3.1.78 \(\int \sec ^5(a+b x) \tan ^3(a+b x) \, dx\) [78]

Optimal. Leaf size=31 \[ -\frac {\sec ^5(a+b x)}{5 b}+\frac {\sec ^7(a+b x)}{7 b} \]

[Out]

-1/5*sec(b*x+a)^5/b+1/7*sec(b*x+a)^7/b

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Rubi [A]
time = 0.02, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2686, 14} \begin {gather*} \frac {\sec ^7(a+b x)}{7 b}-\frac {\sec ^5(a+b x)}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^5*Tan[a + b*x]^3,x]

[Out]

-1/5*Sec[a + b*x]^5/b + Sec[a + b*x]^7/(7*b)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rubi steps

\begin {align*} \int \sec ^5(a+b x) \tan ^3(a+b x) \, dx &=\frac {\text {Subst}\left (\int x^4 \left (-1+x^2\right ) \, dx,x,\sec (a+b x)\right )}{b}\\ &=\frac {\text {Subst}\left (\int \left (-x^4+x^6\right ) \, dx,x,\sec (a+b x)\right )}{b}\\ &=-\frac {\sec ^5(a+b x)}{5 b}+\frac {\sec ^7(a+b x)}{7 b}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 31, normalized size = 1.00 \begin {gather*} -\frac {\sec ^5(a+b x)}{5 b}+\frac {\sec ^7(a+b x)}{7 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^5*Tan[a + b*x]^3,x]

[Out]

-1/5*Sec[a + b*x]^5/b + Sec[a + b*x]^7/(7*b)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(95\) vs. \(2(27)=54\).
time = 0.09, size = 96, normalized size = 3.10

method	result	size
risch	\(-\frac {32 \left (7 \,{\mathrm e}^{9 i \left (b x +a \right )}-6 \,{\mathrm e}^{7 i \left (b x +a \right )}+7 \,{\mathrm e}^{5 i \left (b x +a \right )}\right )}{35 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{7}}\)	\(53\)
derivativedivides	\(\frac {\frac {\sin ^{4}\left (b x +a \right )}{7 \cos \left (b x +a \right )^{7}}+\frac {3 \left (\sin ^{4}\left (b x +a \right )\right )}{35 \cos \left (b x +a \right )^{5}}+\frac {\sin ^{4}\left (b x +a \right )}{35 \cos \left (b x +a \right )^{3}}-\frac {\sin ^{4}\left (b x +a \right )}{35 \cos \left (b x +a \right )}-\frac {\left (2+\sin ^{2}\left (b x +a \right )\right ) \cos \left (b x +a \right )}{35}}{b}\)	\(96\)
default	\(\frac {\frac {\sin ^{4}\left (b x +a \right )}{7 \cos \left (b x +a \right )^{7}}+\frac {3 \left (\sin ^{4}\left (b x +a \right )\right )}{35 \cos \left (b x +a \right )^{5}}+\frac {\sin ^{4}\left (b x +a \right )}{35 \cos \left (b x +a \right )^{3}}-\frac {\sin ^{4}\left (b x +a \right )}{35 \cos \left (b x +a \right )}-\frac {\left (2+\sin ^{2}\left (b x +a \right )\right ) \cos \left (b x +a \right )}{35}}{b}\)	\(96\)
norman	\(\frac {-\frac {4 \left (\tan ^{10}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}-\frac {8 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}+\frac {4}{35 b}-\frac {4 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{5 b}-\frac {4 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}-\frac {8 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{5 b}}{\left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{7}}\)	\(103\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^8*sin(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

1/b*(1/7*sin(b*x+a)^4/cos(b*x+a)^7+3/35*sin(b*x+a)^4/cos(b*x+a)^5+1/35*sin(b*x+a)^4/cos(b*x+a)^3-1/35*sin(b*x+
a)^4/cos(b*x+a)-1/35*(2+sin(b*x+a)^2)*cos(b*x+a))

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Maxima [A]
time = 0.30, size = 25, normalized size = 0.81 \begin {gather*} -\frac {7 \, \cos \left (b x + a\right )^{2} - 5}{35 \, b \cos \left (b x + a\right )^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^8*sin(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/35*(7*cos(b*x + a)^2 - 5)/(b*cos(b*x + a)^7)

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Fricas [A]
time = 0.35, size = 25, normalized size = 0.81 \begin {gather*} -\frac {7 \, \cos \left (b x + a\right )^{2} - 5}{35 \, b \cos \left (b x + a\right )^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^8*sin(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/35*(7*cos(b*x + a)^2 - 5)/(b*cos(b*x + a)^7)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**8*sin(b*x+a)**3,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 6189 deep

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Giac [A]
time = 4.82, size = 25, normalized size = 0.81 \begin {gather*} -\frac {7 \, \cos \left (b x + a\right )^{2} - 5}{35 \, b \cos \left (b x + a\right )^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^8*sin(b*x+a)^3,x, algorithm="giac")

[Out]

-1/35*(7*cos(b*x + a)^2 - 5)/(b*cos(b*x + a)^7)

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Mupad [B]
time = 0.63, size = 25, normalized size = 0.81 \begin {gather*} -\frac {7\,{\cos \left (a+b\,x\right )}^2-5}{35\,b\,{\cos \left (a+b\,x\right )}^7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^3/cos(a + b*x)^8,x)

[Out]

-(7*cos(a + b*x)^2 - 5)/(35*b*cos(a + b*x)^7)

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