3.93 \(\int \frac{1}{(a+b \cos ^{-1}(1+d x^2))^{7/2}} \, dx\)

Optimal. Leaf size=269 \[ -\frac{\sqrt{-d^2 x^4-2 d x^2}}{15 b^3 d x \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}+\frac{x}{15 b^2 \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^{3/2}}+\frac{\sqrt{-d^2 x^4-2 d x^2}}{5 b d x \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^{5/2}}+\frac{2 \sqrt{\pi } \left (\frac{1}{b}\right )^{7/2} \sin \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{15 d x}-\frac{2 \sqrt{\pi } \left (\frac{1}{b}\right )^{7/2} \cos \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{15 d x} \]

[Out]

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

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Rubi [A]  time = 0.060164, antiderivative size = 269, normalized size of antiderivative = 1., 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 = {4829, 4823} \[ -\frac{\sqrt{-d^2 x^4-2 d x^2}}{15 b^3 d x \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}+\frac{x}{15 b^2 \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^{3/2}}+\frac{\sqrt{-d^2 x^4-2 d x^2}}{5 b d x \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^{5/2}}+\frac{2 \sqrt{\pi } \left (\frac{1}{b}\right )^{7/2} \sin \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{15 d x}-\frac{2 \sqrt{\pi } \left (\frac{1}{b}\right )^{7/2} \cos \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{15 d x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCos[1 + d*x^2])^(-7/2),x]

[Out]

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

Rule 4829

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

Rule 4823

Int[((a_.) + ArcCos[1 + (d_.)*(x_)^2]*(b_.))^(-3/2), x_Symbol] :> Simp[Sqrt[-2*d*x^2 - d^2*x^4]/(b*d*x*Sqrt[a
+ b*ArcCos[1 + d*x^2]]), x] + (-Simp[(2*(1/b)^(3/2)*Sqrt[Pi]*Sin[a/(2*b)]*Sin[ArcCos[1 + d*x^2]/2]*FresnelC[Sq
rt[1/(Pi*b)]*Sqrt[a + b*ArcCos[1 + d*x^2]]])/(d*x), x] + Simp[(2*(1/b)^(3/2)*Sqrt[Pi]*Cos[a/(2*b)]*Sin[ArcCos[
1 + d*x^2]/2]*FresnelS[Sqrt[1/(Pi*b)]*Sqrt[a + b*ArcCos[1 + d*x^2]]])/(d*x), x]) /; FreeQ[{a, b, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.546504, size = 308, normalized size = 1.14 \[ -\frac{2 \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \left (a^2 \cos \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right )-\sqrt{\pi } \sqrt{\frac{1}{b}} \sin \left (\frac{a}{2 b}\right ) \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^{5/2} \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )+\sqrt{\pi } \sqrt{\frac{1}{b}} \cos \left (\frac{a}{2 b}\right ) \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^{5/2} S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )+2 a b \cos ^{-1}\left (d x^2+1\right ) \cos \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right )+a b \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right )-3 b^2 \cos \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right )+b^2 \cos ^{-1}\left (d x^2+1\right )^2 \cos \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right )+b^2 \cos ^{-1}\left (d x^2+1\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right )\right )}{15 b^3 d x \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCos[1 + d*x^2])^(-7/2),x]

[Out]

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

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Maple [F]  time = 0.068, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\arccos \left ( d{x}^{2}+1 \right ) \right ) ^{-{\frac{7}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*arccos(d*x^2+1))^(7/2),x)

[Out]

int(1/(a+b*arccos(d*x^2+1))^(7/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \arccos \left (d x^{2} + 1\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arccos(d*x^2+1))^(7/2),x, algorithm="maxima")

[Out]

integrate((b*arccos(d*x^2 + 1) + a)^(-7/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arccos(d*x^2+1))^(7/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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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(1/(a+b*acos(d*x**2+1))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \arccos \left (d x^{2} + 1\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arccos(d*x^2+1))^(7/2),x, algorithm="giac")

[Out]

integrate((b*arccos(d*x^2 + 1) + a)^(-7/2), x)