3.2.70 \(\int \csc ^4(a+b x) \sec ^5(a+b x) \, dx\) [170]

Optimal. Leaf size=89 \[ \frac {35 \tanh ^{-1}(\sin (a+b x))}{8 b}-\frac {35 \csc (a+b x)}{8 b}-\frac {35 \csc ^3(a+b x)}{24 b}+\frac {7 \csc ^3(a+b x) \sec ^2(a+b x)}{8 b}+\frac {\csc ^3(a+b x) \sec ^4(a+b x)}{4 b} \]

[Out]

35/8*arctanh(sin(b*x+a))/b-35/8*csc(b*x+a)/b-35/24*csc(b*x+a)^3/b+7/8*csc(b*x+a)^3*sec(b*x+a)^2/b+1/4*csc(b*x+
a)^3*sec(b*x+a)^4/b

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 89, 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 = {2701, 294, 308, 213} \begin {gather*} -\frac {35 \csc ^3(a+b x)}{24 b}-\frac {35 \csc (a+b x)}{8 b}+\frac {35 \tanh ^{-1}(\sin (a+b x))}{8 b}+\frac {\csc ^3(a+b x) \sec ^4(a+b x)}{4 b}+\frac {7 \csc ^3(a+b x) \sec ^2(a+b x)}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^4*Sec[a + b*x]^5,x]

[Out]

(35*ArcTanh[Sin[a + b*x]])/(8*b) - (35*Csc[a + b*x])/(8*b) - (35*Csc[a + b*x]^3)/(24*b) + (7*Csc[a + b*x]^3*Se
c[a + b*x]^2)/(8*b) + (Csc[a + b*x]^3*Sec[a + b*x]^4)/(4*b)

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2701

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

Rubi steps

\begin {align*} \int \csc ^4(a+b x) \sec ^5(a+b x) \, dx &=-\frac {\text {Subst}\left (\int \frac {x^8}{\left (-1+x^2\right )^3} \, dx,x,\csc (a+b x)\right )}{b}\\ &=\frac {\csc ^3(a+b x) \sec ^4(a+b x)}{4 b}-\frac {7 \text {Subst}\left (\int \frac {x^6}{\left (-1+x^2\right )^2} \, dx,x,\csc (a+b x)\right )}{4 b}\\ &=\frac {7 \csc ^3(a+b x) \sec ^2(a+b x)}{8 b}+\frac {\csc ^3(a+b x) \sec ^4(a+b x)}{4 b}-\frac {35 \text {Subst}\left (\int \frac {x^4}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{8 b}\\ &=\frac {7 \csc ^3(a+b x) \sec ^2(a+b x)}{8 b}+\frac {\csc ^3(a+b x) \sec ^4(a+b x)}{4 b}-\frac {35 \text {Subst}\left (\int \left (1+x^2+\frac {1}{-1+x^2}\right ) \, dx,x,\csc (a+b x)\right )}{8 b}\\ &=-\frac {35 \csc (a+b x)}{8 b}-\frac {35 \csc ^3(a+b x)}{24 b}+\frac {7 \csc ^3(a+b x) \sec ^2(a+b x)}{8 b}+\frac {\csc ^3(a+b x) \sec ^4(a+b x)}{4 b}-\frac {35 \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{8 b}\\ &=\frac {35 \tanh ^{-1}(\sin (a+b x))}{8 b}-\frac {35 \csc (a+b x)}{8 b}-\frac {35 \csc ^3(a+b x)}{24 b}+\frac {7 \csc ^3(a+b x) \sec ^2(a+b x)}{8 b}+\frac {\csc ^3(a+b x) \sec ^4(a+b x)}{4 b}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.01, size = 31, normalized size = 0.35 \begin {gather*} -\frac {\csc ^3(a+b x) \, _2F_1\left (-\frac {3}{2},3;-\frac {1}{2};\sin ^2(a+b x)\right )}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^4*Sec[a + b*x]^5,x]

[Out]

-1/3*(Csc[a + b*x]^3*Hypergeometric2F1[-3/2, 3, -1/2, Sin[a + b*x]^2])/b

________________________________________________________________________________________

Maple [A]
time = 0.09, size = 86, normalized size = 0.97

method	result	size
derivativedivides	\(\frac {\frac {1}{4 \sin \left (b x +a \right )^{3} \cos \left (b x +a \right )^{4}}-\frac {7}{12 \sin \left (b x +a \right )^{3} \cos \left (b x +a \right )^{2}}+\frac {35}{24 \sin \left (b x +a \right ) \cos \left (b x +a \right )^{2}}-\frac {35}{8 \sin \left (b x +a \right )}+\frac {35 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{8}}{b}\)	\(86\)
default	\(\frac {\frac {1}{4 \sin \left (b x +a \right )^{3} \cos \left (b x +a \right )^{4}}-\frac {7}{12 \sin \left (b x +a \right )^{3} \cos \left (b x +a \right )^{2}}+\frac {35}{24 \sin \left (b x +a \right ) \cos \left (b x +a \right )^{2}}-\frac {35}{8 \sin \left (b x +a \right )}+\frac {35 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{8}}{b}\)	\(86\)
risch	\(-\frac {i \left (105 \,{\mathrm e}^{13 i \left (b x +a \right )}+70 \,{\mathrm e}^{11 i \left (b x +a \right )}-329 \,{\mathrm e}^{9 i \left (b x +a \right )}-204 \,{\mathrm e}^{7 i \left (b x +a \right )}-329 \,{\mathrm e}^{5 i \left (b x +a \right )}+70 \,{\mathrm e}^{3 i \left (b x +a \right )}+105 \,{\mathrm e}^{i \left (b x +a \right )}\right )}{12 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{4} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3}}-\frac {35 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{8 b}+\frac {35 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{8 b}\)	\(148\)
norman	\(\frac {-\frac {1}{24 b}-\frac {35 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{24 b}+\frac {63 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}-\frac {35 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}-\frac {35 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}+\frac {63 \left (\tan ^{10}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}-\frac {35 \left (\tan ^{12}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{24 b}-\frac {\tan ^{14}\left (\frac {b x}{2}+\frac {a}{2}\right )}{24 b}}{\left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{4} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{3}}-\frac {35 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{8 b}+\frac {35 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{8 b}\)	\(181\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^5/sin(b*x+a)^4,x,method=_RETURNVERBOSE)

[Out]

1/b*(1/4/sin(b*x+a)^3/cos(b*x+a)^4-7/12/sin(b*x+a)^3/cos(b*x+a)^2+35/24/sin(b*x+a)/cos(b*x+a)^2-35/8/sin(b*x+a
)+35/8*ln(sec(b*x+a)+tan(b*x+a)))

________________________________________________________________________________________

Maxima [A]
time = 0.28, size = 91, normalized size = 1.02 \begin {gather*} -\frac {\frac {2 \, {\left (105 \, \sin \left (b x + a\right )^{6} - 175 \, \sin \left (b x + a\right )^{4} + 56 \, \sin \left (b x + a\right )^{2} + 8\right )}}{\sin \left (b x + a\right )^{7} - 2 \, \sin \left (b x + a\right )^{5} + \sin \left (b x + a\right )^{3}} - 105 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 105 \, \log \left (\sin \left (b x + a\right ) - 1\right )}{48 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5/sin(b*x+a)^4,x, algorithm="maxima")

[Out]

-1/48*(2*(105*sin(b*x + a)^6 - 175*sin(b*x + a)^4 + 56*sin(b*x + a)^2 + 8)/(sin(b*x + a)^7 - 2*sin(b*x + a)^5
+ sin(b*x + a)^3) - 105*log(sin(b*x + a) + 1) + 105*log(sin(b*x + a) - 1))/b

________________________________________________________________________________________

Fricas [A]
time = 0.38, size = 140, normalized size = 1.57 \begin {gather*} -\frac {210 \, \cos \left (b x + a\right )^{6} - 280 \, \cos \left (b x + a\right )^{4} - 105 \, {\left (\cos \left (b x + a\right )^{6} - \cos \left (b x + a\right )^{4}\right )} \log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + 105 \, {\left (\cos \left (b x + a\right )^{6} - \cos \left (b x + a\right )^{4}\right )} \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + 42 \, \cos \left (b x + a\right )^{2} + 12}{48 \, {\left (b \cos \left (b x + a\right )^{6} - b \cos \left (b x + a\right )^{4}\right )} \sin \left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5/sin(b*x+a)^4,x, algorithm="fricas")

[Out]

-1/48*(210*cos(b*x + a)^6 - 280*cos(b*x + a)^4 - 105*(cos(b*x + a)^6 - cos(b*x + a)^4)*log(sin(b*x + a) + 1)*s
in(b*x + a) + 105*(cos(b*x + a)^6 - cos(b*x + a)^4)*log(-sin(b*x + a) + 1)*sin(b*x + a) + 42*cos(b*x + a)^2 +
12)/((b*cos(b*x + a)^6 - b*cos(b*x + a)^4)*sin(b*x + a))

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{5}{\left (a + b x \right )}}{\sin ^{4}{\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)**4,x)

[Out]

Integral(sec(a + b*x)**5/sin(a + b*x)**4, x)

________________________________________________________________________________________

Giac [A]
time = 4.78, size = 85, normalized size = 0.96 \begin {gather*} -\frac {\frac {6 \, {\left (11 \, \sin \left (b x + a\right )^{3} - 13 \, \sin \left (b x + a\right )\right )}}{{\left (\sin \left (b x + a\right )^{2} - 1\right )}^{2}} + \frac {16 \, {\left (9 \, \sin \left (b x + a\right )^{2} + 1\right )}}{\sin \left (b x + a\right )^{3}} - 105 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + 105 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{48 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5/sin(b*x+a)^4,x, algorithm="giac")

[Out]

-1/48*(6*(11*sin(b*x + a)^3 - 13*sin(b*x + a))/(sin(b*x + a)^2 - 1)^2 + 16*(9*sin(b*x + a)^2 + 1)/sin(b*x + a)
^3 - 105*log(abs(sin(b*x + a) + 1)) + 105*log(abs(sin(b*x + a) - 1)))/b

________________________________________________________________________________________

Mupad [B]
time = 0.45, size = 79, normalized size = 0.89 \begin {gather*} \frac {35\,\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{8\,b}-\frac {\frac {35\,{\sin \left (a+b\,x\right )}^6}{8}-\frac {175\,{\sin \left (a+b\,x\right )}^4}{24}+\frac {7\,{\sin \left (a+b\,x\right )}^2}{3}+\frac {1}{3}}{b\,\left ({\sin \left (a+b\,x\right )}^7-2\,{\sin \left (a+b\,x\right )}^5+{\sin \left (a+b\,x\right )}^3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(a + b*x)^5*sin(a + b*x)^4),x)

[Out]

(35*atanh(sin(a + b*x)))/(8*b) - ((7*sin(a + b*x)^2)/3 - (175*sin(a + b*x)^4)/24 + (35*sin(a + b*x)^6)/8 + 1/3
)/(b*(sin(a + b*x)^3 - 2*sin(a + b*x)^5 + sin(a + b*x)^7))

________________________________________________________________________________________