3.4.82 \(\int \frac {(d+e x)^3 (f+g x)^4}{(d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=215 \[ -\frac {g^3 (3 d g+4 e f) \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5}+\frac {2 (d+e x)^2 (e f-9 d g) (d g+e f)^3}{15 d^2 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {(d+e x)^3 (d g+e f)^4}{5 d e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {g^4 \sqrt {d^2-e^2 x^2}}{e^5}+\frac {2 (d+e x) (d g+e f)^2 \left (36 d^2 g^2-8 d e f g+e^2 f^2\right )}{15 d^3 e^5 \sqrt {d^2-e^2 x^2}} \]

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Rubi [A]  time = 0.67, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1635, 641, 217, 203} \begin {gather*} \frac {2 (d+e x) (d g+e f)^2 \left (36 d^2 g^2-8 d e f g+e^2 f^2\right )}{15 d^3 e^5 \sqrt {d^2-e^2 x^2}}-\frac {g^3 (3 d g+4 e f) \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5}+\frac {2 (d+e x)^2 (e f-9 d g) (d g+e f)^3}{15 d^2 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {(d+e x)^3 (d g+e f)^4}{5 d e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {g^4 \sqrt {d^2-e^2 x^2}}{e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^3*(f + g*x)^4)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

((e*f + d*g)^4*(d + e*x)^3)/(5*d*e^5*(d^2 - e^2*x^2)^(5/2)) + (2*(e*f - 9*d*g)*(e*f + d*g)^3*(d + e*x)^2)/(15*
d^2*e^5*(d^2 - e^2*x^2)^(3/2)) + (2*(e*f + d*g)^2*(e^2*f^2 - 8*d*e*f*g + 36*d^2*g^2)*(d + e*x))/(15*d^3*e^5*Sq
rt[d^2 - e^2*x^2]) + (g^4*Sqrt[d^2 - e^2*x^2])/e^5 - (g^3*(4*e*f + 3*d*g)*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e
^5

Rule 203

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 641

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

Rule 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {(d+e x)^3 (f+g x)^4}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {(e f+d g)^4 (d+e x)^3}{5 d e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d+e x)^2 \left (-\frac {2 e^4 f^4-12 d e^3 f^3 g-18 d^2 e^2 f^2 g^2-12 d^3 e f g^3-3 d^4 g^4}{e^4}+\frac {5 d g^2 \left (6 e^2 f^2+4 d e f g+d^2 g^2\right ) x}{e^3}+\frac {5 d g^3 (4 e f+d g) x^2}{e^2}+\frac {5 d g^4 x^3}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d}\\ &=\frac {(e f+d g)^4 (d+e x)^3}{5 d e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 (e f-9 d g) (e f+d g)^3 (d+e x)^2}{15 d^2 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {(d+e x) \left (\frac {2 e^4 f^4-12 d e^3 f^3 g+42 d^2 e^2 f^2 g^2+68 d^3 e f g^3+27 d^4 g^4}{e^4}+\frac {30 d^2 g^3 (2 e f+d g) x}{e^3}+\frac {15 d^2 g^4 x^2}{e^2}\right )}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2}\\ &=\frac {(e f+d g)^4 (d+e x)^3}{5 d e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 (e f-9 d g) (e f+d g)^3 (d+e x)^2}{15 d^2 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (e f+d g)^2 \left (e^2 f^2-8 d e f g+36 d^2 g^2\right ) (d+e x)}{15 d^3 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {\frac {15 d^3 g^3 (4 e f+3 d g)}{e^4}+\frac {15 d^3 g^4 x}{e^3}}{\sqrt {d^2-e^2 x^2}} \, dx}{15 d^3}\\ &=\frac {(e f+d g)^4 (d+e x)^3}{5 d e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 (e f-9 d g) (e f+d g)^3 (d+e x)^2}{15 d^2 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (e f+d g)^2 \left (e^2 f^2-8 d e f g+36 d^2 g^2\right ) (d+e x)}{15 d^3 e^5 \sqrt {d^2-e^2 x^2}}+\frac {g^4 \sqrt {d^2-e^2 x^2}}{e^5}-\frac {\left (g^3 (4 e f+3 d g)\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^4}\\ &=\frac {(e f+d g)^4 (d+e x)^3}{5 d e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 (e f-9 d g) (e f+d g)^3 (d+e x)^2}{15 d^2 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (e f+d g)^2 \left (e^2 f^2-8 d e f g+36 d^2 g^2\right ) (d+e x)}{15 d^3 e^5 \sqrt {d^2-e^2 x^2}}+\frac {g^4 \sqrt {d^2-e^2 x^2}}{e^5}-\frac {\left (g^3 (4 e f+3 d g)\right ) \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4}\\ &=\frac {(e f+d g)^4 (d+e x)^3}{5 d e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {2 (e f-9 d g) (e f+d g)^3 (d+e x)^2}{15 d^2 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (e f+d g)^2 \left (e^2 f^2-8 d e f g+36 d^2 g^2\right ) (d+e x)}{15 d^3 e^5 \sqrt {d^2-e^2 x^2}}+\frac {g^4 \sqrt {d^2-e^2 x^2}}{e^5}-\frac {g^3 (4 e f+3 d g) \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5}\\ \end {align*}

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Mathematica [A]  time = 0.74, size = 168, normalized size = 0.78 \begin {gather*} \frac {\frac {\sqrt {d^2-e^2 x^2} \left (15 d^3 g^4 (d-e x)^3+2 (d-e x)^2 (d g+e f)^2 \left (36 d^2 g^2-8 d e f g+e^2 f^2\right )+3 d^2 (d g+e f)^4+2 d (d-e x) (e f-9 d g) (d g+e f)^3\right )}{d^3 (d-e x)^3}-15 g^3 (3 d g+4 e f) \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{15 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)^3*(f + g*x)^4)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

((Sqrt[d^2 - e^2*x^2]*(3*d^2*(e*f + d*g)^4 + 2*d*(e*f - 9*d*g)*(e*f + d*g)^3*(d - e*x) + 2*(e*f + d*g)^2*(e^2*
f^2 - 8*d*e*f*g + 36*d^2*g^2)*(d - e*x)^2 + 15*d^3*g^4*(d - e*x)^3))/(d^3*(d - e*x)^3) - 15*g^3*(4*e*f + 3*d*g
)*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(15*e^5)

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IntegrateAlgebraic [A]  time = 1.57, size = 294, normalized size = 1.37 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (72 d^6 g^4+88 d^5 e f g^3-171 d^5 e g^4 x+12 d^4 e^2 f^2 g^2-204 d^4 e^2 f g^3 x+117 d^4 e^2 g^4 x^2-12 d^3 e^3 f^3 g-36 d^3 e^3 f^2 g^2 x+128 d^3 e^3 f g^3 x^2-15 d^3 e^3 g^4 x^3+7 d^2 e^4 f^4+36 d^2 e^4 f^3 g x+42 d^2 e^4 f^2 g^2 x^2-6 d e^5 f^4 x-12 d e^5 f^3 g x^2+2 e^6 f^4 x^2\right )}{15 d^3 e^5 (d-e x)^3}-\frac {\sqrt {-e^2} \left (3 d g^4+4 e f g^3\right ) \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((d + e*x)^3*(f + g*x)^4)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(7*d^2*e^4*f^4 - 12*d^3*e^3*f^3*g + 12*d^4*e^2*f^2*g^2 + 88*d^5*e*f*g^3 + 72*d^6*g^4 - 6*
d*e^5*f^4*x + 36*d^2*e^4*f^3*g*x - 36*d^3*e^3*f^2*g^2*x - 204*d^4*e^2*f*g^3*x - 171*d^5*e*g^4*x + 2*e^6*f^4*x^
2 - 12*d*e^5*f^3*g*x^2 + 42*d^2*e^4*f^2*g^2*x^2 + 128*d^3*e^3*f*g^3*x^2 + 117*d^4*e^2*g^4*x^2 - 15*d^3*e^3*g^4
*x^3))/(15*d^3*e^5*(d - e*x)^3) - (Sqrt[-e^2]*(4*e*f*g^3 + 3*d*g^4)*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]]
)/e^6

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fricas [B]  time = 0.57, size = 624, normalized size = 2.90 \begin {gather*} -\frac {7 \, d^{3} e^{4} f^{4} - 12 \, d^{4} e^{3} f^{3} g + 12 \, d^{5} e^{2} f^{2} g^{2} + 88 \, d^{6} e f g^{3} + 72 \, d^{7} g^{4} - {\left (7 \, e^{7} f^{4} - 12 \, d e^{6} f^{3} g + 12 \, d^{2} e^{5} f^{2} g^{2} + 88 \, d^{3} e^{4} f g^{3} + 72 \, d^{4} e^{3} g^{4}\right )} x^{3} + 3 \, {\left (7 \, d e^{6} f^{4} - 12 \, d^{2} e^{5} f^{3} g + 12 \, d^{3} e^{4} f^{2} g^{2} + 88 \, d^{4} e^{3} f g^{3} + 72 \, d^{5} e^{2} g^{4}\right )} x^{2} - 3 \, {\left (7 \, d^{2} e^{5} f^{4} - 12 \, d^{3} e^{4} f^{3} g + 12 \, d^{4} e^{3} f^{2} g^{2} + 88 \, d^{5} e^{2} f g^{3} + 72 \, d^{6} e g^{4}\right )} x + 30 \, {\left (4 \, d^{6} e f g^{3} + 3 \, d^{7} g^{4} - {\left (4 \, d^{3} e^{4} f g^{3} + 3 \, d^{4} e^{3} g^{4}\right )} x^{3} + 3 \, {\left (4 \, d^{4} e^{3} f g^{3} + 3 \, d^{5} e^{2} g^{4}\right )} x^{2} - 3 \, {\left (4 \, d^{5} e^{2} f g^{3} + 3 \, d^{6} e g^{4}\right )} x\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (15 \, d^{3} e^{3} g^{4} x^{3} - 7 \, d^{2} e^{4} f^{4} + 12 \, d^{3} e^{3} f^{3} g - 12 \, d^{4} e^{2} f^{2} g^{2} - 88 \, d^{5} e f g^{3} - 72 \, d^{6} g^{4} - {\left (2 \, e^{6} f^{4} - 12 \, d e^{5} f^{3} g + 42 \, d^{2} e^{4} f^{2} g^{2} + 128 \, d^{3} e^{3} f g^{3} + 117 \, d^{4} e^{2} g^{4}\right )} x^{2} + 3 \, {\left (2 \, d e^{5} f^{4} - 12 \, d^{2} e^{4} f^{3} g + 12 \, d^{3} e^{3} f^{2} g^{2} + 68 \, d^{4} e^{2} f g^{3} + 57 \, d^{5} e g^{4}\right )} x\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{3} e^{8} x^{3} - 3 \, d^{4} e^{7} x^{2} + 3 \, d^{5} e^{6} x - d^{6} e^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(g*x+f)^4/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

-1/15*(7*d^3*e^4*f^4 - 12*d^4*e^3*f^3*g + 12*d^5*e^2*f^2*g^2 + 88*d^6*e*f*g^3 + 72*d^7*g^4 - (7*e^7*f^4 - 12*d
*e^6*f^3*g + 12*d^2*e^5*f^2*g^2 + 88*d^3*e^4*f*g^3 + 72*d^4*e^3*g^4)*x^3 + 3*(7*d*e^6*f^4 - 12*d^2*e^5*f^3*g +
12*d^3*e^4*f^2*g^2 + 88*d^4*e^3*f*g^3 + 72*d^5*e^2*g^4)*x^2 - 3*(7*d^2*e^5*f^4 - 12*d^3*e^4*f^3*g + 12*d^4*e^
3*f^2*g^2 + 88*d^5*e^2*f*g^3 + 72*d^6*e*g^4)*x + 30*(4*d^6*e*f*g^3 + 3*d^7*g^4 - (4*d^3*e^4*f*g^3 + 3*d^4*e^3*
g^4)*x^3 + 3*(4*d^4*e^3*f*g^3 + 3*d^5*e^2*g^4)*x^2 - 3*(4*d^5*e^2*f*g^3 + 3*d^6*e*g^4)*x)*arctan(-(d - sqrt(-e
^2*x^2 + d^2))/(e*x)) - (15*d^3*e^3*g^4*x^3 - 7*d^2*e^4*f^4 + 12*d^3*e^3*f^3*g - 12*d^4*e^2*f^2*g^2 - 88*d^5*e
*f*g^3 - 72*d^6*g^4 - (2*e^6*f^4 - 12*d*e^5*f^3*g + 42*d^2*e^4*f^2*g^2 + 128*d^3*e^3*f*g^3 + 117*d^4*e^2*g^4)*
x^2 + 3*(2*d*e^5*f^4 - 12*d^2*e^4*f^3*g + 12*d^3*e^3*f^2*g^2 + 68*d^4*e^2*f*g^3 + 57*d^5*e*g^4)*x)*sqrt(-e^2*x
^2 + d^2))/(d^3*e^8*x^3 - 3*d^4*e^7*x^2 + 3*d^5*e^6*x - d^6*e^5)

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giac [B]  time = 0.40, size = 411, normalized size = 1.91 \begin {gather*} -{\left (3 \, d g^{4} + 4 \, f g^{3} e\right )} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-5\right )} \mathrm {sgn}\relax (d) + \frac {\sqrt {-x^{2} e^{2} + d^{2}} {\left ({\left ({\left ({\left ({\left ({\left (15 \, g^{4} x e - \frac {2 \, {\left (36 \, d^{5} g^{4} e^{10} + 64 \, d^{4} f g^{3} e^{11} + 21 \, d^{3} f^{2} g^{2} e^{12} - 6 \, d^{2} f^{3} g e^{13} + d f^{4} e^{14}\right )} e^{\left (-10\right )}}{d^{4}}\right )} x - \frac {45 \, {\left (3 \, d^{6} g^{4} e^{9} + 4 \, d^{5} f g^{3} e^{10} + 2 \, d^{4} f^{2} g^{2} e^{11}\right )} e^{\left (-10\right )}}{d^{4}}\right )} x + \frac {5 \, {\left (21 \, d^{7} g^{4} e^{8} + 28 \, d^{6} f g^{3} e^{9} - 6 \, d^{5} f^{2} g^{2} e^{10} - 12 \, d^{4} f^{3} g e^{11} + d^{3} f^{4} e^{12}\right )} e^{\left (-10\right )}}{d^{4}}\right )} x + \frac {5 \, {\left (36 \, d^{8} g^{4} e^{7} + 44 \, d^{7} f g^{3} e^{8} + 6 \, d^{6} f^{2} g^{2} e^{9} - 12 \, d^{5} f^{3} g e^{10} - d^{4} f^{4} e^{11}\right )} e^{\left (-10\right )}}{d^{4}}\right )} x - \frac {15 \, {\left (3 \, d^{9} g^{4} e^{6} + 4 \, d^{8} f g^{3} e^{7} + d^{5} f^{4} e^{10}\right )} e^{\left (-10\right )}}{d^{4}}\right )} x - \frac {{\left (72 \, d^{10} g^{4} e^{5} + 88 \, d^{9} f g^{3} e^{6} + 12 \, d^{8} f^{2} g^{2} e^{7} - 12 \, d^{7} f^{3} g e^{8} + 7 \, d^{6} f^{4} e^{9}\right )} e^{\left (-10\right )}}{d^{4}}\right )}}{15 \, {\left (x^{2} e^{2} - d^{2}\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(g*x+f)^4/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-(3*d*g^4 + 4*f*g^3*e)*arcsin(x*e/d)*e^(-5)*sgn(d) + 1/15*sqrt(-x^2*e^2 + d^2)*((((((15*g^4*x*e - 2*(36*d^5*g^
4*e^10 + 64*d^4*f*g^3*e^11 + 21*d^3*f^2*g^2*e^12 - 6*d^2*f^3*g*e^13 + d*f^4*e^14)*e^(-10)/d^4)*x - 45*(3*d^6*g
^4*e^9 + 4*d^5*f*g^3*e^10 + 2*d^4*f^2*g^2*e^11)*e^(-10)/d^4)*x + 5*(21*d^7*g^4*e^8 + 28*d^6*f*g^3*e^9 - 6*d^5*
f^2*g^2*e^10 - 12*d^4*f^3*g*e^11 + d^3*f^4*e^12)*e^(-10)/d^4)*x + 5*(36*d^8*g^4*e^7 + 44*d^7*f*g^3*e^8 + 6*d^6
*f^2*g^2*e^9 - 12*d^5*f^3*g*e^10 - d^4*f^4*e^11)*e^(-10)/d^4)*x - 15*(3*d^9*g^4*e^6 + 4*d^8*f*g^3*e^7 + d^5*f^
4*e^10)*e^(-10)/d^4)*x - (72*d^10*g^4*e^5 + 88*d^9*f*g^3*e^6 + 12*d^8*f^2*g^2*e^7 - 12*d^7*f^3*g*e^8 + 7*d^6*f
^4*e^9)*e^(-10)/d^4)/(x^2*e^2 - d^2)^3

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maple [B]  time = 0.01, size = 1030, normalized size = 4.79 \begin {gather*} -\frac {e \,g^{4} x^{6}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {3 d \,g^{4} x^{5}}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {4 e f \,g^{3} x^{5}}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {9 d^{2} g^{4} x^{4}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e}+\frac {12 d f \,g^{3} x^{4}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {6 e \,f^{2} g^{2} x^{4}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {d^{3} g^{4} x^{3}}{2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{2}}+\frac {6 d^{2} f \,g^{3} x^{3}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e}+\frac {9 d \,f^{2} g^{2} x^{3}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {2 e \,f^{3} g \,x^{3}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {12 d^{4} g^{4} x^{2}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{3}}-\frac {44 d^{3} f \,g^{3} x^{2}}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{2}}-\frac {2 d^{2} f^{2} g^{2} x^{2}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e}+\frac {4 d \,f^{3} g \,x^{2}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {e \,f^{4} x^{2}}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {3 d^{5} g^{4} x}{10 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{4}}-\frac {18 d^{4} f \,g^{3} x}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{3}}-\frac {21 d^{3} f^{2} g^{2} x}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{2}}+\frac {6 d^{2} f^{3} g x}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e}-\frac {d \,g^{4} x^{3}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{2}}+\frac {4 d \,f^{4} x}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {4 f \,g^{3} x^{3}}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e}+\frac {24 d^{6} g^{4}}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{5}}+\frac {88 d^{5} f \,g^{3}}{15 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{4}}+\frac {4 d^{4} f^{2} g^{2}}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{3}}-\frac {4 d^{3} f^{3} g}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{2}}+\frac {7 d^{2} f^{4}}{15 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e}+\frac {d^{3} g^{4} x}{10 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{4}}+\frac {6 d^{2} f \,g^{3} x}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{3}}+\frac {7 d \,f^{2} g^{2} x}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{2}}+\frac {f^{4} x}{15 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d}-\frac {2 f^{3} g x}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e}+\frac {16 d \,g^{4} x}{5 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{4}}-\frac {3 d \,g^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, e^{4}}+\frac {14 f^{2} g^{2} x}{5 \sqrt {-e^{2} x^{2}+d^{2}}\, d \,e^{2}}-\frac {4 f^{3} g x}{5 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{2} e}+\frac {2 f^{4} x}{15 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{3}}+\frac {32 f \,g^{3} x}{5 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{3}}-\frac {4 f \,g^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*(g*x+f)^4/(-e^2*x^2+d^2)^(7/2),x)

[Out]

4/5*x/(-e^2*x^2+d^2)^(5/2)*d*f^4+1/15/d*x/(-e^2*x^2+d^2)^(3/2)*f^4+2/15/d^3*x/(-e^2*x^2+d^2)^(1/2)*f^4+1/3*x^2
*e/(-e^2*x^2+d^2)^(5/2)*f^4+7/15*d^2/e/(-e^2*x^2+d^2)^(5/2)*f^4+3/5*x^5/(-e^2*x^2+d^2)^(5/2)*d*g^4+24/5/e^5*g^
4*d^6/(-e^2*x^2+d^2)^(5/2)-e*g^4*x^6/(-e^2*x^2+d^2)^(5/2)+6*x^3/e/(-e^2*x^2+d^2)^(5/2)*d^2*f*g^3-18/5*d^4/e^3*
x/(-e^2*x^2+d^2)^(5/2)*f*g^3-44/3*d^3/e^2*x^2/(-e^2*x^2+d^2)^(5/2)*f*g^3-2*d^2/e*x^2/(-e^2*x^2+d^2)^(5/2)*f^2*
g^2+14/5/d/e^2*x/(-e^2*x^2+d^2)^(1/2)*f^2*g^2-4/5/d^2/e*x/(-e^2*x^2+d^2)^(1/2)*f^3*g-21/5*d^3/e^2*x/(-e^2*x^2+
d^2)^(5/2)*f^2*g^2+6/5*d^2/e*x/(-e^2*x^2+d^2)^(5/2)*f^3*g+6/5/e^3*x/(-e^2*x^2+d^2)^(3/2)*d^2*f*g^3+7/5/e^2*x/(
-e^2*x^2+d^2)^(3/2)*d*f^2*g^2+6*x^4*e/(-e^2*x^2+d^2)^(5/2)*f^2*g^2+88/15*d^5/e^4/(-e^2*x^2+d^2)^(5/2)*f*g^3+4/
5*d^4/e^3/(-e^2*x^2+d^2)^(5/2)*f^2*g^2-1/e^2*x^3/(-e^2*x^2+d^2)^(3/2)*d*g^4-4/3/e*x^3/(-e^2*x^2+d^2)^(3/2)*f*g
^3+16/5/e^4*x/(-e^2*x^2+d^2)^(1/2)*d*g^4+32/5/e^3*x/(-e^2*x^2+d^2)^(1/2)*f*g^3-3/e^4/(e^2)^(1/2)*arctan((e^2)^
(1/2)/(-e^2*x^2+d^2)^(1/2)*x)*d*g^4-4/e^3/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)*f*g^3+4/5*x^5
*e/(-e^2*x^2+d^2)^(5/2)*f*g^3+2*x^3*e/(-e^2*x^2+d^2)^(5/2)*f^3*g-3/10*d^5/e^4*x/(-e^2*x^2+d^2)^(5/2)*g^4+12*x^
4/(-e^2*x^2+d^2)^(5/2)*d*f*g^3+9*x^3/(-e^2*x^2+d^2)^(5/2)*d*f^2*g^2+1/2*x^3/e^2/(-e^2*x^2+d^2)^(5/2)*d^3*g^4-1
2/e^3*g^4*d^4*x^2/(-e^2*x^2+d^2)^(5/2)+9/e*g^4*d^2*x^4/(-e^2*x^2+d^2)^(5/2)+1/10/e^4*x/(-e^2*x^2+d^2)^(3/2)*d^
3*g^4-2/5/e*x/(-e^2*x^2+d^2)^(3/2)*f^3*g+4*x^2/(-e^2*x^2+d^2)^(5/2)*d*f^3*g-4/5*d^3/e^2/(-e^2*x^2+d^2)^(5/2)*f
^3*g

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maxima [B]  time = 1.04, size = 1178, normalized size = 5.48

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(g*x+f)^4/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

-e*g^4*x^6/(-e^2*x^2 + d^2)^(5/2) + 6*d^2*g^4*x^4/((-e^2*x^2 + d^2)^(5/2)*e) - 8*d^4*g^4*x^2/((-e^2*x^2 + d^2)
^(5/2)*e^3) + 1/5*d*f^4*x/(-e^2*x^2 + d^2)^(5/2) + 1/15*(4*e^3*f*g^3 + 3*d*e^2*g^4)*x*(15*x^4/((-e^2*x^2 + d^2
)^(5/2)*e^2) - 20*d^2*x^2/((-e^2*x^2 + d^2)^(5/2)*e^4) + 8*d^4/((-e^2*x^2 + d^2)^(5/2)*e^6)) + 3/5*d^2*f^4/((-
e^2*x^2 + d^2)^(5/2)*e) + 4/5*d^3*f^3*g/((-e^2*x^2 + d^2)^(5/2)*e^2) + 16/5*d^6*g^4/((-e^2*x^2 + d^2)^(5/2)*e^
5) + 4/15*f^4*x/((-e^2*x^2 + d^2)^(3/2)*d) + 8/15*f^4*x/(sqrt(-e^2*x^2 + d^2)*d^3) - 1/3*(4*e^3*f*g^3 + 3*d*e^
2*g^4)*x*(3*x^2/((-e^2*x^2 + d^2)^(3/2)*e^2) - 2*d^2/((-e^2*x^2 + d^2)^(3/2)*e^4))/e^2 + 3*(2*e^3*f^2*g^2 + 4*
d*e^2*f*g^3 + d^2*e*g^4)*x^4/((-e^2*x^2 + d^2)^(5/2)*e^2) + 1/2*(4*e^3*f^3*g + 18*d*e^2*f^2*g^2 + 12*d^2*e*f*g
^3 + d^3*g^4)*x^3/((-e^2*x^2 + d^2)^(5/2)*e^2) - 4*(2*e^3*f^2*g^2 + 4*d*e^2*f*g^3 + d^2*e*g^4)*d^2*x^2/((-e^2*
x^2 + d^2)^(5/2)*e^4) + 1/3*(e^3*f^4 + 12*d*e^2*f^3*g + 18*d^2*e*f^2*g^2 + 4*d^3*f*g^3)*x^2/((-e^2*x^2 + d^2)^
(5/2)*e^2) - 3/10*(4*e^3*f^3*g + 18*d*e^2*f^2*g^2 + 12*d^2*e*f*g^3 + d^3*g^4)*d^2*x/((-e^2*x^2 + d^2)^(5/2)*e^
4) + 3/5*(d*e^2*f^4 + 4*d^2*e*f^3*g + 2*d^3*f^2*g^2)*x/((-e^2*x^2 + d^2)^(5/2)*e^2) + 8/5*(2*e^3*f^2*g^2 + 4*d
*e^2*f*g^3 + d^2*e*g^4)*d^4/((-e^2*x^2 + d^2)^(5/2)*e^6) - 2/15*(e^3*f^4 + 12*d*e^2*f^3*g + 18*d^2*e*f^2*g^2 +
4*d^3*f*g^3)*d^2/((-e^2*x^2 + d^2)^(5/2)*e^4) + 4/15*(4*e^3*f*g^3 + 3*d*e^2*g^4)*d^2*x/((-e^2*x^2 + d^2)^(3/2
)*e^6) + 1/10*(4*e^3*f^3*g + 18*d*e^2*f^2*g^2 + 12*d^2*e*f*g^3 + d^3*g^4)*x/((-e^2*x^2 + d^2)^(3/2)*e^4) - 1/5
*(d*e^2*f^4 + 4*d^2*e*f^3*g + 2*d^3*f^2*g^2)*x/((-e^2*x^2 + d^2)^(3/2)*d^2*e^2) - 7/15*(4*e^3*f*g^3 + 3*d*e^2*
g^4)*x/(sqrt(-e^2*x^2 + d^2)*e^6) + 1/5*(4*e^3*f^3*g + 18*d*e^2*f^2*g^2 + 12*d^2*e*f*g^3 + d^3*g^4)*x/(sqrt(-e
^2*x^2 + d^2)*d^2*e^4) - 2/5*(d*e^2*f^4 + 4*d^2*e*f^3*g + 2*d^3*f^2*g^2)*x/(sqrt(-e^2*x^2 + d^2)*d^4*e^2) - (4
*e^3*f*g^3 + 3*d*e^2*g^4)*arcsin(e*x/d)/e^7

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (f+g\,x\right )}^4\,{\left (d+e\,x\right )}^3}{{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f + g*x)^4*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2),x)

[Out]

int(((f + g*x)^4*(d + e*x)^3)/(d^2 - e^2*x^2)^(7/2), x)

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sympy [F(-1)]  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((e*x+d)**3*(g*x+f)**4/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Timed out

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