3.1491 \(\int \frac {(A+B x) (d+e x)^m}{(a+c x^2)^2} \, dx\)

Optimal. Leaf size=361 \[ \frac {(d+e x)^{m+1} \left (a e m (a B e+A c d)-\sqrt {-a} \sqrt {c} \left (A \left (a e^2 (1-m)+c d^2\right )+a B d e m\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{4 a^2 \sqrt {c} (m+1) \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (a e^2+c d^2\right )}+\frac {(d+e x)^{m+1} \left (\sqrt {-a} \sqrt {c} \left (A \left (a e^2 (1-m)+c d^2\right )+a B d e m\right )+a e m (a B e+A c d)\right ) \, _2F_1\left (1,m+1;m+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{4 a^2 \sqrt {c} (m+1) \left (\sqrt {-a} e+\sqrt {c} d\right ) \left (a e^2+c d^2\right )}-\frac {(d+e x)^{m+1} (a (B d-A e)-x (a B e+A c d))}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )} \]

[Out]

-1/2*(e*x+d)^(1+m)*(a*(-A*e+B*d)-(A*c*d+B*a*e)*x)/a/(a*e^2+c*d^2)/(c*x^2+a)+1/4*(e*x+d)^(1+m)*hypergeom([1, 1+
m],[2+m],(e*x+d)*c^(1/2)/(-e*(-a)^(1/2)+d*c^(1/2)))*(a*e*(A*c*d+B*a*e)*m-(A*(c*d^2+a*e^2*(1-m))+a*B*d*e*m)*(-a
)^(1/2)*c^(1/2))/a^2/(a*e^2+c*d^2)/(1+m)/c^(1/2)/(-e*(-a)^(1/2)+d*c^(1/2))+1/4*(e*x+d)^(1+m)*hypergeom([1, 1+m
],[2+m],(e*x+d)*c^(1/2)/(e*(-a)^(1/2)+d*c^(1/2)))*(a*e*(A*c*d+B*a*e)*m+(A*(c*d^2+a*e^2*(1-m))+a*B*d*e*m)*(-a)^
(1/2)*c^(1/2))/a^2/(a*e^2+c*d^2)/(1+m)/c^(1/2)/(e*(-a)^(1/2)+d*c^(1/2))

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Rubi [A]  time = 0.58, antiderivative size = 359, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {823, 831, 68} \[ \frac {(d+e x)^{m+1} \left (a e m (a B e+A c d)-\sqrt {-a} \sqrt {c} \left (a A e^2 (1-m)+a B d e m+A c d^2\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{4 a^2 \sqrt {c} (m+1) \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (a e^2+c d^2\right )}+\frac {(d+e x)^{m+1} \left (\sqrt {-a} \sqrt {c} \left (a A e^2 (1-m)+a B d e m+A c d^2\right )+a e m (a B e+A c d)\right ) \, _2F_1\left (1,m+1;m+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{4 a^2 \sqrt {c} (m+1) \left (\sqrt {-a} e+\sqrt {c} d\right ) \left (a e^2+c d^2\right )}-\frac {(d+e x)^{m+1} (a (B d-A e)-x (a B e+A c d))}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x)^m)/(a + c*x^2)^2,x]

[Out]

-((d + e*x)^(1 + m)*(a*(B*d - A*e) - (A*c*d + a*B*e)*x))/(2*a*(c*d^2 + a*e^2)*(a + c*x^2)) + ((a*e*(A*c*d + a*
B*e)*m - Sqrt[-a]*Sqrt[c]*(A*c*d^2 + a*A*e^2*(1 - m) + a*B*d*e*m))*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 +
m, 2 + m, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(4*a^2*Sqrt[c]*(Sqrt[c]*d - Sqrt[-a]*e)*(c*d^2 + a*e^
2)*(1 + m)) + ((a*e*(A*c*d + a*B*e)*m + Sqrt[-a]*Sqrt[c]*(A*c*d^2 + a*A*e^2*(1 - m) + a*B*d*e*m))*(d + e*x)^(1
+ m)*Hypergeometric2F1[1, 1 + m, 2 + m, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(4*a^2*Sqrt[c]*(Sqrt[c
]*d + Sqrt[-a]*e)*(c*d^2 + a*e^2)*(1 + m))

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 831

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m, (f + g*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !Ration
alQ[m]

Rubi steps

\begin {align*} \int \frac {(A+B x) (d+e x)^m}{\left (a+c x^2\right )^2} \, dx &=-\frac {(d+e x)^{1+m} (a (B d-A e)-(A c d+a B e) x)}{2 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {\int \frac {(d+e x)^m \left (-c \left (A c d^2+a A e^2 (1-m)+a B d e m\right )+c e (A c d+a B e) m x\right )}{a+c x^2} \, dx}{2 a c \left (c d^2+a e^2\right )}\\ &=-\frac {(d+e x)^{1+m} (a (B d-A e)-(A c d+a B e) x)}{2 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {\int \left (\frac {\left (-a \sqrt {c} e (A c d+a B e) m-\sqrt {-a} c \left (A c d^2+a A e^2 (1-m)+a B d e m\right )\right ) (d+e x)^m}{2 a \left (\sqrt {-a}-\sqrt {c} x\right )}+\frac {\left (a \sqrt {c} e (A c d+a B e) m-\sqrt {-a} c \left (A c d^2+a A e^2 (1-m)+a B d e m\right )\right ) (d+e x)^m}{2 a \left (\sqrt {-a}+\sqrt {c} x\right )}\right ) \, dx}{2 a c \left (c d^2+a e^2\right )}\\ &=-\frac {(d+e x)^{1+m} (a (B d-A e)-(A c d+a B e) x)}{2 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {\left (a e (A c d+a B e) m-\sqrt {-a} \sqrt {c} \left (A c d^2+a A e^2 (1-m)+a B d e m\right )\right ) \int \frac {(d+e x)^m}{\sqrt {-a}+\sqrt {c} x} \, dx}{4 a^2 \sqrt {c} \left (c d^2+a e^2\right )}+\frac {\left (a e (A c d+a B e) m+\sqrt {-a} \sqrt {c} \left (A c d^2+a A e^2 (1-m)+a B d e m\right )\right ) \int \frac {(d+e x)^m}{\sqrt {-a}-\sqrt {c} x} \, dx}{4 a^2 \sqrt {c} \left (c d^2+a e^2\right )}\\ &=-\frac {(d+e x)^{1+m} (a (B d-A e)-(A c d+a B e) x)}{2 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac {\left (a e (A c d+a B e) m-\sqrt {-a} \sqrt {c} \left (A c d^2+a A e^2 (1-m)+a B d e m\right )\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{4 a^2 \sqrt {c} \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+m)}+\frac {\left (a e (A c d+a B e) m+\sqrt {-a} \sqrt {c} \left (A c d^2+a A e^2 (1-m)+a B d e m\right )\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{4 a^2 \sqrt {c} \left (\sqrt {c} d+\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.75, size = 310, normalized size = 0.86 \[ \frac {(d+e x)^{m+1} \left (\frac {\sqrt {c} \left (a e m (a B e+A c d)-\sqrt {-a} \sqrt {c} \left (-a A e^2 (m-1)+a B d e m+A c d^2\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{a (m+1) \left (\sqrt {c} d-\sqrt {-a} e\right )}+\frac {\sqrt {c} \left (\sqrt {-a} \sqrt {c} \left (-a A e^2 (m-1)+a B d e m+A c d^2\right )+a e m (a B e+A c d)\right ) \, _2F_1\left (1,m+1;m+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{a (m+1) \left (\sqrt {-a} e+\sqrt {c} d\right )}+\frac {2 c (a (A e-B d+B e x)+A c d x)}{a+c x^2}\right )}{4 a c \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(d + e*x)^m)/(a + c*x^2)^2,x]

[Out]

((d + e*x)^(1 + m)*((2*c*(A*c*d*x + a*(-(B*d) + A*e + B*e*x)))/(a + c*x^2) + (Sqrt[c]*(a*e*(A*c*d + a*B*e)*m -
Sqrt[-a]*Sqrt[c]*(A*c*d^2 - a*A*e^2*(-1 + m) + a*B*d*e*m))*Hypergeometric2F1[1, 1 + m, 2 + m, (Sqrt[c]*(d + e
*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(a*(Sqrt[c]*d - Sqrt[-a]*e)*(1 + m)) + (Sqrt[c]*(a*e*(A*c*d + a*B*e)*m + Sqrt[
-a]*Sqrt[c]*(A*c*d^2 - a*A*e^2*(-1 + m) + a*B*d*e*m))*Hypergeometric2F1[1, 1 + m, 2 + m, (Sqrt[c]*(d + e*x))/(
Sqrt[c]*d + Sqrt[-a]*e)])/(a*(Sqrt[c]*d + Sqrt[-a]*e)*(1 + m))))/(4*a*c*(c*d^2 + a*e^2))

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fricas [F]  time = 1.23, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m}}{c^{2} x^{4} + 2 \, a c x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^m/(c*x^2+a)^2,x, algorithm="fricas")

[Out]

integral((B*x + A)*(e*x + d)^m/(c^2*x^4 + 2*a*c*x^2 + a^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m}}{{\left (c x^{2} + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^m/(c*x^2+a)^2,x, algorithm="giac")

[Out]

integrate((B*x + A)*(e*x + d)^m/(c*x^2 + a)^2, x)

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maple [F]  time = 1.20, size = 0, normalized size = 0.00 \[ \int \frac {\left (B x +A \right ) \left (e x +d \right )^{m}}{\left (c \,x^{2}+a \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^m/(c*x^2+a)^2,x)

[Out]

int((B*x+A)*(e*x+d)^m/(c*x^2+a)^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m}}{{\left (c x^{2} + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^m/(c*x^2+a)^2,x, algorithm="maxima")

[Out]

integrate((B*x + A)*(e*x + d)^m/(c*x^2 + a)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^m}{{\left (c\,x^2+a\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(d + e*x)^m)/(a + c*x^2)^2,x)

[Out]

int(((A + B*x)*(d + e*x)^m)/(a + c*x^2)^2, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**m/(c*x**2+a)**2,x)

[Out]

Timed out

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