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

Optimal. Leaf size=348 \[ -\frac {15 \sqrt {\pi } b^2 x \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) C\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \sin ^{-1}\left (i d x^2+1\right )}}{\sqrt {\pi }}\right )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}+\frac {15 \sqrt {\pi } b^2 x \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \sin ^{-1}\left (i d x^2+1\right )}}{\sqrt {\pi }}\right )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}+15 b^2 x \sqrt {a-i b \sin ^{-1}\left (1+i d x^2\right )}-\frac {5 b \sqrt {d^2 x^4-2 i d x^2} \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{3/2}}{d x}+x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{5/2} \]

[Out]

x*(a-I*b*arcsin(1+I*d*x^2))^(5/2)+15*b^2*x*FresnelS((I/b)^(1/2)*(a-I*b*arcsin(1+I*d*x^2))^(1/2)/Pi^(1/2))*(cos
h(1/2*a/b)-I*sinh(1/2*a/b))*Pi^(1/2)/(cos(1/2*arcsin(1+I*d*x^2))-sin(1/2*arcsin(1+I*d*x^2)))/(I/b)^(1/2)-15*b^
2*x*FresnelC((I/b)^(1/2)*(a-I*b*arcsin(1+I*d*x^2))^(1/2)/Pi^(1/2))*(cosh(1/2*a/b)+I*sinh(1/2*a/b))*Pi^(1/2)/(c
os(1/2*arcsin(1+I*d*x^2))-sin(1/2*arcsin(1+I*d*x^2)))/(I/b)^(1/2)-5*b*(a-I*b*arcsin(1+I*d*x^2))^(3/2)*(-2*I*d*
x^2+d^2*x^4)^(1/2)/d/x+15*b^2*x*(a-I*b*arcsin(1+I*d*x^2))^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 348, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4814, 4811} \[ -\frac {15 \sqrt {\pi } b^2 x \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \sin ^{-1}\left (1+i d x^2\right )}}{\sqrt {\pi }}\right )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}+\frac {15 \sqrt {\pi } b^2 x \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \sin ^{-1}\left (i d x^2+1\right )}}{\sqrt {\pi }}\right )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}+15 b^2 x \sqrt {a-i b \sin ^{-1}\left (1+i d x^2\right )}-\frac {5 b \sqrt {d^2 x^4-2 i d x^2} \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{3/2}}{d x}+x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{5/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

15*b^2*x*Sqrt[a - I*b*ArcSin[1 + I*d*x^2]] - (5*b*Sqrt[(-2*I)*d*x^2 + d^2*x^4]*(a - I*b*ArcSin[1 + I*d*x^2])^(
3/2))/(d*x) + x*(a - I*b*ArcSin[1 + I*d*x^2])^(5/2) + (15*b^2*Sqrt[Pi]*x*FresnelS[(Sqrt[I/b]*Sqrt[a - I*b*ArcS
in[1 + I*d*x^2]])/Sqrt[Pi]]*(Cosh[a/(2*b)] - I*Sinh[a/(2*b)]))/(Sqrt[I/b]*(Cos[ArcSin[1 + I*d*x^2]/2] - Sin[Ar
cSin[1 + I*d*x^2]/2])) - (15*b^2*Sqrt[Pi]*x*FresnelC[(Sqrt[I/b]*Sqrt[a - I*b*ArcSin[1 + I*d*x^2]])/Sqrt[Pi]]*(
Cosh[a/(2*b)] + I*Sinh[a/(2*b)]))/(Sqrt[I/b]*(Cos[ArcSin[1 + I*d*x^2]/2] - Sin[ArcSin[1 + I*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-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{5/2} \, dx &=-\frac {5 b \sqrt {-2 i d x^2+d^2 x^4} \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{3/2}}{d x}+x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{5/2}+\left (15 b^2\right ) \int \sqrt {a-i b \sin ^{-1}\left (1+i d x^2\right )} \, dx\\ &=15 b^2 x \sqrt {a-i b \sin ^{-1}\left (1+i d x^2\right )}-\frac {5 b \sqrt {-2 i d x^2+d^2 x^4} \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{3/2}}{d x}+x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{5/2}+\frac {15 b^2 \sqrt {\pi } x S\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \sin ^{-1}\left (1+i d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}-\frac {15 b^2 \sqrt {\pi } x C\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \sin ^{-1}\left (1+i d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 337, normalized size = 0.97 \[ \frac {15 b^2 x \left (-\sqrt {\pi } \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) C\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \sin ^{-1}\left (i d x^2+1\right )}}{\sqrt {\pi }}\right )+\sqrt {\pi } \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \sin ^{-1}\left (i d x^2+1\right )}}{\sqrt {\pi }}\right )+\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right ) \sqrt {a-i b \sin ^{-1}\left (1+i d x^2\right )}\right )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}+x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{5/2}-\frac {5 b \sqrt {d x^2 \left (d x^2-2 i\right )} \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{3/2}}{d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*b*Sqrt[d*x^2*(-2*I + d*x^2)]*(a - I*b*ArcSin[1 + I*d*x^2])^(3/2))/(d*x) + x*(a - I*b*ArcSin[1 + I*d*x^2])^
(5/2) + (15*b^2*x*(Sqrt[I/b]*Sqrt[a - I*b*ArcSin[1 + I*d*x^2]]*(Cos[ArcSin[1 + I*d*x^2]/2] - Sin[ArcSin[1 + I*
d*x^2]/2]) + Sqrt[Pi]*FresnelS[(Sqrt[I/b]*Sqrt[a - I*b*ArcSin[1 + I*d*x^2]])/Sqrt[Pi]]*(Cosh[a/(2*b)] - I*Sinh
[a/(2*b)]) - Sqrt[Pi]*FresnelC[(Sqrt[I/b]*Sqrt[a - I*b*ArcSin[1 + I*d*x^2]])/Sqrt[Pi]]*(Cosh[a/(2*b)] + I*Sinh
[a/(2*b)])))/(Sqrt[I/b]*(Cos[ArcSin[1 + I*d*x^2]/2] - Sin[ArcSin[1 + I*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*arcsinh(-I+d*x^2))^(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(-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*arcsinh(-I+d*x^2))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was
done assuming [d,t_nostep]=[33,44]Bad conditionned root j= 2 value -63897.972732 ratio 2.32090080903 mindist 1
0.1217065137Warning, choosing root of [1,0,%%%{-6,[2,4]%%%}+%%%{-8,[0,0]%%%},%%%{-8,[3,6]%%%}+%%%{-32,[1,2]%%%
},%%%{-3,[4,8]%%%}+%%%{-24,[2,4]%%%}+%%%{16,[0,0]%%%}] at parameters values [7,-27]Warning, need to choose a b
ranch for the root of a polynomial with parameters. This might be wrong.The choice was done assuming [d,t_nost
ep]=[50,45]Francis algorithm failure for[1.0,0.0,-61509375008,-8.30376562824e+15,-3.1528360132e+20]schur row 1
1.29492e-09Francis algorithm not precise enough for[1.0,0.0,-61509375008,-8.30376562824e+15,-3.1528360132e+20
]Bad conditionned root j= 2 value -101243.096423 ratio 1.67716116885 mindist 7.19967768138Francis algorithm fa
ilure for[1.0,0.0,-137282971022,-2.76877787308e+16,-1.57055117809e+21]Warning, choosing root of [1,0,%%%{-6,[2
,4]%%%}+%%%{-8,[0,0]%%%},%%%{-8,[3,6]%%%}+%%%{-32,[1,2]%%%},%%%{-3,[4,8]%%%}+%%%{-24,[2,4]%%%}+%%%{16,[0,0]%%%
}] at parameters values [63,-49]Evaluation time: 1.57sym2poly/r2sym(const gen & e,const index_m & i,const vect
eur & l) Error: Bad Argument Value

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int \left (a +b \arcsinh \left (d \,x^{2}-i\right )\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(-I+d*x^2))^(5/2),x)

[Out]

int((a+b*arcsinh(-I+d*x^2))^(5/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsinh(d*x^2 - I) + a)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(d*x^2 - 1i))^(5/2),x)

[Out]

int((a + b*asinh(d*x^2 - 1i))^(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((a+b*asinh(-I+d*x**2))**(5/2),x)

[Out]

Timed out

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