3.167 \(\int (f x)^m (d+e x^2) (a+b \csc ^{-1}(c x)) \, dx\)

Optimal. Leaf size=215 \[ \frac{d (f x)^{m+1} \left (a+b \csc ^{-1}(c x)\right )}{f (m+1)}+\frac{e (f x)^{m+3} \left (a+b \csc ^{-1}(c x)\right )}{f^3 (m+3)}+\frac{b x \sqrt{1-c^2 x^2} (f x)^{m+1} \left (c^2 d (m+2) (m+3)+e (m+1)^2\right ) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};c^2 x^2\right )}{c f (m+1)^2 (m+2) (m+3) \sqrt{c^2 x^2} \sqrt{c^2 x^2-1}}+\frac{b e x \sqrt{c^2 x^2-1} (f x)^{m+1}}{c f \left (m^2+5 m+6\right ) \sqrt{c^2 x^2}} \]

[Out]

(b*e*x*(f*x)^(1 + m)*Sqrt[-1 + c^2*x^2])/(c*f*(6 + 5*m + m^2)*Sqrt[c^2*x^2]) + (d*(f*x)^(1 + m)*(a + b*ArcCsc[
c*x]))/(f*(1 + m)) + (e*(f*x)^(3 + m)*(a + b*ArcCsc[c*x]))/(f^3*(3 + m)) + (b*(e*(1 + m)^2 + c^2*d*(2 + m)*(3
+ m))*x*(f*x)^(1 + m)*Sqrt[1 - c^2*x^2]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/(c*f*(1 + m)^2*
(2 + m)*(3 + m)*Sqrt[c^2*x^2]*Sqrt[-1 + c^2*x^2])

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Rubi [A]  time = 0.193828, antiderivative size = 202, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {14, 5239, 12, 459, 365, 364} \[ \frac{d (f x)^{m+1} \left (a+b \csc ^{-1}(c x)\right )}{f (m+1)}+\frac{e (f x)^{m+3} \left (a+b \csc ^{-1}(c x)\right )}{f^3 (m+3)}+\frac{b c x \sqrt{1-c^2 x^2} (f x)^{m+1} \left (\frac{e}{c^2 (m+2) (m+3)}+\frac{d}{(m+1)^2}\right ) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};c^2 x^2\right )}{f \sqrt{c^2 x^2} \sqrt{c^2 x^2-1}}+\frac{b e x \sqrt{c^2 x^2-1} (f x)^{m+1}}{c f \left (m^2+5 m+6\right ) \sqrt{c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*e*x*(f*x)^(1 + m)*Sqrt[-1 + c^2*x^2])/(c*f*(6 + 5*m + m^2)*Sqrt[c^2*x^2]) + (d*(f*x)^(1 + m)*(a + b*ArcCsc[
c*x]))/(f*(1 + m)) + (e*(f*x)^(3 + m)*(a + b*ArcCsc[c*x]))/(f^3*(3 + m)) + (b*c*(d/(1 + m)^2 + e/(c^2*(2 + m)*
(3 + m)))*x*(f*x)^(1 + m)*Sqrt[1 - c^2*x^2]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/(f*Sqrt[c^2
*x^2]*Sqrt[-1 + c^2*x^2])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 5239

Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcCsc[c*x], u, x] + Dist[(b*c*x)/Sqrt[c^2*x^2], Int[SimplifyI
ntegrand[u/(x*Sqrt[c^2*x^2 - 1]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && ((IGtQ[p, 0] &&  !(ILtQ
[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[p, 0] && GtQ[m + 2*p + 3, 0])) || (I
LtQ[(m + 2*p + 1)/2, 0] &&  !ILtQ[(m - 1)/2, 0]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 459

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

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int (f x)^m \left (d+e x^2\right ) \left (a+b \csc ^{-1}(c x)\right ) \, dx &=\frac{d (f x)^{1+m} \left (a+b \csc ^{-1}(c x)\right )}{f (1+m)}+\frac{e (f x)^{3+m} \left (a+b \csc ^{-1}(c x)\right )}{f^3 (3+m)}+\frac{(b c x) \int \frac{(f x)^m \left (d (3+m)+e (1+m) x^2\right )}{(1+m) (3+m) \sqrt{-1+c^2 x^2}} \, dx}{\sqrt{c^2 x^2}}\\ &=\frac{d (f x)^{1+m} \left (a+b \csc ^{-1}(c x)\right )}{f (1+m)}+\frac{e (f x)^{3+m} \left (a+b \csc ^{-1}(c x)\right )}{f^3 (3+m)}+\frac{(b c x) \int \frac{(f x)^m \left (d (3+m)+e (1+m) x^2\right )}{\sqrt{-1+c^2 x^2}} \, dx}{\left (3+4 m+m^2\right ) \sqrt{c^2 x^2}}\\ &=\frac{b e x (f x)^{1+m} \sqrt{-1+c^2 x^2}}{c f \left (6+5 m+m^2\right ) \sqrt{c^2 x^2}}+\frac{d (f x)^{1+m} \left (a+b \csc ^{-1}(c x)\right )}{f (1+m)}+\frac{e (f x)^{3+m} \left (a+b \csc ^{-1}(c x)\right )}{f^3 (3+m)}-\frac{\left (b c \left (-\frac{e (1+m)^2}{c^2 (2+m)}-d (3+m)\right ) x\right ) \int \frac{(f x)^m}{\sqrt{-1+c^2 x^2}} \, dx}{\left (3+4 m+m^2\right ) \sqrt{c^2 x^2}}\\ &=\frac{b e x (f x)^{1+m} \sqrt{-1+c^2 x^2}}{c f \left (6+5 m+m^2\right ) \sqrt{c^2 x^2}}+\frac{d (f x)^{1+m} \left (a+b \csc ^{-1}(c x)\right )}{f (1+m)}+\frac{e (f x)^{3+m} \left (a+b \csc ^{-1}(c x)\right )}{f^3 (3+m)}-\frac{\left (b c \left (-\frac{e (1+m)^2}{c^2 (2+m)}-d (3+m)\right ) x \sqrt{1-c^2 x^2}\right ) \int \frac{(f x)^m}{\sqrt{1-c^2 x^2}} \, dx}{\left (3+4 m+m^2\right ) \sqrt{c^2 x^2} \sqrt{-1+c^2 x^2}}\\ &=\frac{b e x (f x)^{1+m} \sqrt{-1+c^2 x^2}}{c f \left (6+5 m+m^2\right ) \sqrt{c^2 x^2}}+\frac{d (f x)^{1+m} \left (a+b \csc ^{-1}(c x)\right )}{f (1+m)}+\frac{e (f x)^{3+m} \left (a+b \csc ^{-1}(c x)\right )}{f^3 (3+m)}+\frac{b c \left (\frac{e (1+m)^2}{c^2 (2+m)}+d (3+m)\right ) x (f x)^{1+m} \sqrt{1-c^2 x^2} \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};c^2 x^2\right )}{f (1+m) \left (3+4 m+m^2\right ) \sqrt{c^2 x^2} \sqrt{-1+c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.511153, size = 171, normalized size = 0.8 \[ x (f x)^m \left (\frac{\frac{(m+3) \left (d (m+3)+e (m+1) x^2\right ) \left (a+b \csc ^{-1}(c x)\right )}{m+1}-\frac{b c e x^3 \sqrt{1-\frac{1}{c^2 x^2}} \, _2F_1\left (\frac{1}{2},\frac{m+3}{2};\frac{m+5}{2};c^2 x^2\right )}{\sqrt{1-c^2 x^2}}}{(m+3)^2}-\frac{b c d x \sqrt{1-\frac{1}{c^2 x^2}} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};c^2 x^2\right )}{(m+1)^2 \sqrt{1-c^2 x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x*(f*x)^m*(-((b*c*d*Sqrt[1 - 1/(c^2*x^2)]*x*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/((1 + m)^2*
Sqrt[1 - c^2*x^2])) + (((3 + m)*(d*(3 + m) + e*(1 + m)*x^2)*(a + b*ArcCsc[c*x]))/(1 + m) - (b*c*e*Sqrt[1 - 1/(
c^2*x^2)]*x^3*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, c^2*x^2])/Sqrt[1 - c^2*x^2])/(3 + m)^2)

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Maple [F]  time = 3.074, size = 0, normalized size = 0. \begin{align*} \int \left ( fx \right ) ^{m} \left ( e{x}^{2}+d \right ) \left ( a+b{\rm arccsc} \left (cx\right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^m*(e*x^2+d)*(a+b*arccsc(c*x)),x)

[Out]

int((f*x)^m*(e*x^2+d)*(a+b*arccsc(c*x)),x)

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a e x^{2} + a d +{\left (b e x^{2} + b d\right )} \operatorname{arccsc}\left (c x\right )\right )} \left (f x\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*e*x^2 + a*d + (b*e*x^2 + b*d)*arccsc(c*x))*(f*x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)**m*(e*x**2+d)*(a+b*acsc(c*x)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x^{2} + d\right )}{\left (b \operatorname{arccsc}\left (c x\right ) + a\right )} \left (f x\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x^2 + d)*(b*arccsc(c*x) + a)*(f*x)^m, x)