3.1.96 \(\int \sec ^3(a+b x) \tan ^4(a+b x) \, dx\) [96]

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

[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 = 0.05, 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 = {2691, 3853, 3855} \begin {gather*} \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} \end {gather*}

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 2691

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 3853

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 3855

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 = 0.02, size = 99, normalized size = 1.27 \begin {gather*} \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)}{24 b}-\frac {\sec ^5(a+b x) \tan (a+b x)}{6 b}+\frac {\sec ^3(a+b x) \tan ^3(a+b x)}{3 b} \end {gather*}

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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Maple [A]
time = 0.08, size = 94, normalized size = 1.21

method	result	size
derivativedivides	\(\frac {\frac {\sin ^{5}\left (b x +a \right )}{6 \cos \left (b x +a \right )^{6}}+\frac {\sin ^{5}\left (b x +a \right )}{24 \cos \left (b x +a \right )^{4}}-\frac {\sin ^{5}\left (b x +a \right )}{48 \cos \left (b x +a \right )^{2}}-\frac {\left (\sin ^{3}\left (b x +a \right )\right )}{48}-\frac {\sin \left (b x +a \right )}{16}+\frac {\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{16}}{b}\)	\(94\)
default	\(\frac {\frac {\sin ^{5}\left (b x +a \right )}{6 \cos \left (b x +a \right )^{6}}+\frac {\sin ^{5}\left (b x +a \right )}{24 \cos \left (b x +a \right )^{4}}-\frac {\sin ^{5}\left (b x +a \right )}{48 \cos \left (b x +a \right )^{2}}-\frac {\left (\sin ^{3}\left (b x +a \right )\right )}{48}-\frac {\sin \left (b x +a \right )}{16}+\frac {\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{16}}{b}\)	\(94\)
risch	\(-\frac {i \left (3 \,{\mathrm e}^{11 i \left (b x +a \right )}-47 \,{\mathrm e}^{9 i \left (b x +a \right )}+78 \,{\mathrm e}^{7 i \left (b x +a \right )}-78 \,{\mathrm e}^{5 i \left (b x +a \right )}+47 \,{\mathrm e}^{3 i \left (b x +a \right )}-3 \,{\mathrm e}^{i \left (b x +a \right )}\right )}{24 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{6}}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{16 b}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{16 b}\)	\(124\)
norman	\(\frac {-\frac {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{8 b}+\frac {17 \left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{24 b}+\frac {19 \left (\tan ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}+\frac {19 \left (\tan ^{7}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}+\frac {17 \left (\tan ^{9}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{24 b}-\frac {\tan ^{11}\left (\frac {b x}{2}+\frac {a}{2}\right )}{8 b}}{\left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{6}}-\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{16 b}+\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{16 b}\)	\(147\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 91, normalized size = 1.17 \begin {gather*} -\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} \end {gather*}

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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Fricas [A]
time = 0.38, size = 84, normalized size = 1.08 \begin {gather*} \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}} \end {gather*}

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

[Out]

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

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Giac [A]
time = 4.21, size = 73, normalized size = 0.94 \begin {gather*} -\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} \end {gather*}

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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Mupad [B]
time = 7.38, size = 177, normalized size = 2.27 \begin {gather*} \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 )} \end {gather*}

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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