3.4.39 \(\int \frac {(d+e x)^{7/2}}{b x+c x^2} \, dx\)

Optimal. Leaf size=157 \[ \frac {2 e \sqrt {d+e x} \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )}{c^3}+\frac {2 (c d-b e)^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b c^{7/2}}+\frac {2 e (d+e x)^{3/2} (2 c d-b e)}{3 c^2}-\frac {2 d^{7/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b}+\frac {2 e (d+e x)^{5/2}}{5 c} \]

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Rubi [A]  time = 0.39, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {703, 824, 826, 1166, 208} \begin {gather*} \frac {2 e \sqrt {d+e x} \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )}{c^3}+\frac {2 e (d+e x)^{3/2} (2 c d-b e)}{3 c^2}+\frac {2 (c d-b e)^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b c^{7/2}}-\frac {2 d^{7/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b}+\frac {2 e (d+e x)^{5/2}}{5 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(7/2)/(b*x + c*x^2),x]

[Out]

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

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]

Rule 703

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

Rule 824

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

Rule 826

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

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 138, normalized size = 0.88 \begin {gather*} \frac {2 e \sqrt {d+e x} \left (15 b^2 e^2-5 b c e (10 d+e x)+c^2 \left (58 d^2+16 d e x+3 e^2 x^2\right )\right )}{15 c^3}+\frac {2 (c d-b e)^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b c^{7/2}}-\frac {2 d^{7/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*Sqrt[d + e*x]*(15*b^2*e^2 - 5*b*c*e*(10*d + e*x) + c^2*(58*d^2 + 16*d*e*x + 3*e^2*x^2)))/(15*c^3) - (2*d^
(7/2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/b + (2*(c*d - b*e)^(7/2)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]
])/(b*c^(7/2))

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IntegrateAlgebraic [A]  time = 0.27, size = 160, normalized size = 1.02 \begin {gather*} \frac {2 e \sqrt {d+e x} \left (15 b^2 e^2-5 b c e (d+e x)-45 b c d e+45 c^2 d^2+3 c^2 (d+e x)^2+10 c^2 d (d+e x)\right )}{15 c^3}+\frac {2 (b e-c d)^{7/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x} \sqrt {b e-c d}}{c d-b e}\right )}{b c^{7/2}}-\frac {2 d^{7/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)^(7/2)/(b*x + c*x^2),x]

[Out]

(2*e*Sqrt[d + e*x]*(45*c^2*d^2 - 45*b*c*d*e + 15*b^2*e^2 + 10*c^2*d*(d + e*x) - 5*b*c*e*(d + e*x) + 3*c^2*(d +
e*x)^2))/(15*c^3) + (2*(-(c*d) + b*e)^(7/2)*ArcTan[(Sqrt[c]*Sqrt[-(c*d) + b*e]*Sqrt[d + e*x])/(c*d - b*e)])/(
b*c^(7/2)) - (2*d^(7/2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/b

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fricas [A]  time = 1.25, size = 822, normalized size = 5.24 \begin {gather*} \left [\frac {15 \, c^{3} d^{\frac {7}{2}} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) - 15 \, {\left (c^{3} d^{3} - 3 \, b c^{2} d^{2} e + 3 \, b^{2} c d e^{2} - b^{3} e^{3}\right )} \sqrt {\frac {c d - b e}{c}} \log \left (\frac {c e x + 2 \, c d - b e - 2 \, \sqrt {e x + d} c \sqrt {\frac {c d - b e}{c}}}{c x + b}\right ) + 2 \, {\left (3 \, b c^{2} e^{3} x^{2} + 58 \, b c^{2} d^{2} e - 50 \, b^{2} c d e^{2} + 15 \, b^{3} e^{3} + {\left (16 \, b c^{2} d e^{2} - 5 \, b^{2} c e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, b c^{3}}, \frac {15 \, c^{3} d^{\frac {7}{2}} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + 30 \, {\left (c^{3} d^{3} - 3 \, b c^{2} d^{2} e + 3 \, b^{2} c d e^{2} - b^{3} e^{3}\right )} \sqrt {-\frac {c d - b e}{c}} \arctan \left (-\frac {\sqrt {e x + d} c \sqrt {-\frac {c d - b e}{c}}}{c d - b e}\right ) + 2 \, {\left (3 \, b c^{2} e^{3} x^{2} + 58 \, b c^{2} d^{2} e - 50 \, b^{2} c d e^{2} + 15 \, b^{3} e^{3} + {\left (16 \, b c^{2} d e^{2} - 5 \, b^{2} c e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, b c^{3}}, \frac {30 \, c^{3} \sqrt {-d} d^{3} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) - 15 \, {\left (c^{3} d^{3} - 3 \, b c^{2} d^{2} e + 3 \, b^{2} c d e^{2} - b^{3} e^{3}\right )} \sqrt {\frac {c d - b e}{c}} \log \left (\frac {c e x + 2 \, c d - b e - 2 \, \sqrt {e x + d} c \sqrt {\frac {c d - b e}{c}}}{c x + b}\right ) + 2 \, {\left (3 \, b c^{2} e^{3} x^{2} + 58 \, b c^{2} d^{2} e - 50 \, b^{2} c d e^{2} + 15 \, b^{3} e^{3} + {\left (16 \, b c^{2} d e^{2} - 5 \, b^{2} c e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, b c^{3}}, \frac {2 \, {\left (15 \, c^{3} \sqrt {-d} d^{3} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + 15 \, {\left (c^{3} d^{3} - 3 \, b c^{2} d^{2} e + 3 \, b^{2} c d e^{2} - b^{3} e^{3}\right )} \sqrt {-\frac {c d - b e}{c}} \arctan \left (-\frac {\sqrt {e x + d} c \sqrt {-\frac {c d - b e}{c}}}{c d - b e}\right ) + {\left (3 \, b c^{2} e^{3} x^{2} + 58 \, b c^{2} d^{2} e - 50 \, b^{2} c d e^{2} + 15 \, b^{3} e^{3} + {\left (16 \, b c^{2} d e^{2} - 5 \, b^{2} c e^{3}\right )} x\right )} \sqrt {e x + d}\right )}}{15 \, b c^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/15*(15*c^3*d^(7/2)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - 15*(c^3*d^3 - 3*b*c^2*d^2*e + 3*b^2*c*d*e
^2 - b^3*e^3)*sqrt((c*d - b*e)/c)*log((c*e*x + 2*c*d - b*e - 2*sqrt(e*x + d)*c*sqrt((c*d - b*e)/c))/(c*x + b))
+ 2*(3*b*c^2*e^3*x^2 + 58*b*c^2*d^2*e - 50*b^2*c*d*e^2 + 15*b^3*e^3 + (16*b*c^2*d*e^2 - 5*b^2*c*e^3)*x)*sqrt(
e*x + d))/(b*c^3), 1/15*(15*c^3*d^(7/2)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 30*(c^3*d^3 - 3*b*c^2*d
^2*e + 3*b^2*c*d*e^2 - b^3*e^3)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e))
+ 2*(3*b*c^2*e^3*x^2 + 58*b*c^2*d^2*e - 50*b^2*c*d*e^2 + 15*b^3*e^3 + (16*b*c^2*d*e^2 - 5*b^2*c*e^3)*x)*sqrt(
e*x + d))/(b*c^3), 1/15*(30*c^3*sqrt(-d)*d^3*arctan(sqrt(e*x + d)*sqrt(-d)/d) - 15*(c^3*d^3 - 3*b*c^2*d^2*e +
3*b^2*c*d*e^2 - b^3*e^3)*sqrt((c*d - b*e)/c)*log((c*e*x + 2*c*d - b*e - 2*sqrt(e*x + d)*c*sqrt((c*d - b*e)/c))
/(c*x + b)) + 2*(3*b*c^2*e^3*x^2 + 58*b*c^2*d^2*e - 50*b^2*c*d*e^2 + 15*b^3*e^3 + (16*b*c^2*d*e^2 - 5*b^2*c*e^
3)*x)*sqrt(e*x + d))/(b*c^3), 2/15*(15*c^3*sqrt(-d)*d^3*arctan(sqrt(e*x + d)*sqrt(-d)/d) + 15*(c^3*d^3 - 3*b*c
^2*d^2*e + 3*b^2*c*d*e^2 - b^3*e^3)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b
*e)) + (3*b*c^2*e^3*x^2 + 58*b*c^2*d^2*e - 50*b^2*c*d*e^2 + 15*b^3*e^3 + (16*b*c^2*d*e^2 - 5*b^2*c*e^3)*x)*sqr
t(e*x + d))/(b*c^3)]

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giac [A]  time = 0.23, size = 229, normalized size = 1.46 \begin {gather*} \frac {2 \, d^{4} \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-d}}\right )}{b \sqrt {-d}} - \frac {2 \, {\left (c^{4} d^{4} - 4 \, b c^{3} d^{3} e + 6 \, b^{2} c^{2} d^{2} e^{2} - 4 \, b^{3} c d e^{3} + b^{4} e^{4}\right )} \arctan \left (\frac {\sqrt {x e + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{\sqrt {-c^{2} d + b c e} b c^{3}} + \frac {2 \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} c^{4} e + 10 \, {\left (x e + d\right )}^{\frac {3}{2}} c^{4} d e + 45 \, \sqrt {x e + d} c^{4} d^{2} e - 5 \, {\left (x e + d\right )}^{\frac {3}{2}} b c^{3} e^{2} - 45 \, \sqrt {x e + d} b c^{3} d e^{2} + 15 \, \sqrt {x e + d} b^{2} c^{2} e^{3}\right )}}{15 \, c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*d^4*arctan(sqrt(x*e + d)/sqrt(-d))/(b*sqrt(-d)) - 2*(c^4*d^4 - 4*b*c^3*d^3*e + 6*b^2*c^2*d^2*e^2 - 4*b^3*c*d
*e^3 + b^4*e^4)*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c*e))/(sqrt(-c^2*d + b*c*e)*b*c^3) + 2/15*(3*(x*e + d)^
(5/2)*c^4*e + 10*(x*e + d)^(3/2)*c^4*d*e + 45*sqrt(x*e + d)*c^4*d^2*e - 5*(x*e + d)^(3/2)*b*c^3*e^2 - 45*sqrt(
x*e + d)*b*c^3*d*e^2 + 15*sqrt(x*e + d)*b^2*c^2*e^3)/c^5

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maple [B]  time = 0.08, size = 336, normalized size = 2.14 \begin {gather*} -\frac {2 b^{3} e^{4} \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}\, c^{3}}+\frac {8 b^{2} d \,e^{3} \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}\, c^{2}}-\frac {12 b \,d^{2} e^{2} \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}\, c}-\frac {2 c \,d^{4} \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}\, b}+\frac {8 d^{3} e \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}}-\frac {2 d^{\frac {7}{2}} \arctanh \left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b}+\frac {2 \sqrt {e x +d}\, b^{2} e^{3}}{c^{3}}-\frac {6 \sqrt {e x +d}\, b d \,e^{2}}{c^{2}}+\frac {6 \sqrt {e x +d}\, d^{2} e}{c}-\frac {2 \left (e x +d \right )^{\frac {3}{2}} b \,e^{2}}{3 c^{2}}+\frac {4 \left (e x +d \right )^{\frac {3}{2}} d e}{3 c}+\frac {2 \left (e x +d \right )^{\frac {5}{2}} e}{5 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((e*x+d)^(7/2)/(c*x^2+b*x),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(b*e-c*d>0)', see `assume?` for
more details)Is b*e-c*d positive or negative?

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mupad [B]  time = 0.59, size = 2482, normalized size = 15.81

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(7/2)/(b*x + c*x^2),x)

[Out]

((2*e*(b*e - 2*c*d)^2)/c^3 - (2*e*(c*d^2 - b*d*e))/c^2)*(d + e*x)^(1/2) + (2*e*(d + e*x)^(5/2))/(5*c) + (atan(
((((8*(d + e*x)^(1/2)*(b^8*e^10 + 2*c^8*d^8*e^2 - 8*b*c^7*d^7*e^3 + 28*b^2*c^6*d^6*e^4 - 56*b^3*c^5*d^5*e^5 +
70*b^4*c^4*d^4*e^6 - 56*b^5*c^3*d^3*e^7 + 28*b^6*c^2*d^2*e^8 - 8*b^7*c*d*e^9))/c^5 + (((8*(b^5*c^4*d*e^6 - 3*b
^2*c^7*d^4*e^3 + 6*b^3*c^6*d^3*e^4 - 4*b^4*c^5*d^2*e^5))/c^5 + (8*(b^3*c^7*e^3 - 2*b^2*c^8*d*e^2)*(d^7)^(1/2)*
(d + e*x)^(1/2))/(b*c^5))*(d^7)^(1/2))/b)*(d^7)^(1/2)*1i)/b + (((8*(d + e*x)^(1/2)*(b^8*e^10 + 2*c^8*d^8*e^2 -
8*b*c^7*d^7*e^3 + 28*b^2*c^6*d^6*e^4 - 56*b^3*c^5*d^5*e^5 + 70*b^4*c^4*d^4*e^6 - 56*b^5*c^3*d^3*e^7 + 28*b^6*
c^2*d^2*e^8 - 8*b^7*c*d*e^9))/c^5 - (((8*(b^5*c^4*d*e^6 - 3*b^2*c^7*d^4*e^3 + 6*b^3*c^6*d^3*e^4 - 4*b^4*c^5*d^
2*e^5))/c^5 - (8*(b^3*c^7*e^3 - 2*b^2*c^8*d*e^2)*(d^7)^(1/2)*(d + e*x)^(1/2))/(b*c^5))*(d^7)^(1/2))/b)*(d^7)^(
1/2)*1i)/b)/((16*(b^7*d^4*e^10 - 4*c^7*d^11*e^3 + 22*b*c^6*d^10*e^4 - 8*b^6*c*d^5*e^9 - 52*b^2*c^5*d^9*e^5 + 6
9*b^3*c^4*d^8*e^6 - 56*b^4*c^3*d^7*e^7 + 28*b^5*c^2*d^6*e^8))/c^5 + (((8*(d + e*x)^(1/2)*(b^8*e^10 + 2*c^8*d^8
*e^2 - 8*b*c^7*d^7*e^3 + 28*b^2*c^6*d^6*e^4 - 56*b^3*c^5*d^5*e^5 + 70*b^4*c^4*d^4*e^6 - 56*b^5*c^3*d^3*e^7 + 2
8*b^6*c^2*d^2*e^8 - 8*b^7*c*d*e^9))/c^5 + (((8*(b^5*c^4*d*e^6 - 3*b^2*c^7*d^4*e^3 + 6*b^3*c^6*d^3*e^4 - 4*b^4*
c^5*d^2*e^5))/c^5 + (8*(b^3*c^7*e^3 - 2*b^2*c^8*d*e^2)*(d^7)^(1/2)*(d + e*x)^(1/2))/(b*c^5))*(d^7)^(1/2))/b)*(
d^7)^(1/2))/b - (((8*(d + e*x)^(1/2)*(b^8*e^10 + 2*c^8*d^8*e^2 - 8*b*c^7*d^7*e^3 + 28*b^2*c^6*d^6*e^4 - 56*b^3
*c^5*d^5*e^5 + 70*b^4*c^4*d^4*e^6 - 56*b^5*c^3*d^3*e^7 + 28*b^6*c^2*d^2*e^8 - 8*b^7*c*d*e^9))/c^5 - (((8*(b^5*
c^4*d*e^6 - 3*b^2*c^7*d^4*e^3 + 6*b^3*c^6*d^3*e^4 - 4*b^4*c^5*d^2*e^5))/c^5 - (8*(b^3*c^7*e^3 - 2*b^2*c^8*d*e^
2)*(d^7)^(1/2)*(d + e*x)^(1/2))/(b*c^5))*(d^7)^(1/2))/b)*(d^7)^(1/2))/b))*(d^7)^(1/2)*2i)/b + (atan(((((8*(d +
e*x)^(1/2)*(b^8*e^10 + 2*c^8*d^8*e^2 - 8*b*c^7*d^7*e^3 + 28*b^2*c^6*d^6*e^4 - 56*b^3*c^5*d^5*e^5 + 70*b^4*c^4
*d^4*e^6 - 56*b^5*c^3*d^3*e^7 + 28*b^6*c^2*d^2*e^8 - 8*b^7*c*d*e^9))/c^5 + (((8*(b^5*c^4*d*e^6 - 3*b^2*c^7*d^4
*e^3 + 6*b^3*c^6*d^3*e^4 - 4*b^4*c^5*d^2*e^5))/c^5 + (8*(b^3*c^7*e^3 - 2*b^2*c^8*d*e^2)*(-c^7*(b*e - c*d)^7)^(
1/2)*(d + e*x)^(1/2))/(b*c^12))*(-c^7*(b*e - c*d)^7)^(1/2))/(b*c^7))*(-c^7*(b*e - c*d)^7)^(1/2)*1i)/(b*c^7) +
(((8*(d + e*x)^(1/2)*(b^8*e^10 + 2*c^8*d^8*e^2 - 8*b*c^7*d^7*e^3 + 28*b^2*c^6*d^6*e^4 - 56*b^3*c^5*d^5*e^5 + 7
0*b^4*c^4*d^4*e^6 - 56*b^5*c^3*d^3*e^7 + 28*b^6*c^2*d^2*e^8 - 8*b^7*c*d*e^9))/c^5 - (((8*(b^5*c^4*d*e^6 - 3*b^
2*c^7*d^4*e^3 + 6*b^3*c^6*d^3*e^4 - 4*b^4*c^5*d^2*e^5))/c^5 - (8*(b^3*c^7*e^3 - 2*b^2*c^8*d*e^2)*(-c^7*(b*e -
c*d)^7)^(1/2)*(d + e*x)^(1/2))/(b*c^12))*(-c^7*(b*e - c*d)^7)^(1/2))/(b*c^7))*(-c^7*(b*e - c*d)^7)^(1/2)*1i)/(
b*c^7))/((16*(b^7*d^4*e^10 - 4*c^7*d^11*e^3 + 22*b*c^6*d^10*e^4 - 8*b^6*c*d^5*e^9 - 52*b^2*c^5*d^9*e^5 + 69*b^
3*c^4*d^8*e^6 - 56*b^4*c^3*d^7*e^7 + 28*b^5*c^2*d^6*e^8))/c^5 + (((8*(d + e*x)^(1/2)*(b^8*e^10 + 2*c^8*d^8*e^2
- 8*b*c^7*d^7*e^3 + 28*b^2*c^6*d^6*e^4 - 56*b^3*c^5*d^5*e^5 + 70*b^4*c^4*d^4*e^6 - 56*b^5*c^3*d^3*e^7 + 28*b^
6*c^2*d^2*e^8 - 8*b^7*c*d*e^9))/c^5 + (((8*(b^5*c^4*d*e^6 - 3*b^2*c^7*d^4*e^3 + 6*b^3*c^6*d^3*e^4 - 4*b^4*c^5*
d^2*e^5))/c^5 + (8*(b^3*c^7*e^3 - 2*b^2*c^8*d*e^2)*(-c^7*(b*e - c*d)^7)^(1/2)*(d + e*x)^(1/2))/(b*c^12))*(-c^7
*(b*e - c*d)^7)^(1/2))/(b*c^7))*(-c^7*(b*e - c*d)^7)^(1/2))/(b*c^7) - (((8*(d + e*x)^(1/2)*(b^8*e^10 + 2*c^8*d
^8*e^2 - 8*b*c^7*d^7*e^3 + 28*b^2*c^6*d^6*e^4 - 56*b^3*c^5*d^5*e^5 + 70*b^4*c^4*d^4*e^6 - 56*b^5*c^3*d^3*e^7 +
28*b^6*c^2*d^2*e^8 - 8*b^7*c*d*e^9))/c^5 - (((8*(b^5*c^4*d*e^6 - 3*b^2*c^7*d^4*e^3 + 6*b^3*c^6*d^3*e^4 - 4*b^
4*c^5*d^2*e^5))/c^5 - (8*(b^3*c^7*e^3 - 2*b^2*c^8*d*e^2)*(-c^7*(b*e - c*d)^7)^(1/2)*(d + e*x)^(1/2))/(b*c^12))
*(-c^7*(b*e - c*d)^7)^(1/2))/(b*c^7))*(-c^7*(b*e - c*d)^7)^(1/2))/(b*c^7)))*(-c^7*(b*e - c*d)^7)^(1/2)*2i)/(b*
c^7) - (2*e*(b*e - 2*c*d)*(d + e*x)^(3/2))/(3*c^2)

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sympy [A]  time = 93.75, size = 162, normalized size = 1.03 \begin {gather*} \frac {2 e \left (d + e x\right )^{\frac {5}{2}}}{5 c} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (- 2 b e^{2} + 4 c d e\right )}{3 c^{2}} + \frac {\sqrt {d + e x} \left (2 b^{2} e^{3} - 6 b c d e^{2} + 6 c^{2} d^{2} e\right )}{c^{3}} + \frac {2 d^{4} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{b \sqrt {- d}} - \frac {2 \left (b e - c d\right )^{4} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {b e - c d}{c}}} \right )}}{b c^{4} \sqrt {\frac {b e - c d}{c}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*e*(d + e*x)**(5/2)/(5*c) + (d + e*x)**(3/2)*(-2*b*e**2 + 4*c*d*e)/(3*c**2) + sqrt(d + e*x)*(2*b**2*e**3 - 6*
b*c*d*e**2 + 6*c**2*d**2*e)/c**3 + 2*d**4*atan(sqrt(d + e*x)/sqrt(-d))/(b*sqrt(-d)) - 2*(b*e - c*d)**4*atan(sq
rt(d + e*x)/sqrt((b*e - c*d)/c))/(b*c**4*sqrt((b*e - c*d)/c))

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