3.417 \(\int (a+b \sin ^{-1}(1+d x^2))^{5/2} \, dx\)

Optimal. Leaf size=277 \[ -15 b^2 x \sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}+\frac {5 b \sqrt {-d^2 x^4-2 d x^2} \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )^{3/2}}{d x}+\frac {15 \sqrt {\pi } x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) C\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{\left (\frac {1}{b}\right )^{5/2} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}-\frac {15 \sqrt {\pi } x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{\left (\frac {1}{b}\right )^{5/2} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+x \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )^{5/2} \]

[Out]

x*(a+b*arcsin(d*x^2+1))^(5/2)-15*x*FresnelS((1/b)^(1/2)*(a+b*arcsin(d*x^2+1))^(1/2)/Pi^(1/2))*(cos(1/2*a/b)-si
n(1/2*a/b))*Pi^(1/2)/(1/b)^(5/2)/(cos(1/2*arcsin(d*x^2+1))-sin(1/2*arcsin(d*x^2+1)))+15*x*FresnelC((1/b)^(1/2)
*(a+b*arcsin(d*x^2+1))^(1/2)/Pi^(1/2))*(cos(1/2*a/b)+sin(1/2*a/b))*Pi^(1/2)/(1/b)^(5/2)/(cos(1/2*arcsin(d*x^2+
1))-sin(1/2*arcsin(d*x^2+1)))+5*b*(a+b*arcsin(d*x^2+1))^(3/2)*(-d^2*x^4-2*d*x^2)^(1/2)/d/x-15*b^2*x*(a+b*arcsi
n(d*x^2+1))^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4814, 4811} \[ -15 b^2 x \sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}+\frac {5 b \sqrt {-d^2 x^4-2 d x^2} \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )^{3/2}}{d x}+\frac {15 \sqrt {\pi } x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{\left (\frac {1}{b}\right )^{5/2} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}-\frac {15 \sqrt {\pi } x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )}{\left (\frac {1}{b}\right )^{5/2} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+x \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[1 + d*x^2])^(5/2),x]

[Out]

-15*b^2*x*Sqrt[a + b*ArcSin[1 + d*x^2]] + (5*b*Sqrt[-2*d*x^2 - d^2*x^4]*(a + b*ArcSin[1 + d*x^2])^(3/2))/(d*x)
+ x*(a + b*ArcSin[1 + d*x^2])^(5/2) - (15*Sqrt[Pi]*x*FresnelS[(Sqrt[b^(-1)]*Sqrt[a + b*ArcSin[1 + d*x^2]])/Sq
rt[Pi]]*(Cos[a/(2*b)] - Sin[a/(2*b)]))/((b^(-1))^(5/2)*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2]))
+ (15*Sqrt[Pi]*x*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcSin[1 + d*x^2]])/Sqrt[Pi]]*(Cos[a/(2*b)] + Sin[a/(2*b)])
)/((b^(-1))^(5/2)*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2]))

Rule 4811

Int[Sqrt[(a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.)], x_Symbol] :> Simp[x*Sqrt[a + b*ArcSin[c + d*x^2]], x] + (
-Simp[(Sqrt[Pi]*x*(Cos[a/(2*b)] + c*Sin[a/(2*b)])*FresnelC[Sqrt[c/(Pi*b)]*Sqrt[a + b*ArcSin[c + d*x^2]]])/(Sqr
t[c/b]*(Cos[ArcSin[c + d*x^2]/2] - c*Sin[ArcSin[c + d*x^2]/2])), x] + Simp[(Sqrt[Pi]*x*(Cos[a/(2*b)] - c*Sin[a
/(2*b)])*FresnelS[Sqrt[c/(Pi*b)]*Sqrt[a + b*ArcSin[c + d*x^2]]])/(Sqrt[c/b]*(Cos[ArcSin[c + d*x^2]/2] - c*Sin[
ArcSin[c + d*x^2]/2])), x]) /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1]

Rule 4814

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(n_), x_Symbol] :> Simp[x*(a + b*ArcSin[c + d*x^2])^n, x] + (-
Dist[4*b^2*n*(n - 1), Int[(a + b*ArcSin[c + d*x^2])^(n - 2), x], x] + Simp[(2*b*n*Sqrt[-2*c*d*x^2 - d^2*x^4]*(
a + b*ArcSin[c + d*x^2])^(n - 1))/(d*x), x]) /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1] && GtQ[n, 1]

Rubi steps

\begin {align*} \int \left (a+b \sin ^{-1}\left (1+d x^2\right )\right )^{5/2} \, dx &=\frac {5 b \sqrt {-2 d x^2-d^2 x^4} \left (a+b \sin ^{-1}\left (1+d x^2\right )\right )^{3/2}}{d x}+x \left (a+b \sin ^{-1}\left (1+d x^2\right )\right )^{5/2}-\left (15 b^2\right ) \int \sqrt {a+b \sin ^{-1}\left (1+d x^2\right )} \, dx\\ &=-15 b^2 x \sqrt {a+b \sin ^{-1}\left (1+d x^2\right )}+\frac {5 b \sqrt {-2 d x^2-d^2 x^4} \left (a+b \sin ^{-1}\left (1+d x^2\right )\right )^{3/2}}{d x}+x \left (a+b \sin ^{-1}\left (1+d x^2\right )\right )^{5/2}-\frac {15 \sqrt {\pi } x S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \sin ^{-1}\left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{\left (\frac {1}{b}\right )^{5/2} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1+d x^2\right )\right )\right )}+\frac {15 \sqrt {\pi } x C\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \sin ^{-1}\left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{\left (\frac {1}{b}\right )^{5/2} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1+d x^2\right )\right )\right )}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 269, normalized size = 0.97 \[ -\frac {15 x \left (-\sqrt {\pi } \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) C\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )+\sqrt {\pi } \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}}{\sqrt {\pi }}\right )+\sqrt {\frac {1}{b}} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right ) \sqrt {a+b \sin ^{-1}\left (d x^2+1\right )}\right )}{\left (\frac {1}{b}\right )^{5/2} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (d x^2+1\right )\right )\right )}+x \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )^{5/2}+\frac {5 b \sqrt {-d x^2 \left (d x^2+2\right )} \left (a+b \sin ^{-1}\left (d x^2+1\right )\right )^{3/2}}{d x} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[1 + d*x^2])^(5/2),x]

[Out]

(5*b*Sqrt[-(d*x^2*(2 + d*x^2))]*(a + b*ArcSin[1 + d*x^2])^(3/2))/(d*x) + x*(a + b*ArcSin[1 + d*x^2])^(5/2) - (
15*x*(Sqrt[Pi]*FresnelS[(Sqrt[b^(-1)]*Sqrt[a + b*ArcSin[1 + d*x^2]])/Sqrt[Pi]]*(Cos[a/(2*b)] - Sin[a/(2*b)]) -
Sqrt[Pi]*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcSin[1 + d*x^2]])/Sqrt[Pi]]*(Cos[a/(2*b)] + Sin[a/(2*b)]) + Sqrt
[b^(-1)]*Sqrt[a + b*ArcSin[1 + d*x^2]]*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2])))/((b^(-1))^(5/2)
*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2]))

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x^2+1))^(5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \arcsin \left (d x^{2} + 1\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x^2+1))^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(d*x^2 + 1) + a)^(5/2), x)

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \left (a +b \arcsin \left (d \,x^{2}+1\right )\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(d*x^2+1))^(5/2),x)

[Out]

int((a+b*arcsin(d*x^2+1))^(5/2),x)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x^2+1))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found sqrt((-_SAGE_VAR_d*_SAGE_VAR_x^2)-2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(d*x^2 + 1))^(5/2),x)

[Out]

int((a + b*asin(d*x^2 + 1))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \operatorname {asin}{\left (d x^{2} + 1 \right )}\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(d*x**2+1))**(5/2),x)

[Out]

Integral((a + b*asin(d*x**2 + 1))**(5/2), x)

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