3.1.64 \(\int \sec ^3(a+b x) \tan ^2(a+b x) \, dx\) [64]

Optimal. Leaf size=55 \[ -\frac {\tanh ^{-1}(\sin (a+b x))}{8 b}-\frac {\sec (a+b x) \tan (a+b x)}{8 b}+\frac {\sec ^3(a+b x) \tan (a+b x)}{4 b} \]

[Out]

-1/8*arctanh(sin(b*x+a))/b-1/8*sec(b*x+a)*tan(b*x+a)/b+1/4*sec(b*x+a)^3*tan(b*x+a)/b

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Rubi [A]
time = 0.03, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, 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))}{8 b}+\frac {\tan (a+b x) \sec ^3(a+b x)}{4 b}-\frac {\tan (a+b x) \sec (a+b x)}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^3*Tan[a + b*x]^2,x]

[Out]

-1/8*ArcTanh[Sin[a + b*x]]/b - (Sec[a + b*x]*Tan[a + b*x])/(8*b) + (Sec[a + b*x]^3*Tan[a + b*x])/(4*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 ^2(a+b x) \, dx &=\frac {\sec ^3(a+b x) \tan (a+b x)}{4 b}-\frac {1}{4} \int \sec ^3(a+b x) \, dx\\ &=-\frac {\sec (a+b x) \tan (a+b x)}{8 b}+\frac {\sec ^3(a+b x) \tan (a+b x)}{4 b}-\frac {1}{8} \int \sec (a+b x) \, dx\\ &=-\frac {\tanh ^{-1}(\sin (a+b x))}{8 b}-\frac {\sec (a+b x) \tan (a+b x)}{8 b}+\frac {\sec ^3(a+b x) \tan (a+b x)}{4 b}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 55, normalized size = 1.00 \begin {gather*} -\frac {\tanh ^{-1}(\sin (a+b x))}{8 b}-\frac {\sec (a+b x) \tan (a+b x)}{8 b}+\frac {\sec ^3(a+b x) \tan (a+b x)}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^3*Tan[a + b*x]^2,x]

[Out]

-1/8*ArcTanh[Sin[a + b*x]]/b - (Sec[a + b*x]*Tan[a + b*x])/(8*b) + (Sec[a + b*x]^3*Tan[a + b*x])/(4*b)

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Maple [A]
time = 0.07, size = 66, normalized size = 1.20

method	result	size
derivativedivides	\(\frac {\frac {\sin ^{3}\left (b x +a \right )}{4 \cos \left (b x +a \right )^{4}}+\frac {\sin ^{3}\left (b x +a \right )}{8 \cos \left (b x +a \right )^{2}}+\frac {\sin \left (b x +a \right )}{8}-\frac {\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{8}}{b}\)	\(66\)
default	\(\frac {\frac {\sin ^{3}\left (b x +a \right )}{4 \cos \left (b x +a \right )^{4}}+\frac {\sin ^{3}\left (b x +a \right )}{8 \cos \left (b x +a \right )^{2}}+\frac {\sin \left (b x +a \right )}{8}-\frac {\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{8}}{b}\)	\(66\)
risch	\(\frac {i \left ({\mathrm e}^{7 i \left (b x +a \right )}-7 \,{\mathrm e}^{5 i \left (b x +a \right )}+7 \,{\mathrm e}^{3 i \left (b x +a \right )}-{\mathrm e}^{i \left (b x +a \right )}\right )}{4 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{8 b}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{8 b}\)	\(100\)
norman	\(\frac {\frac {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{4 b}+\frac {7 \left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}+\frac {7 \left (\tan ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}+\frac {\tan ^{7}\left (\frac {b x}{2}+\frac {a}{2}\right )}{4 b}}{\left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{4}}+\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{8 b}-\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{8 b}\)	\(115\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^5*sin(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

1/b*(1/4*sin(b*x+a)^3/cos(b*x+a)^4+1/8*sin(b*x+a)^3/cos(b*x+a)^2+1/8*sin(b*x+a)-1/8*ln(sec(b*x+a)+tan(b*x+a)))

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Maxima [A]
time = 0.30, size = 65, normalized size = 1.18 \begin {gather*} \frac {\frac {2 \, {\left (\sin \left (b x + a\right )^{3} + \sin \left (b x + a\right )\right )}}{\sin \left (b x + a\right )^{4} - 2 \, \sin \left (b x + a\right )^{2} + 1} - \log \left (\sin \left (b x + a\right ) + 1\right ) + \log \left (\sin \left (b x + a\right ) - 1\right )}{16 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

1/16*(2*(sin(b*x + a)^3 + sin(b*x + a))/(sin(b*x + a)^4 - 2*sin(b*x + a)^2 + 1) - log(sin(b*x + a) + 1) + log(
sin(b*x + a) - 1))/b

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Fricas [A]
time = 0.42, size = 71, normalized size = 1.29 \begin {gather*} -\frac {\cos \left (b x + a\right )^{4} \log \left (\sin \left (b x + a\right ) + 1\right ) - \cos \left (b x + a\right )^{4} \log \left (-\sin \left (b x + a\right ) + 1\right ) + 2 \, {\left (\cos \left (b x + a\right )^{2} - 2\right )} \sin \left (b x + a\right )}{16 \, b \cos \left (b x + a\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5*sin(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/16*(cos(b*x + a)^4*log(sin(b*x + a) + 1) - cos(b*x + a)^4*log(-sin(b*x + a) + 1) + 2*(cos(b*x + a)^2 - 2)*s
in(b*x + a))/(b*cos(b*x + a)^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sin ^{2}{\left (a + b x \right )} \sec ^{5}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**5*sin(b*x+a)**2,x)

[Out]

Integral(sin(a + b*x)**2*sec(a + b*x)**5, x)

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Giac [A]
time = 4.25, size = 82, normalized size = 1.49 \begin {gather*} \frac {\frac {4 \, {\left (\frac {1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right )\right )}}{{\left (\frac {1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right )\right )}^{2} - 4} - \log \left ({\left | \frac {1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right ) + 2 \right |}\right ) + \log \left ({\left | \frac {1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right ) - 2 \right |}\right )}{32 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5*sin(b*x+a)^2,x, algorithm="giac")

[Out]

1/32*(4*(1/sin(b*x + a) + sin(b*x + a))/((1/sin(b*x + a) + sin(b*x + a))^2 - 4) - log(abs(1/sin(b*x + a) + sin
(b*x + a) + 2)) + log(abs(1/sin(b*x + a) + sin(b*x + a) - 2)))/b

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Mupad [B]
time = 6.54, size = 125, normalized size = 2.27 \begin {gather*} \frac {\frac {{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^7}{4}+\frac {7\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^5}{4}+\frac {7\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^3}{4}+\frac {\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}{4}}{b\,\left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2+1\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )\right )}{4\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^2/cos(a + b*x)^5,x)

[Out]

(tan(a/2 + (b*x)/2)/4 + (7*tan(a/2 + (b*x)/2)^3)/4 + (7*tan(a/2 + (b*x)/2)^5)/4 + tan(a/2 + (b*x)/2)^7/4)/(b*(
6*tan(a/2 + (b*x)/2)^4 - 4*tan(a/2 + (b*x)/2)^2 - 4*tan(a/2 + (b*x)/2)^6 + tan(a/2 + (b*x)/2)^8 + 1)) - atanh(
tan(a/2 + (b*x)/2))/(4*b)

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