Integrand size = 27, antiderivative size = 132 \[ \int \frac {-a+b \tan (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {(i a-b) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{3/2} d}-\frac {(i a+b) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{3/2} d}+\frac {4 a b}{\left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}} \]
(I*a-b)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/(a-I*b)^(3/2)/d-(I*a +b)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/(a+I*b)^(3/2)/d+4*a*b/(a ^2+b^2)/d/(a+b*tan(d*x+c))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.43 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.17 \[ \int \frac {-a+b \tan (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=-\frac {i \cos (c+d x) \left ((a+i b)^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a-i b}\right )-(a-i b)^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a+i b}\right )\right ) (a-b \tan (c+d x))}{(a-i b) (a+i b) d (a \cos (c+d x)-b \sin (c+d x)) \sqrt {a+b \tan (c+d x)}} \]
((-I)*Cos[c + d*x]*((a + I*b)^2*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tan [c + d*x])/(a - I*b)] - (a - I*b)^2*Hypergeometric2F1[-1/2, 1, 1/2, (a + b *Tan[c + d*x])/(a + I*b)])*(a - b*Tan[c + d*x]))/((a - I*b)*(a + I*b)*d*(a *Cos[c + d*x] - b*Sin[c + d*x])*Sqrt[a + b*Tan[c + d*x]])
Time = 0.61 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.98, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {3042, 4012, 25, 3042, 4022, 3042, 4020, 25, 73, 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.
\(\displaystyle \int \frac {b \tan (c+d x)-a}{(a+b \tan (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {b \tan (c+d x)-a}{(a+b \tan (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle \frac {\int -\frac {a^2-2 b \tan (c+d x) a-b^2}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}+\frac {4 a b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {4 a b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\int \frac {a^2-2 b \tan (c+d x) a-b^2}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 a b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\int \frac {a^2-2 b \tan (c+d x) a-b^2}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle \frac {4 a b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\frac {1}{2} (a-i b)^2 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b)^2 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 a b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\frac {1}{2} (a-i b)^2 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b)^2 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle \frac {4 a b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\frac {i (a+i b)^2 \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i (a-i b)^2 \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}}{a^2+b^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {4 a b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\frac {i (a-i b)^2 \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}-\frac {i (a+i b)^2 \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}}{a^2+b^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {4 a b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\frac {(a-i b)^2 \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {(a+i b)^2 \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}}{a^2+b^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {4 a b}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\frac {(a-i b)^2 \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}+\frac {(a+i b)^2 \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}}{a^2+b^2}\) |
-((((a + I*b)^2*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d) + (( a - I*b)^2*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d))/(a^2 + b ^2)) + (4*a*b)/((a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]])
3.4.71.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ (f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x] )^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a , b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 ]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f) Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2 Int[(a + b*Tan[e + f*x])^m*( 1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2 Int[(a + b*Tan[e + f*x])^m *(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !IntegerQ[m]
Leaf count of result is larger than twice the leaf count of optimal. \(2290\) vs. \(2(112)=224\).
Time = 0.12 (sec) , antiderivative size = 2291, normalized size of antiderivative = 17.36
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(2291\) |
default | \(\text {Expression too large to display}\) | \(2291\) |
parts | \(\text {Expression too large to display}\) | \(3685\) |
4*a*b/(a^2+b^2)/d/(a+b*tan(d*x+c))^(1/2)+2/d*b^5/(a^2+b^2)^(5/2)/(2*(a^2+b ^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2 *a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))-2/d*b^5/(a^2+b^2)^(5/2)/(2*(a^2+ b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x +c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))+1/d*b^3/(a^2+b^2)^(3/2)/(2*(a^2 +b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d* x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))+1/4/d*b^3/(a^2+b^2)^2*ln(b*tan (d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^( 1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-1/4/d*b^3/(a^2+b^2)^2*ln((a+b*tan(d*x+ c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2 *(a^2+b^2)^(1/2)+2*a)^(1/2)-1/d*b^3/(a^2+b^2)^(3/2)/(2*(a^2+b^2)^(1/2)-2*a )^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2 *(a^2+b^2)^(1/2)-2*a)^(1/2))+1/d/b/(a^2+b^2)^(3/2)/(2*(a^2+b^2)^(1/2)-2*a) ^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2* (a^2+b^2)^(1/2)-2*a)^(1/2))*a^4-1/d*b^3/(a^2+b^2)^(5/2)/(2*(a^2+b^2)^(1/2) -2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2) )/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^2-1/d/b/(a^2+b^2)^(5/2)/(2*(a^2+b^2)^(1 /2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1 /2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^6-1/4/d/b/(a^2+b^2)^2*ln(b*tan(d*x+c )+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2...
Leaf count of result is larger than twice the leaf count of optimal. 2320 vs. \(2 (106) = 212\).
Time = 0.29 (sec) , antiderivative size = 2320, normalized size of antiderivative = 17.58 \[ \int \frac {-a+b \tan (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
1/2*(8*sqrt(b*tan(d*x + c) + a)*a*b - ((a^2*b + b^3)*d*tan(d*x + c) + (a^3 + a*b^2)*d)*sqrt(-(a^5 - 10*a^3*b^2 + 5*a*b^4 + (a^6 + 3*a^4*b^2 + 3*a^2* b^4 + b^6)*d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2 *b^10 + b^12)*d^4)))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2))*log((5*a^6 *b - 5*a^4*b^3 - 9*a^2*b^5 + b^7)*sqrt(b*tan(d*x + c) + a) + ((a^9 - 6*a^5 *b^4 - 8*a^3*b^6 - 3*a*b^8)*d^3*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4* b^6 - 20*a^2*b^8 + b^10)/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 1 5*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) + (15*a^6*b^2 - 35*a^4*b^4 + 13*a^2*b ^6 - b^8)*d)*sqrt(-(a^5 - 10*a^3*b^2 + 5*a*b^4 + (a^6 + 3*a^4*b^2 + 3*a^2* b^4 + b^6)*d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2 *b^10 + b^12)*d^4)))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2))) + ((a^2*b + b^3)*d*tan(d*x + c) + (a^3 + a*b^2)*d)*sqrt(-(a^5 - 10*a^3*b^2 + 5*a*b^ 4 + (a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^ 4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20 *a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)))/((a^6 + 3*a^4*b^2 + 3*a^ 2*b^4 + b^6)*d^2))*log((5*a^6*b - 5*a^4*b^3 - 9*a^2*b^5 + b^7)*sqrt(b*tan( d*x + c) + a) - ((a^9 - 6*a^5*b^4 - 8*a^3*b^6 - 3*a*b^8)*d^3*sqrt(-(25*a^8 *b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/((a^12 + 6*a^10*b...
\[ \int \frac {-a+b \tan (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=- \int \frac {a}{a \sqrt {a + b \tan {\left (c + d x \right )}} + b \sqrt {a + b \tan {\left (c + d x \right )}} \tan {\left (c + d x \right )}}\, dx - \int \left (- \frac {b \tan {\left (c + d x \right )}}{a \sqrt {a + b \tan {\left (c + d x \right )}} + b \sqrt {a + b \tan {\left (c + d x \right )}} \tan {\left (c + d x \right )}}\right )\, dx \]
-Integral(a/(a*sqrt(a + b*tan(c + d*x)) + b*sqrt(a + b*tan(c + d*x))*tan(c + d*x)), x) - Integral(-b*tan(c + d*x)/(a*sqrt(a + b*tan(c + d*x)) + b*sq rt(a + b*tan(c + d*x))*tan(c + d*x)), x)
Exception generated. \[ \int \frac {-a+b \tan (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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
Timed out. \[ \int \frac {-a+b \tan (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Time = 13.95 (sec) , antiderivative size = 5475, normalized size of antiderivative = 41.48 \[ \int \frac {-a+b \tan (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
log(- (((((24*a*b^4*d^2 - 8*a^3*b^2*d^2)^2/4 - b^4*(16*a^6*d^4 + 16*b^6*d^ 4 + 48*a^2*b^4*d^4 + 48*a^4*b^2*d^4))^(1/2) - 12*a*b^4*d^2 + 4*a^3*b^2*d^2 )/(16*(a^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4)))^(1/2)*(32*b^13 *d^4 + 96*a^2*b^11*d^4 + 64*a^4*b^9*d^4 - 64*a^6*b^7*d^4 - 96*a^8*b^5*d^4 - 32*a^10*b^3*d^4 + (a + b*tan(c + d*x))^(1/2)*((((24*a*b^4*d^2 - 8*a^3*b^ 2*d^2)^2/4 - b^4*(16*a^6*d^4 + 16*b^6*d^4 + 48*a^2*b^4*d^4 + 48*a^4*b^2*d^ 4))^(1/2) - 12*a*b^4*d^2 + 4*a^3*b^2*d^2)/(16*(a^6*d^4 + b^6*d^4 + 3*a^2*b ^4*d^4 + 3*a^4*b^2*d^4)))^(1/2)*(64*a*b^12*d^5 + 320*a^3*b^10*d^5 + 640*a^ 5*b^8*d^5 + 640*a^7*b^6*d^5 + 320*a^9*b^4*d^5 + 64*a^11*b^2*d^5)) + (a + b *tan(c + d*x))^(1/2)*(16*b^12*d^3 + 32*a^2*b^10*d^3 - 32*a^6*b^6*d^3 - 16* a^8*b^4*d^3))*((((24*a*b^4*d^2 - 8*a^3*b^2*d^2)^2/4 - b^4*(16*a^6*d^4 + 16 *b^6*d^4 + 48*a^2*b^4*d^4 + 48*a^4*b^2*d^4))^(1/2) - 12*a*b^4*d^2 + 4*a^3* b^2*d^2)/(16*(a^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4)))^(1/2) - 8*a*b^11*d^2 - 24*a^3*b^9*d^2 - 24*a^5*b^7*d^2 - 8*a^7*b^5*d^2)*((((24*a* b^4*d^2 - 8*a^3*b^2*d^2)^2/4 - b^4*(16*a^6*d^4 + 16*b^6*d^4 + 48*a^2*b^4*d ^4 + 48*a^4*b^2*d^4))^(1/2) - 12*a*b^4*d^2 + 4*a^3*b^2*d^2)/(16*(a^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4)))^(1/2) + log(- ((-(((24*a*b^4*d ^2 - 8*a^3*b^2*d^2)^2/4 - b^4*(16*a^6*d^4 + 16*b^6*d^4 + 48*a^2*b^4*d^4 + 48*a^4*b^2*d^4))^(1/2) + 12*a*b^4*d^2 - 4*a^3*b^2*d^2)/(16*(a^6*d^4 + b^6* d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4)))^(1/2)*(32*b^13*d^4 + 96*a^2*b^11...