Integrand size = 25, antiderivative size = 150 \[ \int (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {(a-i b)^{3/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{3/2} (i A-B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 (A b+a B) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 B (a+b \tan (c+d x))^{3/2}}{3 d} \] Output:
-(a-I*b)^(3/2)*(I*A+B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d+(a+ I*b)^(3/2)*(I*A-B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/d+2*(A*b+ B*a)*(a+b*tan(d*x+c))^(1/2)/d+2/3*B*(a+b*tan(d*x+c))^(3/2)/d
Time = 0.36 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.93 \[ \int (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {-3 i (a-i b)^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )+3 i (a+i b)^{3/2} (A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+2 \sqrt {a+b \tan (c+d x)} (3 A b+4 a B+b B \tan (c+d x))}{3 d} \] Input:
Integrate[(a + b*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x]
Output:
((-3*I)*(a - I*b)^(3/2)*(A - I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] + (3*I)*(a + I*b)^(3/2)*(A + I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]] /Sqrt[a + I*b]] + 2*Sqrt[a + b*Tan[c + d*x]]*(3*A*b + 4*a*B + b*B*Tan[c + d*x]))/(3*d)
Time = 0.77 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.87, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4011, 3042, 4011, 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.
\(\displaystyle \int (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))dx\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int \sqrt {a+b \tan (c+d x)} (a A-b B+(A b+a B) \tan (c+d x))dx+\frac {2 B (a+b \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {a+b \tan (c+d x)} (a A-b B+(A b+a B) \tan (c+d x))dx+\frac {2 B (a+b \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int \frac {A a^2-2 b B a-A b^2+\left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 (a B+A b) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 B (a+b \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A a^2-2 b B a-A b^2+\left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 (a B+A b) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 B (a+b \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle \frac {1}{2} (a-i b)^2 (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b)^2 (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 (a B+A b) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 B (a+b \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} (a-i b)^2 (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b)^2 (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 (a B+A b) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 B (a+b \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle \frac {i (a-i b)^2 (A-i B) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i (a+i b)^2 (A+i B) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 (a B+A b) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 B (a+b \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {i (a-i b)^2 (A-i B) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {i (a+i b)^2 (A+i B) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 (a B+A b) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 B (a+b \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(a-i b)^2 (A-i B) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {(a+i b)^2 (A+i B) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {2 (a B+A b) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 B (a+b \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(a-i b)^{3/2} (A-i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{3/2} (A+i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}+\frac {2 (a B+A b) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 B (a+b \tan (c+d x))^{3/2}}{3 d}\) |
Input:
Int[(a + b*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x]
Output:
((a - I*b)^(3/2)*(A - I*B)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/d + ((a + I *b)^(3/2)*(A + I*B)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/d + (2*(A*b + a*B) *Sqrt[a + b*Tan[c + d*x]])/d + (2*B*(a + b*Tan[c + d*x])^(3/2))/(3*d)
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[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int [(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] , x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]
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. \(1656\) vs. \(2(126)=252\).
Time = 0.14 (sec) , antiderivative size = 1657, normalized size of antiderivative = 11.05
method | result | size |
parts | \(\text {Expression too large to display}\) | \(1657\) |
derivativedivides | \(\text {Expression too large to display}\) | \(1665\) |
default | \(\text {Expression too large to display}\) | \(1665\) |
Input:
int((a+b*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
A*(2*b*(a+b*tan(d*x+c))^(1/2)/d+1/4/d/b*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)*(a^2+b^2)^(1/2)*a-1/4/d/b*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)*a^2+1/4/d*b*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/(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))*(a^2+b^2)^(1/2)-2/d*b/(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))*a-1/4/b/d*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))*(a^2+b ^2)^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a+1/4/b/d*ln(b*tan(d*x+c)+a+(a+b*t an(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)*a^2-1/4*b/d*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 )-b/d/(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+b^2)^(1/2)+2 *b/d/(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)+B*(2/3/d*(a+b*t an(d*x+c))^(3/2)+2/d*(a+b*tan(d*x+c))^(1/2)*a-1/4/d*ln((a+b*tan(d*x+c))...
Leaf count of result is larger than twice the leaf count of optimal. 3058 vs. \(2 (120) = 240\).
Time = 0.29 (sec) , antiderivative size = 3058, normalized size of antiderivative = 20.39 \[ \int (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \] Input:
integrate((a+b*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")
Output:
Too large to include
\[ \int (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx \] Input:
integrate((a+b*tan(d*x+c))**(3/2)*(A+B*tan(d*x+c)),x)
Output:
Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**(3/2), x)
Exception generated. \[ \int (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Exception raised: ValueError} \] Input:
integrate((a+b*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")
Output:
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
Exception generated. \[ \int (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+b*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{%%%{1,[0,14,5]%%%}+%%%{6,[0,12,5]%%%}+%%%{15,[0,10,5]%%%}+ %%%{20,[0
Time = 16.25 (sec) , antiderivative size = 2823, normalized size of antiderivative = 18.82 \[ \int (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \] Input:
int((A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(3/2),x)
Output:
log((((16*b^2*(((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - A^2*a^3*d^2 + 3*A^2 *a*b^2*d^2)/d^4)^(1/2)*(A*b^3 + A*a^2*b - a*d*(((-A^4*b^2*d^4*(3*a^2 - b^2 )^2)^(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^2*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x) )^(1/2)))/d - (16*A^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^4 + b^4 - 6*a^2*b^ 2))/d^2)*(((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^ 2*d^2)/d^4)^(1/2))/2 - (16*A^3*a*b^3*(a^2 + b^2)^2)/d^3)*((6*A^4*a^2*b^4*d ^4 - A^4*b^6*d^4 - 9*A^4*a^4*b^2*d^4)^(1/2)/(4*d^4) - (A^2*a^3)/(4*d^2) + (3*A^2*a*b^2)/(4*d^2))^(1/2) - log((((16*b^2*(((-A^4*b^2*d^4*(3*a^2 - b^2) ^2)^(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^2*d^2)/d^4)^(1/2)*(A*b^3 + A*a^2*b + a *d*(((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^2*d^2) /d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/d + (16*A^2*b^2*(a + b*tan(c + d* x))^(1/2)*(a^4 + b^4 - 6*a^2*b^2))/d^2)*(((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^( 1/2) - A^2*a^3*d^2 + 3*A^2*a*b^2*d^2)/d^4)^(1/2))/2 - (16*A^3*a*b^3*(a^2 + b^2)^2)/d^3)*(((6*A^4*a^2*b^4*d^4 - A^4*b^6*d^4 - 9*A^4*a^4*b^2*d^4)^(1/2 ) - A^2*a^3*d^2 + 3*A^2*a*b^2*d^2)/(4*d^4))^(1/2) - log((((16*b^2*(-((-A^4 *b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + A^2*a^3*d^2 - 3*A^2*a*b^2*d^2)/d^4)^(1/2 )*(A*b^3 + A*a^2*b + a*d*(-((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + A^2*a^3 *d^2 - 3*A^2*a*b^2*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/d + (16*A^ 2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^4 + b^4 - 6*a^2*b^2))/d^2)*(-((-A^4*b^ 2*d^4*(3*a^2 - b^2)^2)^(1/2) + A^2*a^3*d^2 - 3*A^2*a*b^2*d^2)/d^4)^(1/2...
\[ \int (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\left (\int \sqrt {a +\tan \left (d x +c \right ) b}d x \right ) a^{2}+\left (\int \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{2}d x \right ) b^{2}+2 \left (\int \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )d x \right ) a b \] Input:
int((a+b*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x)
Output:
int(sqrt(tan(c + d*x)*b + a),x)*a**2 + int(sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**2,x)*b**2 + 2*int(sqrt(tan(c + d*x)*b + a)*tan(c + d*x),x)*a*b