3.1.48 \(\int (a \sec ^2(x))^{5/2} \, dx\) [48]

Optimal. Leaf size=65 \[ \frac {3}{8} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \tan (x)}{\sqrt {a \sec ^2(x)}}\right )+\frac {3}{8} a^2 \sqrt {a \sec ^2(x)} \tan (x)+\frac {1}{4} a \left (a \sec ^2(x)\right )^{3/2} \tan (x) \]

[Out]

3/8*a^(5/2)*arctanh(a^(1/2)*tan(x)/(a*sec(x)^2)^(1/2))+1/4*a*(a*sec(x)^2)^(3/2)*tan(x)+3/8*a^2*(a*sec(x)^2)^(1
/2)*tan(x)

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Rubi [A]
time = 0.02, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4207, 201, 223, 212} \begin {gather*} \frac {3}{8} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \tan (x)}{\sqrt {a \sec ^2(x)}}\right )+\frac {3}{8} a^2 \tan (x) \sqrt {a \sec ^2(x)}+\frac {1}{4} a \tan (x) \left (a \sec ^2(x)\right )^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*Sec[x]^2)^(5/2),x]

[Out]

(3*a^(5/2)*ArcTanh[(Sqrt[a]*Tan[x])/Sqrt[a*Sec[x]^2]])/8 + (3*a^2*Sqrt[a*Sec[x]^2]*Tan[x])/8 + (a*(a*Sec[x]^2)
^(3/2)*Tan[x])/4

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 4207

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[b*(ff/
f), Subst[Int[(b + b*ff^2*x^2)^(p - 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{b, e, f, p}, x] &&  !IntegerQ[p
]

Rubi steps

\begin {align*} \int \left (a \sec ^2(x)\right )^{5/2} \, dx &=a \text {Subst}\left (\int \left (a+a x^2\right )^{3/2} \, dx,x,\tan (x)\right )\\ &=\frac {1}{4} a \left (a \sec ^2(x)\right )^{3/2} \tan (x)+\frac {1}{4} \left (3 a^2\right ) \text {Subst}\left (\int \sqrt {a+a x^2} \, dx,x,\tan (x)\right )\\ &=\frac {3}{8} a^2 \sqrt {a \sec ^2(x)} \tan (x)+\frac {1}{4} a \left (a \sec ^2(x)\right )^{3/2} \tan (x)+\frac {1}{8} \left (3 a^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+a x^2}} \, dx,x,\tan (x)\right )\\ &=\frac {3}{8} a^2 \sqrt {a \sec ^2(x)} \tan (x)+\frac {1}{4} a \left (a \sec ^2(x)\right )^{3/2} \tan (x)+\frac {1}{8} \left (3 a^3\right ) \text {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {\tan (x)}{\sqrt {a \sec ^2(x)}}\right )\\ &=\frac {3}{8} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \tan (x)}{\sqrt {a \sec ^2(x)}}\right )+\frac {3}{8} a^2 \sqrt {a \sec ^2(x)} \tan (x)+\frac {1}{4} a \left (a \sec ^2(x)\right )^{3/2} \tan (x)\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 72, normalized size = 1.11 \begin {gather*} \frac {1}{16} \cos ^5(x) \left (a \sec ^2(x)\right )^{5/2} \left (-6 \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+6 \log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )+\frac {1}{2} \sec ^4(x) (11 \sin (x)+3 \sin (3 x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*Sec[x]^2)^(5/2),x]

[Out]

(Cos[x]^5*(a*Sec[x]^2)^(5/2)*(-6*Log[Cos[x/2] - Sin[x/2]] + 6*Log[Cos[x/2] + Sin[x/2]] + (Sec[x]^4*(11*Sin[x]
+ 3*Sin[3*x]))/2))/16

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Maple [A]
time = 0.27, size = 66, normalized size = 1.02

method	result	size
default	\(\frac {\left (3 \left (\cos ^{4}\left (x \right )\right ) \ln \left (-\frac {\cos \left (x \right )-1-\sin \left (x \right )}{\sin \left (x \right )}\right )-3 \left (\cos ^{4}\left (x \right )\right ) \ln \left (-\frac {\cos \left (x \right )-1+\sin \left (x \right )}{\sin \left (x \right )}\right )+3 \left (\cos ^{2}\left (x \right )\right ) \sin \left (x \right )+2 \sin \left (x \right )\right ) \cos \left (x \right ) \left (\frac {a}{\cos \left (x \right )^{2}}\right )^{\frac {5}{2}}}{8}\)	\(66\)
risch	\(-\frac {i a^{2} \sqrt {\frac {a \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{2}}}\, \left (3 \,{\mathrm e}^{6 i x}+11 \,{\mathrm e}^{4 i x}-11 \,{\mathrm e}^{2 i x}-3\right )}{4 \left ({\mathrm e}^{2 i x}+1\right )^{3}}+\frac {3 a^{2} \sqrt {\frac {a \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{2}}}\, \ln \left ({\mathrm e}^{i x}+i\right ) \cos \left (x \right )}{4}-\frac {3 a^{2} \sqrt {\frac {a \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{2}}}\, \ln \left ({\mathrm e}^{i x}-i\right ) \cos \left (x \right )}{4}\)	\(126\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sec(x)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/8*(3*cos(x)^4*ln(-(cos(x)-1-sin(x))/sin(x))-3*cos(x)^4*ln(-(cos(x)-1+sin(x))/sin(x))+3*cos(x)^2*sin(x)+2*sin
(x))*cos(x)*(a/cos(x)^2)^(5/2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 1111 vs. \(2 (49) = 98\).
time = 0.85, size = 1111, normalized size = 17.09 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(x)^2)^(5/2),x, algorithm="maxima")

[Out]

1/16*(176*a^2*cos(3*x)*sin(2*x) + 48*a^2*cos(x)*sin(2*x) - 48*a^2*cos(2*x)*sin(x) - 12*a^2*sin(x) + 4*(3*a^2*s
in(7*x) + 11*a^2*sin(5*x) - 11*a^2*sin(3*x) - 3*a^2*sin(x))*cos(8*x) - 24*(2*a^2*sin(6*x) + 3*a^2*sin(4*x) + 2
*a^2*sin(2*x))*cos(7*x) + 16*(11*a^2*sin(5*x) - 11*a^2*sin(3*x) - 3*a^2*sin(x))*cos(6*x) - 88*(3*a^2*sin(4*x)
+ 2*a^2*sin(2*x))*cos(5*x) - 24*(11*a^2*sin(3*x) + 3*a^2*sin(x))*cos(4*x) + 3*(a^2*cos(8*x)^2 + 16*a^2*cos(6*x
)^2 + 36*a^2*cos(4*x)^2 + 16*a^2*cos(2*x)^2 + a^2*sin(8*x)^2 + 16*a^2*sin(6*x)^2 + 36*a^2*sin(4*x)^2 + 48*a^2*
sin(4*x)*sin(2*x) + 16*a^2*sin(2*x)^2 + 8*a^2*cos(2*x) + a^2 + 2*(4*a^2*cos(6*x) + 6*a^2*cos(4*x) + 4*a^2*cos(
2*x) + a^2)*cos(8*x) + 8*(6*a^2*cos(4*x) + 4*a^2*cos(2*x) + a^2)*cos(6*x) + 12*(4*a^2*cos(2*x) + a^2)*cos(4*x)
+ 4*(2*a^2*sin(6*x) + 3*a^2*sin(4*x) + 2*a^2*sin(2*x))*sin(8*x) + 16*(3*a^2*sin(4*x) + 2*a^2*sin(2*x))*sin(6*
x))*log(cos(x)^2 + sin(x)^2 + 2*sin(x) + 1) - 3*(a^2*cos(8*x)^2 + 16*a^2*cos(6*x)^2 + 36*a^2*cos(4*x)^2 + 16*a
^2*cos(2*x)^2 + a^2*sin(8*x)^2 + 16*a^2*sin(6*x)^2 + 36*a^2*sin(4*x)^2 + 48*a^2*sin(4*x)*sin(2*x) + 16*a^2*sin
(2*x)^2 + 8*a^2*cos(2*x) + a^2 + 2*(4*a^2*cos(6*x) + 6*a^2*cos(4*x) + 4*a^2*cos(2*x) + a^2)*cos(8*x) + 8*(6*a^
2*cos(4*x) + 4*a^2*cos(2*x) + a^2)*cos(6*x) + 12*(4*a^2*cos(2*x) + a^2)*cos(4*x) + 4*(2*a^2*sin(6*x) + 3*a^2*s
in(4*x) + 2*a^2*sin(2*x))*sin(8*x) + 16*(3*a^2*sin(4*x) + 2*a^2*sin(2*x))*sin(6*x))*log(cos(x)^2 + sin(x)^2 -
2*sin(x) + 1) - 4*(3*a^2*cos(7*x) + 11*a^2*cos(5*x) - 11*a^2*cos(3*x) - 3*a^2*cos(x))*sin(8*x) + 12*(4*a^2*cos
(6*x) + 6*a^2*cos(4*x) + 4*a^2*cos(2*x) + a^2)*sin(7*x) - 16*(11*a^2*cos(5*x) - 11*a^2*cos(3*x) - 3*a^2*cos(x)
)*sin(6*x) + 44*(6*a^2*cos(4*x) + 4*a^2*cos(2*x) + a^2)*sin(5*x) + 24*(11*a^2*cos(3*x) + 3*a^2*cos(x))*sin(4*x
) - 44*(4*a^2*cos(2*x) + a^2)*sin(3*x))*sqrt(a)/(2*(4*cos(6*x) + 6*cos(4*x) + 4*cos(2*x) + 1)*cos(8*x) + cos(8
*x)^2 + 8*(6*cos(4*x) + 4*cos(2*x) + 1)*cos(6*x) + 16*cos(6*x)^2 + 12*(4*cos(2*x) + 1)*cos(4*x) + 36*cos(4*x)^
2 + 16*cos(2*x)^2 + 4*(2*sin(6*x) + 3*sin(4*x) + 2*sin(2*x))*sin(8*x) + sin(8*x)^2 + 16*(3*sin(4*x) + 2*sin(2*
x))*sin(6*x) + 16*sin(6*x)^2 + 36*sin(4*x)^2 + 48*sin(4*x)*sin(2*x) + 16*sin(2*x)^2 + 8*cos(2*x) + 1)

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Fricas [A]
time = 2.96, size = 56, normalized size = 0.86 \begin {gather*} -\frac {{\left (3 \, a^{2} \cos \left (x\right )^{4} \log \left (-\frac {\sin \left (x\right ) - 1}{\sin \left (x\right ) + 1}\right ) - 2 \, {\left (3 \, a^{2} \cos \left (x\right )^{2} + 2 \, a^{2}\right )} \sin \left (x\right )\right )} \sqrt {\frac {a}{\cos \left (x\right )^{2}}}}{16 \, \cos \left (x\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(x)^2)^(5/2),x, algorithm="fricas")

[Out]

-1/16*(3*a^2*cos(x)^4*log(-(sin(x) - 1)/(sin(x) + 1)) - 2*(3*a^2*cos(x)^2 + 2*a^2)*sin(x))*sqrt(a/cos(x)^2)/co
s(x)^3

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \sec ^{2}{\left (x \right )}\right )^{\frac {5}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(x)**2)**(5/2),x)

[Out]

Integral((a*sec(x)**2)**(5/2), x)

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Giac [A]
time = 0.42, size = 67, normalized size = 1.03 \begin {gather*} \frac {1}{16} \, {\left (3 \, a^{2} \log \left (\sin \left (x\right ) + 1\right ) \mathrm {sgn}\left (\cos \left (x\right )\right ) - 3 \, a^{2} \log \left (-\sin \left (x\right ) + 1\right ) \mathrm {sgn}\left (\cos \left (x\right )\right ) - \frac {2 \, {\left (3 \, a^{2} \mathrm {sgn}\left (\cos \left (x\right )\right ) \sin \left (x\right )^{3} - 5 \, a^{2} \mathrm {sgn}\left (\cos \left (x\right )\right ) \sin \left (x\right )\right )}}{{\left (\sin \left (x\right )^{2} - 1\right )}^{2}}\right )} \sqrt {a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(x)^2)^(5/2),x, algorithm="giac")

[Out]

1/16*(3*a^2*log(sin(x) + 1)*sgn(cos(x)) - 3*a^2*log(-sin(x) + 1)*sgn(cos(x)) - 2*(3*a^2*sgn(cos(x))*sin(x)^3 -
5*a^2*sgn(cos(x))*sin(x))/(sin(x)^2 - 1)^2)*sqrt(a)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int {\left (\frac {a}{{\cos \left (x\right )}^2}\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a/cos(x)^2)^(5/2),x)

[Out]

int((a/cos(x)^2)^(5/2), x)

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