Integrand size = 23, antiderivative size = 160 \[ \int \frac {\cosh ^6(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\frac {x}{b^3}-\frac {\sqrt {a-b} \left (8 a^2+4 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} b^3 d}-\frac {(a-b) \tanh (c+d x)}{4 a b d \left (a-(a-b) \tanh ^2(c+d x)\right )^2}-\frac {(a-b) (4 a+3 b) \tanh (c+d x)}{8 a^2 b^2 d \left (a-(a-b) \tanh ^2(c+d x)\right )} \]
x/b^3-1/8*(8*a^2+4*a*b+3*b^2)*arctanh((a-b)^(1/2)*tanh(d*x+c)/a^(1/2))*(a- b)^(1/2)/a^(5/2)/b^3/d-1/4*(a-b)*tanh(d*x+c)/a/b/d/(a-(a-b)*tanh(d*x+c)^2) ^2-1/8*(a-b)*(4*a+3*b)*tanh(d*x+c)/a^2/b^2/d/(a-(a-b)*tanh(d*x+c)^2)
Time = 1.30 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.02 \[ \int \frac {\cosh ^6(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\frac {8 (c+d x)-\frac {\left (8 a^3-4 a^2 b-a b^2-3 b^3\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {a}}\right )}{a^{5/2} \sqrt {a-b}}+\frac {4 (a-b)^2 b \sinh (2 (c+d x))}{a (2 a-b+b \cosh (2 (c+d x)))^2}+\frac {3 b \left (-2 a^2+a b+b^2\right ) \sinh (2 (c+d x))}{a^2 (2 a-b+b \cosh (2 (c+d x)))}}{8 b^3 d} \]
(8*(c + d*x) - ((8*a^3 - 4*a^2*b - a*b^2 - 3*b^3)*ArcTanh[(Sqrt[a - b]*Tan h[c + d*x])/Sqrt[a]])/(a^(5/2)*Sqrt[a - b]) + (4*(a - b)^2*b*Sinh[2*(c + d *x)])/(a*(2*a - b + b*Cosh[2*(c + d*x)])^2) + (3*b*(-2*a^2 + a*b + b^2)*Si nh[2*(c + d*x)])/(a^2*(2*a - b + b*Cosh[2*(c + d*x)])))/(8*b^3*d)
Time = 0.42 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.13, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 3670, 316, 25, 402, 25, 397, 219, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\cosh ^6(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (i c+i d x)^6}{\left (a-b \sin (i c+i d x)^2\right )^3}dx\) |
\(\Big \downarrow \) 3670 |
\(\displaystyle \frac {\int \frac {1}{\left (1-\tanh ^2(c+d x)\right ) \left (a-(a-b) \tanh ^2(c+d x)\right )^3}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 316 |
\(\displaystyle \frac {-\frac {\int -\frac {3 (a-b) \tanh ^2(c+d x)+a+3 b}{\left (1-\tanh ^2(c+d x)\right ) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}d\tanh (c+d x)}{4 a b}-\frac {(a-b) \tanh (c+d x)}{4 a b \left (a-(a-b) \tanh ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\int \frac {3 (a-b) \tanh ^2(c+d x)+a+3 b}{\left (1-\tanh ^2(c+d x)\right ) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}d\tanh (c+d x)}{4 a b}-\frac {(a-b) \tanh (c+d x)}{4 a b \left (a-(a-b) \tanh ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 402 |
\(\displaystyle \frac {\frac {\frac {\left (-\frac {4 a}{b}+\frac {3 b}{a}+1\right ) \tanh (c+d x)}{2 \left (a-(a-b) \tanh ^2(c+d x)\right )}-\frac {\int -\frac {4 a^2+b a+3 b^2+(a-b) (4 a+3 b) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (a-(a-b) \tanh ^2(c+d x)\right )}d\tanh (c+d x)}{2 a b}}{4 a b}-\frac {(a-b) \tanh (c+d x)}{4 a b \left (a-(a-b) \tanh ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\frac {\int \frac {4 a^2+b a+3 b^2+(a-b) (4 a+3 b) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (a-(a-b) \tanh ^2(c+d x)\right )}d\tanh (c+d x)}{2 a b}+\frac {\left (-\frac {4 a}{b}+\frac {3 b}{a}+1\right ) \tanh (c+d x)}{2 \left (a-(a-b) \tanh ^2(c+d x)\right )}}{4 a b}-\frac {(a-b) \tanh (c+d x)}{4 a b \left (a-(a-b) \tanh ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 397 |
\(\displaystyle \frac {\frac {\frac {\frac {8 a^2 \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{b}-\frac {(a-b) \left (8 a^2+4 a b+3 b^2\right ) \int \frac {1}{a-(a-b) \tanh ^2(c+d x)}d\tanh (c+d x)}{b}}{2 a b}+\frac {\left (-\frac {4 a}{b}+\frac {3 b}{a}+1\right ) \tanh (c+d x)}{2 \left (a-(a-b) \tanh ^2(c+d x)\right )}}{4 a b}-\frac {(a-b) \tanh (c+d x)}{4 a b \left (a-(a-b) \tanh ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\frac {\frac {8 a^2 \text {arctanh}(\tanh (c+d x))}{b}-\frac {(a-b) \left (8 a^2+4 a b+3 b^2\right ) \int \frac {1}{a-(a-b) \tanh ^2(c+d x)}d\tanh (c+d x)}{b}}{2 a b}+\frac {\left (-\frac {4 a}{b}+\frac {3 b}{a}+1\right ) \tanh (c+d x)}{2 \left (a-(a-b) \tanh ^2(c+d x)\right )}}{4 a b}-\frac {(a-b) \tanh (c+d x)}{4 a b \left (a-(a-b) \tanh ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {\frac {\frac {8 a^2 \text {arctanh}(\tanh (c+d x))}{b}-\frac {\sqrt {a-b} \left (8 a^2+4 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} b}}{2 a b}+\frac {\left (-\frac {4 a}{b}+\frac {3 b}{a}+1\right ) \tanh (c+d x)}{2 \left (a-(a-b) \tanh ^2(c+d x)\right )}}{4 a b}-\frac {(a-b) \tanh (c+d x)}{4 a b \left (a-(a-b) \tanh ^2(c+d x)\right )^2}}{d}\) |
(-1/4*((a - b)*Tanh[c + d*x])/(a*b*(a - (a - b)*Tanh[c + d*x]^2)^2) + (((8 *a^2*ArcTanh[Tanh[c + d*x]])/b - (Sqrt[a - b]*(8*a^2 + 4*a*b + 3*b^2)*ArcT anh[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[a]])/(Sqrt[a]*b))/(2*a*b) + ((1 - (4* a)/b + (3*b)/a)*Tanh[c + d*x])/(2*(a - (a - b)*Tanh[c + d*x]^2)))/(4*a*b)) /d
3.4.39.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Su bst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]
Leaf count of result is larger than twice the leaf count of optimal. \(428\) vs. \(2(146)=292\).
Time = 0.15 (sec) , antiderivative size = 429, normalized size of antiderivative = 2.68
\[\frac {\frac {\frac {2 \left (-\frac {b \left (4 a^{2}+a b -5 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{8 a}+\frac {\left (4 a^{3}-23 a^{2} b +7 a \,b^{2}+12 b^{3}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 a^{2}}+\frac {\left (4 a^{3}-23 a^{2} b +7 a \,b^{2}+12 b^{3}\right ) b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8 a^{2}}-\frac {b \left (4 a^{2}+a b -5 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a}\right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a \right )^{2}}+\frac {\left (8 a^{3}-4 a^{2} b -a \,b^{2}-3 b^{3}\right ) \left (\frac {\left (\sqrt {-b \left (a -b \right )}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}-\frac {\left (\sqrt {-b \left (a -b \right )}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{4 a}}{b^{3}}+\frac {\ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{3}}}{d}\]
1/d*(2/b^3*((-1/8*b*(4*a^2+a*b-5*b^2)/a*tanh(1/2*d*x+1/2*c)^7+1/8*(4*a^3-2 3*a^2*b+7*a*b^2+12*b^3)/a^2*b*tanh(1/2*d*x+1/2*c)^5+1/8*(4*a^3-23*a^2*b+7* a*b^2+12*b^3)/a^2*b*tanh(1/2*d*x+1/2*c)^3-1/8*b*(4*a^2+a*b-5*b^2)/a*tanh(1 /2*d*x+1/2*c))/(tanh(1/2*d*x+1/2*c)^4*a-2*tanh(1/2*d*x+1/2*c)^2*a+4*b*tanh (1/2*d*x+1/2*c)^2+a)^2+1/8/a*(8*a^3-4*a^2*b-a*b^2-3*b^3)*(1/2*((-b*(a-b))^ (1/2)+b)/a/(-b*(a-b))^(1/2)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2)*arctan(a* tanh(1/2*d*x+1/2*c)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2))-1/2*((-b*(a-b))^ (1/2)-b)/a/(-b*(a-b))^(1/2)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2)*arctanh(a *tanh(1/2*d*x+1/2*c)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2))))+1/b^3*ln(1+ta nh(1/2*d*x+1/2*c))-1/b^3*ln(tanh(1/2*d*x+1/2*c)-1))
Leaf count of result is larger than twice the leaf count of optimal. 2623 vs. \(2 (148) = 296\).
Time = 0.32 (sec) , antiderivative size = 5511, normalized size of antiderivative = 34.44 \[ \int \frac {\cosh ^6(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {\cosh ^6(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\cosh ^6(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more details)Is
\[ \int \frac {\cosh ^6(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\int { \frac {\cosh \left (d x + c\right )^{6}}{{\left (b \sinh \left (d x + c\right )^{2} + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\cosh ^6(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^3} \, dx=\int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^6}{{\left (b\,{\mathrm {sinh}\left (c+d\,x\right )}^2+a\right )}^3} \,d x \]