3.196 \(\int \frac {x^m (a+b \sinh ^{-1}(c x))}{(d+c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=402 \[ \frac {b c (2-m) m \sqrt {c^2 x^2+1} x^{m+2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;-c^2 x^2\right )}{3 d^2 \left (m^2+3 m+2\right ) \sqrt {c^2 d x^2+d}}-\frac {(2-m) m \sqrt {c^2 x^2+1} x^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};-c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 (m+1) \sqrt {c^2 d x^2+d}}+\frac {(2-m) x^{m+1} \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt {c^2 d x^2+d}}+\frac {x^{m+1} \left (a+b \sinh ^{-1}(c x)\right )}{3 d \left (c^2 d x^2+d\right )^{3/2}}-\frac {b c (2-m) \sqrt {c^2 x^2+1} x^{m+2} \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-c^2 x^2\right )}{3 d^2 (m+2) \sqrt {c^2 d x^2+d}}-\frac {b c \sqrt {c^2 x^2+1} x^{m+2} \, _2F_1\left (2,\frac {m+2}{2};\frac {m+4}{2};-c^2 x^2\right )}{3 d^2 (m+2) \sqrt {c^2 d x^2+d}} \]

[Out]

1/3*x^(1+m)*(a+b*arcsinh(c*x))/d/(c^2*d*x^2+d)^(3/2)+1/3*(2-m)*x^(1+m)*(a+b*arcsinh(c*x))/d^2/(c^2*d*x^2+d)^(1
/2)-1/3*(2-m)*m*x^(1+m)*(a+b*arcsinh(c*x))*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],-c^2*x^2)*(c^2*x^2+1)^(1/2)/
d^2/(1+m)/(c^2*d*x^2+d)^(1/2)-1/3*b*c*(2-m)*x^(2+m)*hypergeom([1, 1+1/2*m],[2+1/2*m],-c^2*x^2)*(c^2*x^2+1)^(1/
2)/d^2/(2+m)/(c^2*d*x^2+d)^(1/2)-1/3*b*c*x^(2+m)*hypergeom([2, 1+1/2*m],[2+1/2*m],-c^2*x^2)*(c^2*x^2+1)^(1/2)/
d^2/(2+m)/(c^2*d*x^2+d)^(1/2)+1/3*b*c*(2-m)*m*x^(2+m)*HypergeometricPFQ([1, 1+1/2*m, 1+1/2*m],[3/2+1/2*m, 2+1/
2*m],-c^2*x^2)*(c^2*x^2+1)^(1/2)/d^2/(m^2+3*m+2)/(c^2*d*x^2+d)^(1/2)

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Rubi [A]  time = 0.47, antiderivative size = 402, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {5755, 5764, 5762, 364} \[ \frac {b c (2-m) m \sqrt {c^2 x^2+1} x^{m+2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;-c^2 x^2\right )}{3 d^2 \left (m^2+3 m+2\right ) \sqrt {c^2 d x^2+d}}-\frac {(2-m) m \sqrt {c^2 x^2+1} x^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};-c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 (m+1) \sqrt {c^2 d x^2+d}}+\frac {(2-m) x^{m+1} \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt {c^2 d x^2+d}}+\frac {x^{m+1} \left (a+b \sinh ^{-1}(c x)\right )}{3 d \left (c^2 d x^2+d\right )^{3/2}}-\frac {b c (2-m) \sqrt {c^2 x^2+1} x^{m+2} \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-c^2 x^2\right )}{3 d^2 (m+2) \sqrt {c^2 d x^2+d}}-\frac {b c \sqrt {c^2 x^2+1} x^{m+2} \, _2F_1\left (2,\frac {m+2}{2};\frac {m+4}{2};-c^2 x^2\right )}{3 d^2 (m+2) \sqrt {c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*(a + b*ArcSinh[c*x]))/(3*d*(d + c^2*d*x^2)^(3/2)) + ((2 - m)*x^(1 + m)*(a + b*ArcSinh[c*x]))/(3*d^2
*Sqrt[d + c^2*d*x^2]) - ((2 - m)*m*x^(1 + m)*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])*Hypergeometric2F1[1/2, (1
+ m)/2, (3 + m)/2, -(c^2*x^2)])/(3*d^2*(1 + m)*Sqrt[d + c^2*d*x^2]) - (b*c*(2 - m)*x^(2 + m)*Sqrt[1 + c^2*x^2]
*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -(c^2*x^2)])/(3*d^2*(2 + m)*Sqrt[d + c^2*d*x^2]) - (b*c*x^(2 + m)*
Sqrt[1 + c^2*x^2]*Hypergeometric2F1[2, (2 + m)/2, (4 + m)/2, -(c^2*x^2)])/(3*d^2*(2 + m)*Sqrt[d + c^2*d*x^2])
+ (b*c*(2 - m)*m*x^(2 + m)*Sqrt[1 + c^2*x^2]*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, -(
c^2*x^2)])/(3*d^2*(2 + 3*m + m^2)*Sqrt[d + c^2*d*x^2])

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])

Rule 5755

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp
[((f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(2*d*f*(p + 1)), x] + (Dist[(m + 2*p + 3)/(2*d*(p
+ 1)), Int[(f*x)^m*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^F
racPart[p])/(2*f*(p + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[
c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] &&  !GtQ
[m, 1] && (IntegerQ[m] || IntegerQ[p] || EqQ[n, 1])

Rule 5762

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[((f*x
)^(m + 1)*(a + b*ArcSinh[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -(c^2*x^2)])/(Sqrt[d]*f*(m + 1)),
x] - Simp[(b*c*(f*x)^(m + 2)*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, -(c^2*x^2)])/(Sqrt
[d]*f^2*(m + 1)*(m + 2)), x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && GtQ[d, 0] &&  !IntegerQ[m]

Rule 5764

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist
[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2], Int[((f*x)^m*(a + b*ArcSinh[c*x])^n)/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a
, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] &&  !GtQ[d, 0] && (IntegerQ[m] || EqQ[n, 1])

Rubi steps

\begin {align*} \int \frac {x^m \left (a+b \sinh ^{-1}(c x)\right )}{\left (d+c^2 d x^2\right )^{5/2}} \, dx &=\frac {x^{1+m} \left (a+b \sinh ^{-1}(c x)\right )}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {(2-m) \int \frac {x^m \left (a+b \sinh ^{-1}(c x)\right )}{\left (d+c^2 d x^2\right )^{3/2}} \, dx}{3 d}-\frac {\left (b c \sqrt {1+c^2 x^2}\right ) \int \frac {x^{1+m}}{\left (1+c^2 x^2\right )^2} \, dx}{3 d^2 \sqrt {d+c^2 d x^2}}\\ &=\frac {x^{1+m} \left (a+b \sinh ^{-1}(c x)\right )}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {(2-m) x^{1+m} \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt {d+c^2 d x^2}}-\frac {b c x^{2+m} \sqrt {1+c^2 x^2} \, _2F_1\left (2,\frac {2+m}{2};\frac {4+m}{2};-c^2 x^2\right )}{3 d^2 (2+m) \sqrt {d+c^2 d x^2}}-\frac {((2-m) m) \int \frac {x^m \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {d+c^2 d x^2}} \, dx}{3 d^2}-\frac {\left (b c (2-m) \sqrt {1+c^2 x^2}\right ) \int \frac {x^{1+m}}{1+c^2 x^2} \, dx}{3 d^2 \sqrt {d+c^2 d x^2}}\\ &=\frac {x^{1+m} \left (a+b \sinh ^{-1}(c x)\right )}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {(2-m) x^{1+m} \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt {d+c^2 d x^2}}-\frac {b c (2-m) x^{2+m} \sqrt {1+c^2 x^2} \, _2F_1\left (1,\frac {2+m}{2};\frac {4+m}{2};-c^2 x^2\right )}{3 d^2 (2+m) \sqrt {d+c^2 d x^2}}-\frac {b c x^{2+m} \sqrt {1+c^2 x^2} \, _2F_1\left (2,\frac {2+m}{2};\frac {4+m}{2};-c^2 x^2\right )}{3 d^2 (2+m) \sqrt {d+c^2 d x^2}}-\frac {\left ((2-m) m \sqrt {1+c^2 x^2}\right ) \int \frac {x^m \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{3 d^2 \sqrt {d+c^2 d x^2}}\\ &=\frac {x^{1+m} \left (a+b \sinh ^{-1}(c x)\right )}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {(2-m) x^{1+m} \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt {d+c^2 d x^2}}-\frac {(2-m) m x^{1+m} \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};-c^2 x^2\right )}{3 d^2 (1+m) \sqrt {d+c^2 d x^2}}-\frac {b c (2-m) x^{2+m} \sqrt {1+c^2 x^2} \, _2F_1\left (1,\frac {2+m}{2};\frac {4+m}{2};-c^2 x^2\right )}{3 d^2 (2+m) \sqrt {d+c^2 d x^2}}-\frac {b c x^{2+m} \sqrt {1+c^2 x^2} \, _2F_1\left (2,\frac {2+m}{2};\frac {4+m}{2};-c^2 x^2\right )}{3 d^2 (2+m) \sqrt {d+c^2 d x^2}}+\frac {b c (2-m) m x^{2+m} \sqrt {1+c^2 x^2} \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};-c^2 x^2\right )}{3 d^2 \left (2+3 m+m^2\right ) \sqrt {d+c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.42, size = 286, normalized size = 0.71 \[ \frac {x^{m+1} \left ((2-m) \left (c^2 x^2+1\right ) \left (-m \sqrt {c^2 x^2+1} \left ((m+2) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};-c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )-b c x \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;-c^2 x^2\right )\right )+(m+1) (m+2) \left (a+b \sinh ^{-1}(c x)\right )-b c (m+1) x \sqrt {c^2 x^2+1} \, _2F_1\left (1,\frac {m}{2}+1;\frac {m}{2}+2;-c^2 x^2\right )\right )+(m+1) (m+2) \left (a+b \sinh ^{-1}(c x)\right )-b c (m+1) x \left (c^2 x^2+1\right )^{3/2} \, _2F_1\left (2,\frac {m}{2}+1;\frac {m}{2}+2;-c^2 x^2\right )\right )}{3 d^2 (m+1) (m+2) \left (c^2 x^2+1\right ) \sqrt {c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*((1 + m)*(2 + m)*(a + b*ArcSinh[c*x]) - b*c*(1 + m)*x*(1 + c^2*x^2)^(3/2)*Hypergeometric2F1[2, 1 +
m/2, 2 + m/2, -(c^2*x^2)] + (2 - m)*(1 + c^2*x^2)*((1 + m)*(2 + m)*(a + b*ArcSinh[c*x]) - b*c*(1 + m)*x*Sqrt[1
+ c^2*x^2]*Hypergeometric2F1[1, 1 + m/2, 2 + m/2, -(c^2*x^2)] - m*Sqrt[1 + c^2*x^2]*((2 + m)*(a + b*ArcSinh[c
*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -(c^2*x^2)] - b*c*x*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2},
{3/2 + m/2, 2 + m/2}, -(c^2*x^2)]))))/(3*d^2*(1 + m)*(2 + m)*(1 + c^2*x^2)*Sqrt[d + c^2*d*x^2])

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c^{2} d x^{2} + d} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{m}}{c^{6} d^{3} x^{6} + 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} + d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c^2*d*x^2 + d)*(b*arcsinh(c*x) + a)*x^m/(c^6*d^3*x^6 + 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 + d^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{m}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsinh(c*x) + a)*x^m/(c^2*d*x^2 + d)^(5/2), x)

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maple [F]  time = 0.54, size = 0, normalized size = 0.00 \[ \int \frac {x^{m} \left (a +b \arcsinh \left (c x \right )\right )}{\left (c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(5/2),x)

[Out]

int(x^m*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{m}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

integrate((b*arcsinh(c*x) + a)*x^m/(c^2*d*x^2 + d)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^m\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (d\,c^2\,x^2+d\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^m*(a + b*asinh(c*x)))/(d + c^2*d*x^2)^(5/2),x)

[Out]

int((x^m*(a + b*asinh(c*x)))/(d + c^2*d*x^2)^(5/2), 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(x**m*(a+b*asinh(c*x))/(c**2*d*x**2+d)**(5/2),x)

[Out]

Timed out

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