3.220 \(\int (a-a \sec ^2(c+d x))^{7/2} \, dx\)

Optimal. Leaf size=134 \[ -\frac {a^3 \tan (c+d x) \sqrt {-a \tan ^2(c+d x)}}{2 d}-\frac {a^3 \tan ^5(c+d x) \sqrt {-a \tan ^2(c+d x)}}{6 d}+\frac {a^3 \tan ^3(c+d x) \sqrt {-a \tan ^2(c+d x)}}{4 d}-\frac {a^3 \cot (c+d x) \sqrt {-a \tan ^2(c+d x)} \log (\cos (c+d x))}{d} \]

[Out]

-a^3*cot(d*x+c)*ln(cos(d*x+c))*(-a*tan(d*x+c)^2)^(1/2)/d-1/2*a^3*(-a*tan(d*x+c)^2)^(1/2)*tan(d*x+c)/d+1/4*a^3*
(-a*tan(d*x+c)^2)^(1/2)*tan(d*x+c)^3/d-1/6*a^3*(-a*tan(d*x+c)^2)^(1/2)*tan(d*x+c)^5/d

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Rubi [A]  time = 0.06, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {4121, 3658, 3473, 3475} \[ -\frac {a^3 \tan ^5(c+d x) \sqrt {-a \tan ^2(c+d x)}}{6 d}+\frac {a^3 \tan ^3(c+d x) \sqrt {-a \tan ^2(c+d x)}}{4 d}-\frac {a^3 \tan (c+d x) \sqrt {-a \tan ^2(c+d x)}}{2 d}-\frac {a^3 \cot (c+d x) \sqrt {-a \tan ^2(c+d x)} \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a - a*Sec[c + d*x]^2)^(7/2),x]

[Out]

-((a^3*Cot[c + d*x]*Log[Cos[c + d*x]]*Sqrt[-(a*Tan[c + d*x]^2)])/d) - (a^3*Tan[c + d*x]*Sqrt[-(a*Tan[c + d*x]^
2)])/(2*d) + (a^3*Tan[c + d*x]^3*Sqrt[-(a*Tan[c + d*x]^2)])/(4*d) - (a^3*Tan[c + d*x]^5*Sqrt[-(a*Tan[c + d*x]^
2)])/(6*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 )^{7/2} \, dx &=\int \left (-a \tan ^2(c+d x)\right )^{7/2} \, dx\\ &=-\left (\left (a^3 \cot (c+d x) \sqrt {-a \tan ^2(c+d x)}\right ) \int \tan ^7(c+d x) \, dx\right )\\ &=-\frac {a^3 \tan ^5(c+d x) \sqrt {-a \tan ^2(c+d x)}}{6 d}+\left (a^3 \cot (c+d x) \sqrt {-a \tan ^2(c+d x)}\right ) \int \tan ^5(c+d x) \, dx\\ &=\frac {a^3 \tan ^3(c+d x) \sqrt {-a \tan ^2(c+d x)}}{4 d}-\frac {a^3 \tan ^5(c+d x) \sqrt {-a \tan ^2(c+d x)}}{6 d}-\left (a^3 \cot (c+d x) \sqrt {-a \tan ^2(c+d x)}\right ) \int \tan ^3(c+d x) \, dx\\ &=-\frac {a^3 \tan (c+d x) \sqrt {-a \tan ^2(c+d x)}}{2 d}+\frac {a^3 \tan ^3(c+d x) \sqrt {-a \tan ^2(c+d x)}}{4 d}-\frac {a^3 \tan ^5(c+d x) \sqrt {-a \tan ^2(c+d x)}}{6 d}+\left (a^3 \cot (c+d x) \sqrt {-a \tan ^2(c+d x)}\right ) \int \tan (c+d x) \, dx\\ &=-\frac {a^3 \cot (c+d x) \log (\cos (c+d x)) \sqrt {-a \tan ^2(c+d x)}}{d}-\frac {a^3 \tan (c+d x) \sqrt {-a \tan ^2(c+d x)}}{2 d}+\frac {a^3 \tan ^3(c+d x) \sqrt {-a \tan ^2(c+d x)}}{4 d}-\frac {a^3 \tan ^5(c+d x) \sqrt {-a \tan ^2(c+d x)}}{6 d}\\ \end {align*}

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Mathematica [A]  time = 2.16, size = 70, normalized size = 0.52 \[ \frac {\cot ^7(c+d x) \left (-a \tan ^2(c+d x)\right )^{7/2} \left (2 \tan ^6(c+d x)-3 \tan ^4(c+d x)+6 \tan ^2(c+d x)+12 \log (\cos (c+d x))\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - a*Sec[c + d*x]^2)^(7/2),x]

[Out]

(Cot[c + d*x]^7*(-(a*Tan[c + d*x]^2))^(7/2)*(12*Log[Cos[c + d*x]] + 6*Tan[c + d*x]^2 - 3*Tan[c + d*x]^4 + 2*Ta
n[c + d*x]^6))/(12*d)

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fricas [A]  time = 0.45, size = 100, normalized size = 0.75 \[ -\frac {{\left (12 \, a^{3} \cos \left (d x + c\right )^{6} \log \left (-\cos \left (d x + c\right )\right ) + 18 \, a^{3} \cos \left (d x + c\right )^{4} - 9 \, a^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3}\right )} \sqrt {\frac {a \cos \left (d x + c\right )^{2} - a}{\cos \left (d x + c\right )^{2}}}}{12 \, d \cos \left (d x + c\right )^{5} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*sec(d*x+c)^2)^(7/2),x, algorithm="fricas")

[Out]

-1/12*(12*a^3*cos(d*x + c)^6*log(-cos(d*x + c)) + 18*a^3*cos(d*x + c)^4 - 9*a^3*cos(d*x + c)^2 + 2*a^3)*sqrt((
a*cos(d*x + c)^2 - a)/cos(d*x + c)^2)/(d*cos(d*x + c)^5*sin(d*x + c))

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giac [A]  time = 0.71, size = 217, normalized size = 1.62 \[ -\frac {6 \, \sqrt {-a} a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} + 2\right ) - 6 \, \sqrt {-a} a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} - 2\right ) + \frac {11 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}\right )}^{3} \sqrt {-a} a^{3} - 90 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}\right )}^{2} \sqrt {-a} a^{3} + 276 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}\right )} \sqrt {-a} a^{3} - 408 \, \sqrt {-a} a^{3}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} - 2\right )}^{3}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*sec(d*x+c)^2)^(7/2),x, algorithm="giac")

[Out]

-1/12*(6*sqrt(-a)*a^3*log(tan(1/2*d*x + 1/2*c)^2 + 1/tan(1/2*d*x + 1/2*c)^2 + 2) - 6*sqrt(-a)*a^3*log(tan(1/2*
d*x + 1/2*c)^2 + 1/tan(1/2*d*x + 1/2*c)^2 - 2) + (11*(tan(1/2*d*x + 1/2*c)^2 + 1/tan(1/2*d*x + 1/2*c)^2)^3*sqr
t(-a)*a^3 - 90*(tan(1/2*d*x + 1/2*c)^2 + 1/tan(1/2*d*x + 1/2*c)^2)^2*sqrt(-a)*a^3 + 276*(tan(1/2*d*x + 1/2*c)^
2 + 1/tan(1/2*d*x + 1/2*c)^2)*sqrt(-a)*a^3 - 408*sqrt(-a)*a^3)/(tan(1/2*d*x + 1/2*c)^2 + 1/tan(1/2*d*x + 1/2*c
)^2 - 2)^3)/d

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maple [A]  time = 2.33, size = 168, normalized size = 1.25 \[ \frac {\left (12 \left (\cos ^{6}\left (d x +c \right )\right ) \ln \left (-\frac {-\sin \left (d x +c \right )-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-12 \left (\cos ^{6}\left (d x +c \right )\right ) \ln \left (\frac {2}{1+\cos \left (d x +c \right )}\right )+12 \left (\cos ^{6}\left (d x +c \right )\right ) \ln \left (-\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-11 \left (\cos ^{6}\left (d x +c \right )\right )+18 \left (\cos ^{4}\left (d x +c \right )\right )-9 \left (\cos ^{2}\left (d x +c \right )\right )+2\right ) \cos \left (d x +c \right ) \left (-\frac {a \left (\sin ^{2}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{2}}\right )^{\frac {7}{2}}}{12 d \sin \left (d x +c \right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a-a*sec(d*x+c)^2)^(7/2),x)

[Out]

1/12/d*(12*cos(d*x+c)^6*ln(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))-12*cos(d*x+c)^6*ln(2/(1+cos(d*x+c)))+12*cos
(d*x+c)^6*ln(-(-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))-11*cos(d*x+c)^6+18*cos(d*x+c)^4-9*cos(d*x+c)^2+2)*cos(d*x
+c)*(-a*sin(d*x+c)^2/cos(d*x+c)^2)^(7/2)/sin(d*x+c)^7

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maxima [A]  time = 0.45, size = 81, normalized size = 0.60 \[ -\frac {2 \, \sqrt {-a} a^{3} \tan \left (d x + c\right )^{6} - 3 \, \sqrt {-a} a^{3} \tan \left (d x + c\right )^{4} + 6 \, \sqrt {-a} a^{3} \tan \left (d x + c\right )^{2} - 6 \, \sqrt {-a} a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*sec(d*x+c)^2)^(7/2),x, algorithm="maxima")

[Out]

-1/12*(2*sqrt(-a)*a^3*tan(d*x + c)^6 - 3*sqrt(-a)*a^3*tan(d*x + c)^4 + 6*sqrt(-a)*a^3*tan(d*x + c)^2 - 6*sqrt(
-a)*a^3*log(tan(d*x + c)^2 + 1))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a-\frac {a}{{\cos \left (c+d\,x\right )}^2}\right )}^{7/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - a/cos(c + d*x)^2)^(7/2),x)

[Out]

int((a - a/cos(c + d*x)^2)^(7/2), 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-a*sec(d*x+c)**2)**(7/2),x)

[Out]

Timed out

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