3.96 \(\int \sec ^3(a+b x) \tan ^4(a+b x) \, dx\)

Optimal. Leaf size=78 \[ \frac {\tanh ^{-1}(\sin (a+b x))}{16 b}+\frac {\tan ^3(a+b x) \sec ^3(a+b x)}{6 b}-\frac {\tan (a+b x) \sec ^3(a+b x)}{8 b}+\frac {\tan (a+b x) \sec (a+b x)}{16 b} \]

[Out]

1/16*arctanh(sin(b*x+a))/b+1/16*sec(b*x+a)*tan(b*x+a)/b-1/8*sec(b*x+a)^3*tan(b*x+a)/b+1/6*sec(b*x+a)^3*tan(b*x
+a)^3/b

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Rubi [A]  time = 0.07, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2611, 3768, 3770} \[ \frac {\tanh ^{-1}(\sin (a+b x))}{16 b}+\frac {\tan ^3(a+b x) \sec ^3(a+b x)}{6 b}-\frac {\tan (a+b x) \sec ^3(a+b x)}{8 b}+\frac {\tan (a+b x) \sec (a+b x)}{16 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^3*Tan[a + b*x]^4,x]

[Out]

ArcTanh[Sin[a + b*x]]/(16*b) + (Sec[a + b*x]*Tan[a + b*x])/(16*b) - (Sec[a + b*x]^3*Tan[a + b*x])/(8*b) + (Sec
[a + b*x]^3*Tan[a + b*x]^3)/(6*b)

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

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

Rubi steps

\begin {align*} \int \sec ^3(a+b x) \tan ^4(a+b x) \, dx &=\frac {\sec ^3(a+b x) \tan ^3(a+b x)}{6 b}-\frac {1}{2} \int \sec ^3(a+b x) \tan ^2(a+b x) \, dx\\ &=-\frac {\sec ^3(a+b x) \tan (a+b x)}{8 b}+\frac {\sec ^3(a+b x) \tan ^3(a+b x)}{6 b}+\frac {1}{8} \int \sec ^3(a+b x) \, dx\\ &=\frac {\sec (a+b x) \tan (a+b x)}{16 b}-\frac {\sec ^3(a+b x) \tan (a+b x)}{8 b}+\frac {\sec ^3(a+b x) \tan ^3(a+b x)}{6 b}+\frac {1}{16} \int \sec (a+b x) \, dx\\ &=\frac {\tanh ^{-1}(\sin (a+b x))}{16 b}+\frac {\sec (a+b x) \tan (a+b x)}{16 b}-\frac {\sec ^3(a+b x) \tan (a+b x)}{8 b}+\frac {\sec ^3(a+b x) \tan ^3(a+b x)}{6 b}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 99, normalized size = 1.27 \[ \frac {\tanh ^{-1}(\sin (a+b x))}{16 b}-\frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}+\frac {\tan ^3(a+b x) \sec ^3(a+b x)}{3 b}+\frac {\tan (a+b x) \sec ^3(a+b x)}{24 b}+\frac {\tan (a+b x) \sec (a+b x)}{16 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^3*Tan[a + b*x]^4,x]

[Out]

ArcTanh[Sin[a + b*x]]/(16*b) + (Sec[a + b*x]*Tan[a + b*x])/(16*b) + (Sec[a + b*x]^3*Tan[a + b*x])/(24*b) - (Se
c[a + b*x]^5*Tan[a + b*x])/(6*b) + (Sec[a + b*x]^3*Tan[a + b*x]^3)/(3*b)

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fricas [A]  time = 0.44, size = 84, normalized size = 1.08 \[ \frac {3 \, \cos \left (b x + a\right )^{6} \log \left (\sin \left (b x + a\right ) + 1\right ) - 3 \, \cos \left (b x + a\right )^{6} \log \left (-\sin \left (b x + a\right ) + 1\right ) + 2 \, {\left (3 \, \cos \left (b x + a\right )^{4} - 14 \, \cos \left (b x + a\right )^{2} + 8\right )} \sin \left (b x + a\right )}{96 \, b \cos \left (b x + a\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^7*sin(b*x+a)^4,x, algorithm="fricas")

[Out]

1/96*(3*cos(b*x + a)^6*log(sin(b*x + a) + 1) - 3*cos(b*x + a)^6*log(-sin(b*x + a) + 1) + 2*(3*cos(b*x + a)^4 -
14*cos(b*x + a)^2 + 8)*sin(b*x + a))/(b*cos(b*x + a)^6)

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giac [A]  time = 0.53, size = 73, normalized size = 0.94 \[ -\frac {\frac {2 \, {\left (3 \, \sin \left (b x + a\right )^{5} + 8 \, \sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )\right )}}{{\left (\sin \left (b x + a\right )^{2} - 1\right )}^{3}} - 3 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + 3 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{96 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^7*sin(b*x+a)^4,x, algorithm="giac")

[Out]

-1/96*(2*(3*sin(b*x + a)^5 + 8*sin(b*x + a)^3 - 3*sin(b*x + a))/(sin(b*x + a)^2 - 1)^3 - 3*log(abs(sin(b*x + a
) + 1)) + 3*log(abs(sin(b*x + a) - 1)))/b

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maple [A]  time = 0.04, size = 108, normalized size = 1.38 \[ \frac {\sin ^{5}\left (b x +a \right )}{6 b \cos \left (b x +a \right )^{6}}+\frac {\sin ^{5}\left (b x +a \right )}{24 b \cos \left (b x +a \right )^{4}}-\frac {\sin ^{5}\left (b x +a \right )}{48 b \cos \left (b x +a \right )^{2}}-\frac {\sin ^{3}\left (b x +a \right )}{48 b}-\frac {\sin \left (b x +a \right )}{16 b}+\frac {\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{16 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^7*sin(b*x+a)^4,x)

[Out]

1/6/b*sin(b*x+a)^5/cos(b*x+a)^6+1/24/b*sin(b*x+a)^5/cos(b*x+a)^4-1/48/b*sin(b*x+a)^5/cos(b*x+a)^2-1/48*sin(b*x
+a)^3/b-1/16*sin(b*x+a)/b+1/16/b*ln(sec(b*x+a)+tan(b*x+a))

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maxima [A]  time = 0.53, size = 91, normalized size = 1.17 \[ -\frac {\frac {2 \, {\left (3 \, \sin \left (b x + a\right )^{5} + 8 \, \sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )\right )}}{\sin \left (b x + a\right )^{6} - 3 \, \sin \left (b x + a\right )^{4} + 3 \, \sin \left (b x + a\right )^{2} - 1} - 3 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 3 \, \log \left (\sin \left (b x + a\right ) - 1\right )}{96 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^7*sin(b*x+a)^4,x, algorithm="maxima")

[Out]

-1/96*(2*(3*sin(b*x + a)^5 + 8*sin(b*x + a)^3 - 3*sin(b*x + a))/(sin(b*x + a)^6 - 3*sin(b*x + a)^4 + 3*sin(b*x
+ a)^2 - 1) - 3*log(sin(b*x + a) + 1) + 3*log(sin(b*x + a) - 1))/b

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mupad [B]  time = 7.38, size = 177, normalized size = 2.27 \[ \frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )\right )}{8\,b}+\frac {-\frac {{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^{11}}{8}+\frac {17\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^9}{24}+\frac {19\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^7}{4}+\frac {19\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^5}{4}+\frac {17\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^3}{24}-\frac {\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}{8}}{b\,\left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^4/cos(a + b*x)^7,x)

[Out]

atanh(tan(a/2 + (b*x)/2))/(8*b) + ((17*tan(a/2 + (b*x)/2)^3)/24 - tan(a/2 + (b*x)/2)/8 + (19*tan(a/2 + (b*x)/2
)^5)/4 + (19*tan(a/2 + (b*x)/2)^7)/4 + (17*tan(a/2 + (b*x)/2)^9)/24 - tan(a/2 + (b*x)/2)^11/8)/(b*(15*tan(a/2
+ (b*x)/2)^4 - 6*tan(a/2 + (b*x)/2)^2 - 20*tan(a/2 + (b*x)/2)^6 + 15*tan(a/2 + (b*x)/2)^8 - 6*tan(a/2 + (b*x)/
2)^10 + tan(a/2 + (b*x)/2)^12 + 1))

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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(sec(b*x+a)**7*sin(b*x+a)**4,x)

[Out]

Timed out

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