3.221 \(\int (a-a \sec ^2(c+d x))^{5/2} \, dx\)

Optimal. Leaf size=101 \[ -\frac {a^2 \tan (c+d x) \sqrt {-a \tan ^2(c+d x)}}{2 d}+\frac {a^2 \tan ^3(c+d x) \sqrt {-a \tan ^2(c+d x)}}{4 d}-\frac {a^2 \cot (c+d x) \sqrt {-a \tan ^2(c+d x)} \log (\cos (c+d x))}{d} \]

[Out]

-a^2*cot(d*x+c)*ln(cos(d*x+c))*(-a*tan(d*x+c)^2)^(1/2)/d-1/2*a^2*(-a*tan(d*x+c)^2)^(1/2)*tan(d*x+c)/d+1/4*a^2*
(-a*tan(d*x+c)^2)^(1/2)*tan(d*x+c)^3/d

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Rubi [A]  time = 0.05, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {4121, 3658, 3473, 3475} \[ \frac {a^2 \tan ^3(c+d x) \sqrt {-a \tan ^2(c+d x)}}{4 d}-\frac {a^2 \tan (c+d x) \sqrt {-a \tan ^2(c+d x)}}{2 d}-\frac {a^2 \cot (c+d x) \sqrt {-a \tan ^2(c+d x)} \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a - a*Sec[c + d*x]^2)^(5/2),x]

[Out]

-((a^2*Cot[c + d*x]*Log[Cos[c + d*x]]*Sqrt[-(a*Tan[c + d*x]^2)])/d) - (a^2*Tan[c + d*x]*Sqrt[-(a*Tan[c + d*x]^
2)])/(2*d) + (a^2*Tan[c + d*x]^3*Sqrt[-(a*Tan[c + d*x]^2)])/(4*d)

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 4121

Int[(u_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(b*tan[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

Rubi steps

\begin {align*} \int \left (a-a \sec ^2(c+d x)\right )^{5/2} \, dx &=\int \left (-a \tan ^2(c+d x)\right )^{5/2} \, dx\\ &=\left (a^2 \cot (c+d x) \sqrt {-a \tan ^2(c+d x)}\right ) \int \tan ^5(c+d x) \, dx\\ &=\frac {a^2 \tan ^3(c+d x) \sqrt {-a \tan ^2(c+d x)}}{4 d}-\left (a^2 \cot (c+d x) \sqrt {-a \tan ^2(c+d x)}\right ) \int \tan ^3(c+d x) \, dx\\ &=-\frac {a^2 \tan (c+d x) \sqrt {-a \tan ^2(c+d x)}}{2 d}+\frac {a^2 \tan ^3(c+d x) \sqrt {-a \tan ^2(c+d x)}}{4 d}+\left (a^2 \cot (c+d x) \sqrt {-a \tan ^2(c+d x)}\right ) \int \tan (c+d x) \, dx\\ &=-\frac {a^2 \cot (c+d x) \log (\cos (c+d x)) \sqrt {-a \tan ^2(c+d x)}}{d}-\frac {a^2 \tan (c+d x) \sqrt {-a \tan ^2(c+d x)}}{2 d}+\frac {a^2 \tan ^3(c+d x) \sqrt {-a \tan ^2(c+d x)}}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.54, size = 60, normalized size = 0.59 \[ -\frac {\cot ^5(c+d x) \left (-a \tan ^2(c+d x)\right )^{5/2} \left (-\tan ^4(c+d x)+2 \tan ^2(c+d x)+4 \log (\cos (c+d x))\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - a*Sec[c + d*x]^2)^(5/2),x]

[Out]

-1/4*(Cot[c + d*x]^5*(-(a*Tan[c + d*x]^2))^(5/2)*(4*Log[Cos[c + d*x]] + 2*Tan[c + d*x]^2 - Tan[c + d*x]^4))/d

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fricas [A]  time = 0.89, size = 87, normalized size = 0.86 \[ -\frac {{\left (4 \, a^{2} \cos \left (d x + c\right )^{4} \log \left (-\cos \left (d x + c\right )\right ) + 4 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right )^{2} - a}{\cos \left (d x + c\right )^{2}}}}{4 \, d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*a^2*cos(d*x + c)^4*log(-cos(d*x + c)) + 4*a^2*cos(d*x + c)^2 - a^2)*sqrt((a*cos(d*x + c)^2 - a)/cos(d*
x + c)^2)/(d*cos(d*x + c)^3*sin(d*x + c))

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giac [A]  time = 0.49, size = 182, normalized size = 1.80 \[ -\frac {2 \, \sqrt {-a} a^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} + 2\right ) - 2 \, \sqrt {-a} a^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} - 2\right ) + \frac {3 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}\right )}^{2} \sqrt {-a} a^{2} - 20 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}\right )} \sqrt {-a} a^{2} + 44 \, \sqrt {-a} a^{2}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} - 2\right )}^{2}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*sqrt(-a)*a^2*log(tan(1/2*d*x + 1/2*c)^2 + 1/tan(1/2*d*x + 1/2*c)^2 + 2) - 2*sqrt(-a)*a^2*log(tan(1/2*d
*x + 1/2*c)^2 + 1/tan(1/2*d*x + 1/2*c)^2 - 2) + (3*(tan(1/2*d*x + 1/2*c)^2 + 1/tan(1/2*d*x + 1/2*c)^2)^2*sqrt(
-a)*a^2 - 20*(tan(1/2*d*x + 1/2*c)^2 + 1/tan(1/2*d*x + 1/2*c)^2)*sqrt(-a)*a^2 + 44*sqrt(-a)*a^2)/(tan(1/2*d*x
+ 1/2*c)^2 + 1/tan(1/2*d*x + 1/2*c)^2 - 2)^2)/d

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maple [A]  time = 1.99, size = 158, normalized size = 1.56 \[ -\frac {\left (4 \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\frac {-\sin \left (d x +c \right )-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )+4 \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-4 \ln \left (\frac {2}{1+\cos \left (d x +c \right )}\right ) \left (\cos ^{4}\left (d x +c \right )\right )-3 \left (\cos ^{4}\left (d x +c \right )\right )+4 \left (\cos ^{2}\left (d x +c \right )\right )-1\right ) \cos \left (d x +c \right ) \left (-\frac {a \left (\sin ^{2}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{2}}\right )^{\frac {5}{2}}}{4 d \sin \left (d x +c \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a-a*sec(d*x+c)^2)^(5/2),x)

[Out]

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

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maxima [A]  time = 0.45, size = 62, normalized size = 0.61 \[ \frac {\sqrt {-a} a^{2} \tan \left (d x + c\right )^{4} - 2 \, \sqrt {-a} a^{2} \tan \left (d x + c\right )^{2} + 2 \, \sqrt {-a} a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(-a)*a^2*tan(d*x + c)^4 - 2*sqrt(-a)*a^2*tan(d*x + c)^2 + 2*sqrt(-a)*a^2*log(tan(d*x + c)^2 + 1))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a-\frac {a}{{\cos \left (c+d\,x\right )}^2}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - a/cos(c + d*x)^2)^(5/2),x)

[Out]

int((a - a/cos(c + d*x)^2)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*sec(d*x+c)**2)**(5/2),x)

[Out]

Integral((-a*sec(c + d*x)**2 + a)**(5/2), x)

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