3.3.78 \(\int \frac {c e+d e x}{(a+b \text {ArcSin}(c+d x))^{7/2}} \, dx\) [278]

Optimal. Leaf size=252 \[ -\frac {2 e (c+d x) \sqrt {1-(c+d x)^2}}{5 b d (a+b \text {ArcSin}(c+d x))^{5/2}}-\frac {4 e}{15 b^2 d (a+b \text {ArcSin}(c+d x))^{3/2}}+\frac {8 e (c+d x)^2}{15 b^2 d (a+b \text {ArcSin}(c+d x))^{3/2}}+\frac {32 e (c+d x) \sqrt {1-(c+d x)^2}}{15 b^3 d \sqrt {a+b \text {ArcSin}(c+d x)}}-\frac {32 e \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )}{15 b^{7/2} d}-\frac {32 e \sqrt {\pi } S\left (\frac {2 \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )}{15 b^{7/2} d} \]

[Out]

-4/15*e/b^2/d/(a+b*arcsin(d*x+c))^(3/2)+8/15*e*(d*x+c)^2/b^2/d/(a+b*arcsin(d*x+c))^(3/2)-32/15*e*cos(2*a/b)*Fr
esnelC(2*(a+b*arcsin(d*x+c))^(1/2)/b^(1/2)/Pi^(1/2))*Pi^(1/2)/b^(7/2)/d-32/15*e*FresnelS(2*(a+b*arcsin(d*x+c))
^(1/2)/b^(1/2)/Pi^(1/2))*sin(2*a/b)*Pi^(1/2)/b^(7/2)/d-2/5*e*(d*x+c)*(1-(d*x+c)^2)^(1/2)/b/d/(a+b*arcsin(d*x+c
))^(5/2)+32/15*e*(d*x+c)*(1-(d*x+c)^2)^(1/2)/b^3/d/(a+b*arcsin(d*x+c))^(1/2)

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Rubi [A]
time = 0.34, antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {4889, 12, 4729, 4807, 4727, 3387, 3386, 3432, 3385, 3433, 4737} \begin {gather*} -\frac {32 \sqrt {\pi } e \cos \left (\frac {2 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {\pi } \sqrt {b}}\right )}{15 b^{7/2} d}-\frac {32 \sqrt {\pi } e \sin \left (\frac {2 a}{b}\right ) S\left (\frac {2 \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )}{15 b^{7/2} d}+\frac {32 e \sqrt {1-(c+d x)^2} (c+d x)}{15 b^3 d \sqrt {a+b \text {ArcSin}(c+d x)}}+\frac {8 e (c+d x)^2}{15 b^2 d (a+b \text {ArcSin}(c+d x))^{3/2}}-\frac {4 e}{15 b^2 d (a+b \text {ArcSin}(c+d x))^{3/2}}-\frac {2 e \sqrt {1-(c+d x)^2} (c+d x)}{5 b d (a+b \text {ArcSin}(c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)/(a + b*ArcSin[c + d*x])^(7/2),x]

[Out]

(-2*e*(c + d*x)*Sqrt[1 - (c + d*x)^2])/(5*b*d*(a + b*ArcSin[c + d*x])^(5/2)) - (4*e)/(15*b^2*d*(a + b*ArcSin[c
+ d*x])^(3/2)) + (8*e*(c + d*x)^2)/(15*b^2*d*(a + b*ArcSin[c + d*x])^(3/2)) + (32*e*(c + d*x)*Sqrt[1 - (c + d
*x)^2])/(15*b^3*d*Sqrt[a + b*ArcSin[c + d*x]]) - (32*e*Sqrt[Pi]*Cos[(2*a)/b]*FresnelC[(2*Sqrt[a + b*ArcSin[c +
d*x]])/(Sqrt[b]*Sqrt[Pi])])/(15*b^(7/2)*d) - (32*e*Sqrt[Pi]*FresnelS[(2*Sqrt[a + b*ArcSin[c + d*x]])/(Sqrt[b]
*Sqrt[Pi])]*Sin[(2*a)/b])/(15*b^(7/2)*d)

Rule 12

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

Rule 3385

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Cos[f*(x^2/d)],
x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3386

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Sin[f*(x^2/d)], x], x,
Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3387

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) +
f*x]/Sqrt[c + d*x], x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/Sqrt[c + d*x], x], x] /; FreeQ[{c
, d, e, f}, x] && ComplexFreeQ[f] && NeQ[d*e - c*f, 0]

Rule 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3433

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 4727

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[x^m*Sqrt[1 - c^2*x^2]*((a + b*ArcSin
[c*x])^(n + 1)/(b*c*(n + 1))), x] - Dist[1/(b^2*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[x^(n + 1), Sin[
-a/b + x/b]^(m - 1)*(m - (m + 1)*Sin[-a/b + x/b]^2), x], x], x, a + b*ArcSin[c*x]], x] /; FreeQ[{a, b, c}, x]
&& IGtQ[m, 0] && GeQ[n, -2] && LtQ[n, -1]

Rule 4729

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[x^m*Sqrt[1 - c^2*x^2]*((a + b*ArcSin
[c*x])^(n + 1)/(b*c*(n + 1))), x] + (Dist[c*((m + 1)/(b*(n + 1))), Int[x^(m + 1)*((a + b*ArcSin[c*x])^(n + 1)/
Sqrt[1 - c^2*x^2]), x], x] - Dist[m/(b*c*(n + 1)), Int[x^(m - 1)*((a + b*ArcSin[c*x])^(n + 1)/Sqrt[1 - c^2*x^2
]), x], x]) /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && LtQ[n, -2]

Rule 4737

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*Si
mp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSin[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c
^2*d + e, 0] && NeQ[n, -1]

Rule 4807

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

Rule 4889

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + f*(x/d))^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rubi steps

\begin {align*} \int \frac {c e+d e x}{\left (a+b \sin ^{-1}(c+d x)\right )^{7/2}} \, dx &=\frac {\text {Subst}\left (\int \frac {e x}{\left (a+b \sin ^{-1}(x)\right )^{7/2}} \, dx,x,c+d x\right )}{d}\\ &=\frac {e \text {Subst}\left (\int \frac {x}{\left (a+b \sin ^{-1}(x)\right )^{7/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 e (c+d x) \sqrt {1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}+\frac {(2 e) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \left (a+b \sin ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{5 b d}-\frac {(4 e) \text {Subst}\left (\int \frac {x^2}{\sqrt {1-x^2} \left (a+b \sin ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{5 b d}\\ &=-\frac {2 e (c+d x) \sqrt {1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac {4 e}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {8 e (c+d x)^2}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac {(16 e) \text {Subst}\left (\int \frac {x}{\left (a+b \sin ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{15 b^2 d}\\ &=-\frac {2 e (c+d x) \sqrt {1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac {4 e}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {8 e (c+d x)^2}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {32 e (c+d x) \sqrt {1-(c+d x)^2}}{15 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {(32 e) \text {Subst}\left (\int \frac {\cos (2 x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{15 b^3 d}\\ &=-\frac {2 e (c+d x) \sqrt {1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac {4 e}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {8 e (c+d x)^2}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {32 e (c+d x) \sqrt {1-(c+d x)^2}}{15 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {\left (32 e \cos \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {2 a}{b}+2 x\right )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{15 b^3 d}-\frac {\left (32 e \sin \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {2 a}{b}+2 x\right )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{15 b^3 d}\\ &=-\frac {2 e (c+d x) \sqrt {1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac {4 e}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {8 e (c+d x)^2}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {32 e (c+d x) \sqrt {1-(c+d x)^2}}{15 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {\left (64 e \cos \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \cos \left (\frac {2 x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c+d x)}\right )}{15 b^4 d}-\frac {\left (64 e \sin \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \sin \left (\frac {2 x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c+d x)}\right )}{15 b^4 d}\\ &=-\frac {2 e (c+d x) \sqrt {1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac {4 e}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {8 e (c+d x)^2}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {32 e (c+d x) \sqrt {1-(c+d x)^2}}{15 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {32 e \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) C\left (\frac {2 \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )}{15 b^{7/2} d}-\frac {32 e \sqrt {\pi } S\left (\frac {2 \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )}{15 b^{7/2} d}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.57, size = 254, normalized size = 1.01 \begin {gather*} -\frac {e \left ((a+b \text {ArcSin}(c+d x)) \left (e^{-\frac {2 i a}{b}} \left (2 e^{\frac {2 i (a+b \text {ArcSin}(c+d x))}{b}} (4 i a+b+4 i b \text {ArcSin}(c+d x))+8 \sqrt {2} b \left (-\frac {i (a+b \text {ArcSin}(c+d x))}{b}\right )^{3/2} \text {Gamma}\left (\frac {1}{2},-\frac {2 i (a+b \text {ArcSin}(c+d x))}{b}\right )\right )+2 e^{-2 i \text {ArcSin}(c+d x)} \left (-4 i a+b-4 i b \text {ArcSin}(c+d x)+4 \sqrt {2} b e^{\frac {2 i (a+b \text {ArcSin}(c+d x))}{b}} \left (\frac {i (a+b \text {ArcSin}(c+d x))}{b}\right )^{3/2} \text {Gamma}\left (\frac {1}{2},\frac {2 i (a+b \text {ArcSin}(c+d x))}{b}\right )\right )\right )+3 b^2 \sin (2 \text {ArcSin}(c+d x))\right )}{15 b^3 d (a+b \text {ArcSin}(c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)/(a + b*ArcSin[c + d*x])^(7/2),x]

[Out]

-1/15*(e*((a + b*ArcSin[c + d*x])*((2*E^(((2*I)*(a + b*ArcSin[c + d*x]))/b)*((4*I)*a + b + (4*I)*b*ArcSin[c +
d*x]) + 8*Sqrt[2]*b*(((-I)*(a + b*ArcSin[c + d*x]))/b)^(3/2)*Gamma[1/2, ((-2*I)*(a + b*ArcSin[c + d*x]))/b])/E
^(((2*I)*a)/b) + (2*((-4*I)*a + b - (4*I)*b*ArcSin[c + d*x] + 4*Sqrt[2]*b*E^(((2*I)*(a + b*ArcSin[c + d*x]))/b
)*((I*(a + b*ArcSin[c + d*x]))/b)^(3/2)*Gamma[1/2, ((2*I)*(a + b*ArcSin[c + d*x]))/b]))/E^((2*I)*ArcSin[c + d*
x])) + 3*b^2*Sin[2*ArcSin[c + d*x]]))/(b^3*d*(a + b*ArcSin[c + d*x])^(5/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(642\) vs. \(2(208)=416\).
time = 0.28, size = 643, normalized size = 2.55

method	result	size
default	\(-\frac {e \left (16 \arcsin \left (d x +c \right )^{2} \sqrt {a +b \arcsin \left (d x +c \right )}\, \cos \left (\frac {2 a}{b}\right ) \FresnelC \left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) \sqrt {-\frac {2}{b}}\, \sqrt {2}\, \sqrt {\pi }\, b^{2}-16 \arcsin \left (d x +c \right )^{2} \sqrt {a +b \arcsin \left (d x +c \right )}\, \sin \left (\frac {2 a}{b}\right ) \mathrm {S}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) \sqrt {-\frac {2}{b}}\, \sqrt {2}\, \sqrt {\pi }\, b^{2}+32 \arcsin \left (d x +c \right ) \sqrt {a +b \arcsin \left (d x +c \right )}\, \cos \left (\frac {2 a}{b}\right ) \FresnelC \left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) \sqrt {-\frac {2}{b}}\, \sqrt {2}\, \sqrt {\pi }\, a b -32 \arcsin \left (d x +c \right ) \sqrt {a +b \arcsin \left (d x +c \right )}\, \sin \left (\frac {2 a}{b}\right ) \mathrm {S}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) \sqrt {-\frac {2}{b}}\, \sqrt {2}\, \sqrt {\pi }\, a b +16 \sqrt {a +b \arcsin \left (d x +c \right )}\, \cos \left (\frac {2 a}{b}\right ) \FresnelC \left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) \sqrt {-\frac {2}{b}}\, \sqrt {2}\, \sqrt {\pi }\, a^{2}-16 \sqrt {a +b \arcsin \left (d x +c \right )}\, \sin \left (\frac {2 a}{b}\right ) \mathrm {S}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) \sqrt {-\frac {2}{b}}\, \sqrt {2}\, \sqrt {\pi }\, a^{2}+16 \arcsin \left (d x +c \right )^{2} \sin \left (-\frac {2 \left (a +b \arcsin \left (d x +c \right )\right )}{b}+\frac {2 a}{b}\right ) b^{2}+32 \arcsin \left (d x +c \right ) \sin \left (-\frac {2 \left (a +b \arcsin \left (d x +c \right )\right )}{b}+\frac {2 a}{b}\right ) a b +4 \arcsin \left (d x +c \right ) \cos \left (-\frac {2 \left (a +b \arcsin \left (d x +c \right )\right )}{b}+\frac {2 a}{b}\right ) b^{2}+16 \sin \left (-\frac {2 \left (a +b \arcsin \left (d x +c \right )\right )}{b}+\frac {2 a}{b}\right ) a^{2}-3 \sin \left (-\frac {2 \left (a +b \arcsin \left (d x +c \right )\right )}{b}+\frac {2 a}{b}\right ) b^{2}+4 \cos \left (-\frac {2 \left (a +b \arcsin \left (d x +c \right )\right )}{b}+\frac {2 a}{b}\right ) a b \right )}{15 d \,b^{3} \left (a +b \arcsin \left (d x +c \right )\right )^{\frac {5}{2}}}\)	\(643\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)/(a+b*arcsin(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

-1/15/d*e/b^3*(16*arcsin(d*x+c)^2*(-2/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*cos(2*a/b)*FresnelC(2*2^(1/2)/Pi^(1/2
)/(-2/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*2^(1/2)*Pi^(1/2)*b^2-16*arcsin(d*x+c)^2*(a+b*arcsin(d*x+c))^(1/2)*
sin(2*a/b)*FresnelS(2*2^(1/2)/Pi^(1/2)/(-2/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*(-2/b)^(1/2)*2^(1/2)*Pi^(1/2)
*b^2+32*arcsin(d*x+c)*(-2/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*cos(2*a/b)*FresnelC(2*2^(1/2)/Pi^(1/2)/(-2/b)^(1/
2)*(a+b*arcsin(d*x+c))^(1/2)/b)*2^(1/2)*Pi^(1/2)*a*b-32*arcsin(d*x+c)*(a+b*arcsin(d*x+c))^(1/2)*sin(2*a/b)*Fre
snelS(2*2^(1/2)/Pi^(1/2)/(-2/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*(-2/b)^(1/2)*2^(1/2)*Pi^(1/2)*a*b+16*(-2/b)
^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*cos(2*a/b)*FresnelC(2*2^(1/2)/Pi^(1/2)/(-2/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)
/b)*2^(1/2)*Pi^(1/2)*a^2-16*(a+b*arcsin(d*x+c))^(1/2)*sin(2*a/b)*FresnelS(2*2^(1/2)/Pi^(1/2)/(-2/b)^(1/2)*(a+b
*arcsin(d*x+c))^(1/2)/b)*(-2/b)^(1/2)*2^(1/2)*Pi^(1/2)*a^2+16*arcsin(d*x+c)^2*sin(-2*(a+b*arcsin(d*x+c))/b+2*a
/b)*b^2+32*arcsin(d*x+c)*sin(-2*(a+b*arcsin(d*x+c))/b+2*a/b)*a*b+4*arcsin(d*x+c)*cos(-2*(a+b*arcsin(d*x+c))/b+
2*a/b)*b^2+16*sin(-2*(a+b*arcsin(d*x+c))/b+2*a/b)*a^2-3*sin(-2*(a+b*arcsin(d*x+c))/b+2*a/b)*b^2+4*cos(-2*(a+b*
arcsin(d*x+c))/b+2*a/b)*a*b)/(a+b*arcsin(d*x+c))^(5/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)/(a+b*arcsin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((d*x*e + c*e)/(b*arcsin(d*x + c) + a)^(7/2), x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)/(a+b*arcsin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} e \left (\int \frac {c}{a^{3} \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} + 3 a^{2} b \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} \operatorname {asin}{\left (c + d x \right )} + 3 a b^{2} \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} \operatorname {asin}^{2}{\left (c + d x \right )} + b^{3} \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} \operatorname {asin}^{3}{\left (c + d x \right )}}\, dx + \int \frac {d x}{a^{3} \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} + 3 a^{2} b \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} \operatorname {asin}{\left (c + d x \right )} + 3 a b^{2} \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} \operatorname {asin}^{2}{\left (c + d x \right )} + b^{3} \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} \operatorname {asin}^{3}{\left (c + d x \right )}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)/(a+b*asin(d*x+c))**(7/2),x)

[Out]

e*(Integral(c/(a**3*sqrt(a + b*asin(c + d*x)) + 3*a**2*b*sqrt(a + b*asin(c + d*x))*asin(c + d*x) + 3*a*b**2*sq
rt(a + b*asin(c + d*x))*asin(c + d*x)**2 + b**3*sqrt(a + b*asin(c + d*x))*asin(c + d*x)**3), x) + Integral(d*x
/(a**3*sqrt(a + b*asin(c + d*x)) + 3*a**2*b*sqrt(a + b*asin(c + d*x))*asin(c + d*x) + 3*a*b**2*sqrt(a + b*asin
(c + d*x))*asin(c + d*x)**2 + b**3*sqrt(a + b*asin(c + d*x))*asin(c + d*x)**3), x))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)/(a+b*arcsin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)/(b*arcsin(d*x + c) + a)^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {c\,e+d\,e\,x}{{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)/(a + b*asin(c + d*x))^(7/2),x)

[Out]

int((c*e + d*e*x)/(a + b*asin(c + d*x))^(7/2), x)

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