)^2 (e+f x)^{5/2}}{5 d^3}-\frac{2 b (b d e+b c f-2 a d f) (e+f x)^{7/2}}{7 d^2 f^2}+\frac{2 b^2 (e+f x)^{9/2}}{9 d f^2}+\frac{\left ((b c-a d)^2 (d e-c f)\right ) \int \frac{(e+f x)^{3/2}}{c+d x} \, dx}{d^3}\\ &=\frac{2 (b c-a d)^2 (d e-c f) (e+f x)^{3/2}}{3 d^4}+\frac{2 (b c-a d)^2 (e+f x)^{5/2}}{5 d^3}-\frac{2 b (b d e+b c f-2 a d f) (e+f x)^{7/2}}{7 d^2 f^2}+\frac{2 b^2 (e+f x)^{9/2}}{9 d f^2}+\frac{\left ((b c-a d)^2 (d e-c f)^2\right ) \int \frac{\sqrt{e+f x}}{c+d x} \, dx}{d^4}\\ &=\frac{2 (b c-a d)^2 (d e-c f)^2 \sqrt{e+f x}}{d^5}+\frac{2 (b c-a d)^2 (d e-c f) (e+f x)^{3/2}}{3 d^4}+\frac{2 (b c-a d)^2 (e+f x)^{5/2}}{5 d^3}-\frac{2 b (b d e+b c f-2 a d f) (e+f x)^{7/2}}{7 d^2 f^2}+\frac{2 b^2 (e+f x)^{9/2}}{9 d f^2}+\frac{\left ((b c-a d)^2 (d e-c f)^3\right ) \int \frac{1}{(c+d x) \sqrt{e+f x}} \, dx}{d^5}\\ &=\frac{2 (b c-a d)^2 (d e-c f)^2 \sqrt{e+f x}}{d^5}+\frac{2 (b c-a d)^2 (d e-c f) (e+f x)^{3/2}}{3 d^4}+\frac{2 (b c-a d)^2 (e+f x)^{5/2}}{5 d^3}-\frac{2 b (b d e+b c f-2 a d f) (e+f x)^{7/2}}{7 d^2 f^2}+\frac{2 b^2 (e+f x)^{9/2}}{9 d f^2}+\frac{\left (2 (b c-a d)^2 (d e-c f)^3\right ) \operatorname{Subst}\left (\int \frac{1}{c-\frac{d e}{f}+\frac{d x^2}{f}} \, dx,x,\sqrt{e+f x}\right )}{d^5 f}\\ &=\frac{2 (b c-a d)^2 (d e-c f)^2 \sqrt{e+f x}}{d^5}+\frac{2 (b c-a d)^2 (d e-c f) (e+f x)^{3/2}}{3 d^4}+\frac{2 (b c-a d)^2 (e+f x)^{5/2}}{5 d^3}-\frac{2 b (b d e+b c f-2 a d f) (e+f x)^{7/2}}{7 d^2 f^2}+\frac{2 b^2 (e+f x)^{9/2}}{9 d f^2}-\frac{2 (b c-a d)^2 (d e-c f)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{d^{11/2}}\\ \end{align*}

Mathematica [A]  time = 0.454719, size = 175, normalized size = 0.84 \[ \frac{2 \left (\frac{105 (b c-a d)^2 (d e-c f) \left (\sqrt{d} \sqrt{e+f x} (-3 c f+4 d e+d f x)-3 (d e-c f)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )\right )}{d^{5/2}}-\frac{45 b d (e+f x)^{7/2} (-2 a d f+b c f+b d e)}{f^2}+63 (e+f x)^{5/2} (b c-a d)^2+\frac{35 b^2 d^2 (e+f x)^{9/2}}{f^2}\right )}{315 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(63*(b*c - a*d)^2*(e + f*x)^(5/2) - (45*b*d*(b*d*e + b*c*f - 2*a*d*f)*(e + f*x)^(7/2))/f^2 + (35*b^2*d^2*(e
+ f*x)^(9/2))/f^2 + (105*(b*c - a*d)^2*(d*e - c*f)*(Sqrt[d]*Sqrt[e + f*x]*(4*d*e - 3*c*f + d*f*x) - 3*(d*e -
c*f)^(3/2)*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]]))/d^(5/2)))/(315*d^3)

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Maple [B]  time = 0.013, size = 972, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2*(f*x+e)^(5/2)/(d*x+c),x)

[Out]

-2/3*f/d^4*(f*x+e)^(3/2)*b^2*c^3-4/5/d^2*(f*x+e)^(5/2)*a*b*c+2/3/d^3*(f*x+e)^(3/2)*b^2*c^2*e+2/d^3*b^2*c^2*e^2
*(f*x+e)^(1/2)+4/7/f/d*(f*x+e)^(7/2)*a*b+4/3*f/d^3*(f*x+e)^(3/2)*a*b*c^2-4*f/d^2*a^2*c*e*(f*x+e)^(1/2)+2*f^2/d
^3*a^2*c^2*(f*x+e)^(1/2)+2/d*a^2*e^2*(f*x+e)^(1/2)+2/5/d^3*(f*x+e)^(5/2)*b^2*c^2+2/((c*f-d*e)*d)^(1/2)*arctan(
(f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a^2*e^3+2/3/d*(f*x+e)^(3/2)*a^2*e-4*f^2/d^4*a*b*c^3*(f*x+e)^(1/2)-4*f/d^4
*b^2*c^3*e*(f*x+e)^(1/2)-2*f^3/d^3/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a^2*c^3-2*f
^3/d^5/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*b^2*c^5-4/d^2*a*b*c*e^2*(f*x+e)^(1/2)+2
/d^2/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*b^2*c^2*e^3-4/3/d^2*(f*x+e)^(3/2)*a*b*c*e
+2/5/d*(f*x+e)^(5/2)*a^2+8*f/d^3*a*b*c^2*e*(f*x+e)^(1/2)+6*f^2/d^4/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/
((c*f-d*e)*d)^(1/2))*b^2*c^4*e-6*f/d^3/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*b^2*c^3
*e^2-4/d/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a*b*c*e^3+6*f^2/d^2/((c*f-d*e)*d)^(1/
2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a^2*c^2*e-6*f/d/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*
f-d*e)*d)^(1/2))*a^2*c*e^2-12*f^2/d^3/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a*b*c^3*
e+12*f/d^2/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a*b*c^2*e^2+4*f^3/d^4/((c*f-d*e)*d)
^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a*b*c^4-2/7/f/d^2*(f*x+e)^(7/2)*b^2*c-2/7/f^2/d*(f*x+e)^(7/
2)*b^2*e+2*f^2/d^5*b^2*c^4*(f*x+e)^(1/2)-2/3*f/d^2*(f*x+e)^(3/2)*a^2*c+2/9*b^2*(f*x+e)^(9/2)/d/f^2

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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((b*x+a)^2*(f*x+e)^(5/2)/(d*x+c),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.57661, size = 2234, normalized size = 10.74 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/315*(315*((b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*e^2*f^2 - 2*(b^2*c^3*d - 2*a*b*c^2*d^2 + a^2*c*d^3)*e*f^3 +
(b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2)*f^4)*sqrt((d*e - c*f)/d)*log((d*f*x + 2*d*e - c*f - 2*sqrt(f*x + e)*d*s
qrt((d*e - c*f)/d))/(d*x + c)) + 2*(35*b^2*d^4*f^4*x^4 - 10*b^2*d^4*e^4 - 45*(b^2*c*d^3 - 2*a*b*d^4)*e^3*f + 4
83*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*e^2*f^2 - 735*(b^2*c^3*d - 2*a*b*c^2*d^2 + a^2*c*d^3)*e*f^3 + 315*(b^
2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2)*f^4 + 5*(19*b^2*d^4*e*f^3 - 9*(b^2*c*d^3 - 2*a*b*d^4)*f^4)*x^3 + 3*(25*b^2*
d^4*e^2*f^2 - 45*(b^2*c*d^3 - 2*a*b*d^4)*e*f^3 + 21*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*f^4)*x^2 + (5*b^2*d^
4*e^3*f - 135*(b^2*c*d^3 - 2*a*b*d^4)*e^2*f^2 + 231*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*e*f^3 - 105*(b^2*c^3
*d - 2*a*b*c^2*d^2 + a^2*c*d^3)*f^4)*x)*sqrt(f*x + e))/(d^5*f^2), -2/315*(315*((b^2*c^2*d^2 - 2*a*b*c*d^3 + a^
2*d^4)*e^2*f^2 - 2*(b^2*c^3*d - 2*a*b*c^2*d^2 + a^2*c*d^3)*e*f^3 + (b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2)*f^4)*
sqrt(-(d*e - c*f)/d)*arctan(-sqrt(f*x + e)*d*sqrt(-(d*e - c*f)/d)/(d*e - c*f)) - (35*b^2*d^4*f^4*x^4 - 10*b^2*
d^4*e^4 - 45*(b^2*c*d^3 - 2*a*b*d^4)*e^3*f + 483*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*e^2*f^2 - 735*(b^2*c^3*
d - 2*a*b*c^2*d^2 + a^2*c*d^3)*e*f^3 + 315*(b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2)*f^4 + 5*(19*b^2*d^4*e*f^3 - 9
*(b^2*c*d^3 - 2*a*b*d^4)*f^4)*x^3 + 3*(25*b^2*d^4*e^2*f^2 - 45*(b^2*c*d^3 - 2*a*b*d^4)*e*f^3 + 21*(b^2*c^2*d^2
- 2*a*b*c*d^3 + a^2*d^4)*f^4)*x^2 + (5*b^2*d^4*e^3*f - 135*(b^2*c*d^3 - 2*a*b*d^4)*e^2*f^2 + 231*(b^2*c^2*d^2
- 2*a*b*c*d^3 + a^2*d^4)*e*f^3 - 105*(b^2*c^3*d - 2*a*b*c^2*d^2 + a^2*c*d^3)*f^4)*x)*sqrt(f*x + e))/(d^5*f^2)
]

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Sympy [A]  time = 91.0127, size = 374, normalized size = 1.8 \begin{align*} \frac{2 b^{2} \left (e + f x\right )^{\frac{9}{2}}}{9 d f^{2}} + \frac{\left (e + f x\right )^{\frac{7}{2}} \left (4 a b d f - 2 b^{2} c f - 2 b^{2} d e\right )}{7 d^{2} f^{2}} + \frac{\left (e + f x\right )^{\frac{5}{2}} \left (2 a^{2} d^{2} - 4 a b c d + 2 b^{2} c^{2}\right )}{5 d^{3}} + \frac{\left (e + f x\right )^{\frac{3}{2}} \left (- 2 a^{2} c d^{2} f + 2 a^{2} d^{3} e + 4 a b c^{2} d f - 4 a b c d^{2} e - 2 b^{2} c^{3} f + 2 b^{2} c^{2} d e\right )}{3 d^{4}} + \frac{\sqrt{e + f x} \left (2 a^{2} c^{2} d^{2} f^{2} - 4 a^{2} c d^{3} e f + 2 a^{2} d^{4} e^{2} - 4 a b c^{3} d f^{2} + 8 a b c^{2} d^{2} e f - 4 a b c d^{3} e^{2} + 2 b^{2} c^{4} f^{2} - 4 b^{2} c^{3} d e f + 2 b^{2} c^{2} d^{2} e^{2}\right )}{d^{5}} - \frac{2 \left (a d - b c\right )^{2} \left (c f - d e\right )^{3} \operatorname{atan}{\left (\frac{\sqrt{e + f x}}{\sqrt{\frac{c f - d e}{d}}} \right )}}{d^{6} \sqrt{\frac{c f - d e}{d}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**2*(f*x+e)**(5/2)/(d*x+c),x)

[Out]

2*b**2*(e + f*x)**(9/2)/(9*d*f**2) + (e + f*x)**(7/2)*(4*a*b*d*f - 2*b**2*c*f - 2*b**2*d*e)/(7*d**2*f**2) + (e
+ f*x)**(5/2)*(2*a**2*d**2 - 4*a*b*c*d + 2*b**2*c**2)/(5*d**3) + (e + f*x)**(3/2)*(-2*a**2*c*d**2*f + 2*a**2*
d**3*e + 4*a*b*c**2*d*f - 4*a*b*c*d**2*e - 2*b**2*c**3*f + 2*b**2*c**2*d*e)/(3*d**4) + sqrt(e + f*x)*(2*a**2*c
**2*d**2*f**2 - 4*a**2*c*d**3*e*f + 2*a**2*d**4*e**2 - 4*a*b*c**3*d*f**2 + 8*a*b*c**2*d**2*e*f - 4*a*b*c*d**3*
e**2 + 2*b**2*c**4*f**2 - 4*b**2*c**3*d*e*f + 2*b**2*c**2*d**2*e**2)/d**5 - 2*(a*d - b*c)**2*(c*f - d*e)**3*at
an(sqrt(e + f*x)/sqrt((c*f - d*e)/d))/(d**6*sqrt((c*f - d*e)/d))

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Giac [B]  time = 2.47214, size = 909, normalized size = 4.37 \begin{align*} -\frac{2 \,{\left (b^{2} c^{5} f^{3} - 2 \, a b c^{4} d f^{3} + a^{2} c^{3} d^{2} f^{3} - 3 \, b^{2} c^{4} d f^{2} e + 6 \, a b c^{3} d^{2} f^{2} e - 3 \, a^{2} c^{2} d^{3} f^{2} e + 3 \, b^{2} c^{3} d^{2} f e^{2} - 6 \, a b c^{2} d^{3} f e^{2} + 3 \, a^{2} c d^{4} f e^{2} - b^{2} c^{2} d^{3} e^{3} + 2 \, a b c d^{4} e^{3} - a^{2} d^{5} e^{3}\right )} \arctan \left (\frac{\sqrt{f x + e} d}{\sqrt{c d f - d^{2} e}}\right )}{\sqrt{c d f - d^{2} e} d^{5}} + \frac{2 \,{\left (35 \,{\left (f x + e\right )}^{\frac{9}{2}} b^{2} d^{8} f^{16} - 45 \,{\left (f x + e\right )}^{\frac{7}{2}} b^{2} c d^{7} f^{17} + 90 \,{\left (f x + e\right )}^{\frac{7}{2}} a b d^{8} f^{17} + 63 \,{\left (f x + e\right )}^{\frac{5}{2}} b^{2} c^{2} d^{6} f^{18} - 126 \,{\left (f x + e\right )}^{\frac{5}{2}} a b c d^{7} f^{18} + 63 \,{\left (f x + e\right )}^{\frac{5}{2}} a^{2} d^{8} f^{18} - 105 \,{\left (f x + e\right )}^{\frac{3}{2}} b^{2} c^{3} d^{5} f^{19} + 210 \,{\left (f x + e\right )}^{\frac{3}{2}} a b c^{2} d^{6} f^{19} - 105 \,{\left (f x + e\right )}^{\frac{3}{2}} a^{2} c d^{7} f^{19} + 315 \, \sqrt{f x + e} b^{2} c^{4} d^{4} f^{20} - 630 \, \sqrt{f x + e} a b c^{3} d^{5} f^{20} + 315 \, \sqrt{f x + e} a^{2} c^{2} d^{6} f^{20} - 45 \,{\left (f x + e\right )}^{\frac{7}{2}} b^{2} d^{8} f^{16} e + 105 \,{\left (f x + e\right )}^{\frac{3}{2}} b^{2} c^{2} d^{6} f^{18} e - 210 \,{\left (f x + e\right )}^{\frac{3}{2}} a b c d^{7} f^{18} e + 105 \,{\left (f x + e\right )}^{\frac{3}{2}} a^{2} d^{8} f^{18} e - 630 \, \sqrt{f x + e} b^{2} c^{3} d^{5} f^{19} e + 1260 \, \sqrt{f x + e} a b c^{2} d^{6} f^{19} e - 630 \, \sqrt{f x + e} a^{2} c d^{7} f^{19} e + 315 \, \sqrt{f x + e} b^{2} c^{2} d^{6} f^{18} e^{2} - 630 \, \sqrt{f x + e} a b c d^{7} f^{18} e^{2} + 315 \, \sqrt{f x + e} a^{2} d^{8} f^{18} e^{2}\right )}}{315 \, d^{9} f^{18}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(b^2*c^5*f^3 - 2*a*b*c^4*d*f^3 + a^2*c^3*d^2*f^3 - 3*b^2*c^4*d*f^2*e + 6*a*b*c^3*d^2*f^2*e - 3*a^2*c^2*d^3*
f^2*e + 3*b^2*c^3*d^2*f*e^2 - 6*a*b*c^2*d^3*f*e^2 + 3*a^2*c*d^4*f*e^2 - b^2*c^2*d^3*e^3 + 2*a*b*c*d^4*e^3 - a^
2*d^5*e^3)*arctan(sqrt(f*x + e)*d/sqrt(c*d*f - d^2*e))/(sqrt(c*d*f - d^2*e)*d^5) + 2/315*(35*(f*x + e)^(9/2)*b
^2*d^8*f^16 - 45*(f*x + e)^(7/2)*b^2*c*d^7*f^17 + 90*(f*x + e)^(7/2)*a*b*d^8*f^17 + 63*(f*x + e)^(5/2)*b^2*c^2
*d^6*f^18 - 126*(f*x + e)^(5/2)*a*b*c*d^7*f^18 + 63*(f*x + e)^(5/2)*a^2*d^8*f^18 - 105*(f*x + e)^(3/2)*b^2*c^3
*d^5*f^19 + 210*(f*x + e)^(3/2)*a*b*c^2*d^6*f^19 - 105*(f*x + e)^(3/2)*a^2*c*d^7*f^19 + 315*sqrt(f*x + e)*b^2*
c^4*d^4*f^20 - 630*sqrt(f*x + e)*a*b*c^3*d^5*f^20 + 315*sqrt(f*x + e)*a^2*c^2*d^6*f^20 - 45*(f*x + e)^(7/2)*b^
2*d^8*f^16*e + 105*(f*x + e)^(3/2)*b^2*c^2*d^6*f^18*e - 210*(f*x + e)^(3/2)*a*b*c*d^7*f^18*e + 105*(f*x + e)^(
3/2)*a^2*d^8*f^18*e - 630*sqrt(f*x + e)*b^2*c^3*d^5*f^19*e + 1260*sqrt(f*x + e)*a*b*c^2*d^6*f^19*e - 630*sqrt(
f*x + e)*a^2*c*d^7*f^19*e + 315*sqrt(f*x + e)*b^2*c^2*d^6*f^18*e^2 - 630*sqrt(f*x + e)*a*b*c*d^7*f^18*e^2 + 31
5*sqrt(f*x + e)*a^2*d^8*f^18*e^2)/(d^9*f^18)