3.3.48 \(\int \frac {B+A x}{(17-18 x+5 x^2) \sqrt {13-22 x+10 x^2}} \, dx\) [248]
Optimal. Leaf size=80 \[ -\frac {(2 A+B) \tan ^{-1}\left (\frac {\sqrt {35} (2-x)}{\sqrt {13-22 x+10 x^2}}\right )}{\sqrt {35}}-\frac {(A+B) \tanh ^{-1}\left (\frac {\sqrt {35} (1-x)}{2 \sqrt {13-22 x+10 x^2}}\right )}{2 \sqrt {35}} \]
[Out]
-1/35*(2*A+B)*arctan((2-x)*35^(1/2)/(10*x^2-22*x+13)^(1/2))*35^(1/2)-1/70*(A+B)*arctanh(1/2*(1-x)*35^(1/2)/(10
*x^2-22*x+13)^(1/2))*35^(1/2)
________________________________________________________________________________________
Rubi [A]
time = 0.09, antiderivative size = 89, normalized size of antiderivative = 1.11, number of steps
used = 5, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1049, 1043,
212, 210} \begin {gather*} -\frac {(2 A+B) \text {ArcTan}\left (\frac {\sqrt {35} (2-x)}{\sqrt {10 x^2-22 x+13}}\right )}{\sqrt {35}}-\frac {(A+B) \tanh ^{-1}\left (\frac {\sqrt {35} (-x (A+B)+A+B)}{2 \sqrt {10 x^2-22 x+13} (A+B)}\right )}{2 \sqrt {35}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(B + A*x)/((17 - 18*x + 5*x^2)*Sqrt[13 - 22*x + 10*x^2]),x]
[Out]
-(((2*A + B)*ArcTan[(Sqrt[35]*(2 - x))/Sqrt[13 - 22*x + 10*x^2]])/Sqrt[35]) - ((A + B)*ArcTanh[(Sqrt[35]*(A +
B - (A + B)*x))/(2*(A + B)*Sqrt[13 - 22*x + 10*x^2])])/(2*Sqrt[35])
Rule 210
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])
Rule 212
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] && (GtQ[a, 0] || LtQ[b, 0])
Rule 1043
Int[((g_.) + (h_.)*(x_))/(((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symb
ol] :> Dist[-2*g*(g*b - 2*a*h), Subst[Int[1/Simp[g*(g*b - 2*a*h)*(b^2 - 4*a*c) - (b*d - a*e)*x^2, x], x], x, S
imp[g*b - 2*a*h - (b*h - 2*g*c)*x, x]/Sqrt[d + e*x + f*x^2]], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && NeQ[
b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] && NeQ[b*d - a*e, 0] && EqQ[h^2*(b*d - a*e) - 2*g*h*(c*d - a*f) + g^2*(
c*e - b*f), 0]
Rule 1049
Int[((g_.) + (h_.)*(x_))/(((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symb
ol] :> With[{q = Rt[(c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f), 2]}, Dist[1/(2*q), Int[Simp[h*(b*d - a*e) - g*(c*
d - a*f - q) - (g*(c*e - b*f) - h*(c*d - a*f + q))*x, x]/((a + b*x + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x] - D
ist[1/(2*q), Int[Simp[h*(b*d - a*e) - g*(c*d - a*f + q) - (g*(c*e - b*f) - h*(c*d - a*f - q))*x, x]/((a + b*x
+ c*x^2)*Sqrt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e
^2 - 4*d*f, 0] && NeQ[b*d - a*e, 0] && NegQ[b^2 - 4*a*c]
Rubi steps
\begin {align*} \int \frac {B+A x}{\left (17-18 x+5 x^2\right ) \sqrt {13-22 x+10 x^2}} \, dx &=\frac {1}{70} \int \frac {140 (A+B)-70 (A+B) x}{\left (17-18 x+5 x^2\right ) \sqrt {13-22 x+10 x^2}} \, dx-\frac {1}{70} \int \frac {70 (2 A+B)-70 (2 A+B) x}{\left (17-18 x+5 x^2\right ) \sqrt {13-22 x+10 x^2}} \, dx\\ &=\left (560 (A+B)^2\right ) \text {Subst}\left (\int \frac {1}{313600 (A+B)^2-140 x^2} \, dx,x,\frac {-140 (A+B)+140 (A+B) x}{\sqrt {13-22 x+10 x^2}}\right )+\left (2240 (2 A+B)^2\right ) \text {Subst}\left (\int \frac {1}{-1254400 (2 A+B)^2-140 x^2} \, dx,x,\frac {1120 (2 A+B)-560 (2 A+B) x}{\sqrt {13-22 x+10 x^2}}\right )\\ &=-\frac {(2 A+B) \tan ^{-1}\left (\frac {\sqrt {35} (2-x)}{\sqrt {13-22 x+10 x^2}}\right )}{\sqrt {35}}-\frac {(A+B) \tanh ^{-1}\left (\frac {\sqrt {35} (A+B-(A+B) x)}{2 (A+B) \sqrt {13-22 x+10 x^2}}\right )}{2 \sqrt {35}}\\ \end {align*}
________________________________________________________________________________________
Mathematica [C] Result contains complex when optimal does not.
time = 0.90, size = 124, normalized size = 1.55 \begin {gather*} \frac {((1+4 i) A+(1+2 i) B) \tanh ^{-1}\left (\frac {(4-i) \sqrt {10}-(2-i) \sqrt {10} x+(2-i) \sqrt {13-22 x+10 x^2}}{\sqrt {35}}\right )+((1-4 i) A+(1-2 i) B) \tanh ^{-1}\left (\frac {(4+i) \sqrt {10}-(2+i) \sqrt {10} x+(2+i) \sqrt {13-22 x+10 x^2}}{\sqrt {35}}\right )}{2 \sqrt {35}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(B + A*x)/((17 - 18*x + 5*x^2)*Sqrt[13 - 22*x + 10*x^2]),x]
[Out]
(((1 + 4*I)*A + (1 + 2*I)*B)*ArcTanh[((4 - I)*Sqrt[10] - (2 - I)*Sqrt[10]*x + (2 - I)*Sqrt[13 - 22*x + 10*x^2]
)/Sqrt[35]] + ((1 - 4*I)*A + (1 - 2*I)*B)*ArcTanh[((4 + I)*Sqrt[10] - (2 + I)*Sqrt[10]*x + (2 + I)*Sqrt[13 - 2
2*x + 10*x^2])/Sqrt[35]])/(2*Sqrt[35])
________________________________________________________________________________________
Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(191\) vs.
\(2(64)=128\).
time = 0.21, size = 192, normalized size = 2.40
| | |
| method |
result |
size |
| | |
| default |
\(\frac {\sqrt {\frac {\left (-2+x \right )^{2}}{\left (1-x \right )^{2}}+9}\, \sqrt {35}\, \left (\arctanh \left (\frac {2 \sqrt {\frac {\left (-2+x \right )^{2}}{\left (1-x \right )^{2}}+9}\, \sqrt {35}}{35}\right ) A -4 \arctan \left (\frac {\sqrt {35}\, \left (-2+x \right )}{\sqrt {\frac {\left (-2+x \right )^{2}}{\left (1-x \right )^{2}}+9}\, \left (1-x \right )}\right ) A +\arctanh \left (\frac {2 \sqrt {\frac {\left (-2+x \right )^{2}}{\left (1-x \right )^{2}}+9}\, \sqrt {35}}{35}\right ) B -2 \arctan \left (\frac {\sqrt {35}\, \left (-2+x \right )}{\sqrt {\frac {\left (-2+x \right )^{2}}{\left (1-x \right )^{2}}+9}\, \left (1-x \right )}\right ) B \right )}{70 \sqrt {\frac {\frac {\left (-2+x \right )^{2}}{\left (1-x \right )^{2}}+9}{\left (\frac {-2+x}{1-x}+1\right )^{2}}}\, \left (\frac {-2+x}{1-x}+1\right )}\) |
\(192\) |
| | |
|
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A*x+B)/(5*x^2-18*x+17)/(10*x^2-22*x+13)^(1/2),x,method=_RETURNVERBOSE)
[Out]
1/70*((-2+x)^2/(1-x)^2+9)^(1/2)*35^(1/2)*(arctanh(2/35*((-2+x)^2/(1-x)^2+9)^(1/2)*35^(1/2))*A-4*arctan(35^(1/2
)/((-2+x)^2/(1-x)^2+9)^(1/2)*(-2+x)/(1-x))*A+arctanh(2/35*((-2+x)^2/(1-x)^2+9)^(1/2)*35^(1/2))*B-2*arctan(35^(
1/2)/((-2+x)^2/(1-x)^2+9)^(1/2)*(-2+x)/(1-x))*B)/(((-2+x)^2/(1-x)^2+9)/((-2+x)/(1-x)+1)^2)^(1/2)/((-2+x)/(1-x)
+1)
________________________________________________________________________________________
Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A*x+B)/(5*x^2-18*x+17)/(10*x^2-22*x+13)^(1/2),x, algorithm="maxima")
[Out]
integrate((A*x + B)/(sqrt(10*x^2 - 22*x + 13)*(5*x^2 - 18*x + 17)), x)
________________________________________________________________________________________
Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 20890 vs.
\(2 (61) = 122\).
time = 37.96, size = 20890, normalized size = 261.12 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A*x+B)/(5*x^2-18*x+17)/(10*x^2-22*x+13)^(1/2),x, algorithm="fricas")
[Out]
1/1120*(16*sqrt(35)*sqrt(1/2)*(289*A^4 + 612*A^3*B + 494*A^2*B^2 + 180*A*B^3 + 25*B^4)^(3/4)*sqrt(4*A^4 + 12*A
^3*B + 13*A^2*B^2 + 6*A*B^3 + B^4)*sqrt((289*A^4 + 612*A^3*B + 494*A^2*B^2 + 180*A*B^3 + 25*B^4 + sqrt(289*A^4
+ 612*A^3*B + 494*A^2*B^2 + 180*A*B^3 + 25*B^4)*(15*A^2 + 14*A*B + 3*B^2))/(4*A^4 + 12*A^3*B + 13*A^2*B^2 + 6
*A*B^3 + B^4))*arctan(1/36*(51*sqrt(1/2)*sqrt(10*x^2 - 22*x + 13)*(35*(289*A^4 + 612*A^3*B + 494*A^2*B^2 + 180
*A*B^3 + 25*B^4)^(3/4)*(sqrt(35)*(10*(37746766*A^5 + 119135263*A^4*B + 147622344*A^3*B^2 + 90353742*A^2*B^3 +
27435770*A*B^4 + 3315875*B^5)*x^7 - 6*(525099162*A^5 + 1658839381*A^4*B + 2057521128*A^3*B^2 + 1260580554*A^2*
B^3 + 383153070*A*B^4 + 46353425*B^5)*x^6 + (11194351862*A^5 + 35402804819*A^4*B + 43962536712*A^3*B^2 + 26966
081046*A^2*B^3 + 8205884626*A*B^4 + 993887335*B^5)*x^5 - 1206372700*A^5 - 3851354182*A^4*B - 4830122076*A^3*B^
2 - 2992103088*A^2*B^3 - 919382984*A*B^4 - 112420490*B^5 - 6*(3668124034*A^5 + 11615757785*A^4*B + 14444112720
*A^3*B^2 + 8872112490*A^2*B^3 + 2703527006*A*B^4 + 327893485*B^5)*x^4 + 8*(3238361416*A^5 + 10270669150*A^4*B
+ 12792326235*A^3*B^2 + 7870355925*A^2*B^3 + 2402142389*A*B^4 + 291804965*B^5)*x^3 - 26*(703185926*A^5 + 22342
79915*A^4*B + 2788200480*A^3*B^2 + 1718711310*A^2*B^3 + 525568234*A*B^4 + 63963415*B^5)*x^2 + 83317*(86054*A^5
+ 274019*A^4*B + 342732*A^3*B^2 + 211746*A^2*B^3 + 64894*A*B^4 + 7915*B^5)*x)*sqrt(289*A^4 + 612*A^3*B + 494*
A^2*B^2 + 180*A*B^3 + 25*B^4)*sqrt(4*A^4 + 12*A^3*B + 13*A^2*B^2 + 6*A*B^3 + B^4) - sqrt(35)*(10*(391824466*A^
7 + 1550927221*A^6*B + 2593048804*A^5*B^2 + 2382266887*A^4*B^3 + 1301402690*A^3*B^4 + 423130067*A^2*B^5 + 7582
0040*A*B^6 + 5771825*B^7)*x^7 - 9541773124*A^7 - 36548013346*A^6*B - 58731302500*A^5*B^2 - 51440404210*A^4*B^3
- 26510469596*A^3*B^4 - 8015319494*A^2*B^5 - 1307961980*A*B^6 - 87770150*B^7 - 6*(5346233542*A^7 + 2111878976
7*A^6*B + 35226004828*A^5*B^2 + 32274364509*A^4*B^3 + 17575325750*A^3*B^4 + 5693158449*A^2*B^5 + 1015651880*A*
B^6 + 76903275*B^7)*x^6 + (111338441258*A^7 + 438711020081*A^6*B + 729622269428*A^5*B^2 + 666207416699*A^4*B^3
+ 361345637914*A^3*B^4 + 116500131895*A^2*B^5 + 20666192200*A*B^6 + 1553972125*B^7)*x^5 - 6*(35456378270*A^7
+ 139271419123*A^6*B + 230765518196*A^5*B^2 + 209795040553*A^4*B^3 + 113210170654*A^3*B^4 + 36277447349*A^2*B^
5 + 6387737680*A*B^6 + 475897775*B^7)*x^4 + 8*(30225816400*A^7 + 118252662082*A^6*B + 195010544479*A^5*B^2 + 1
76296794397*A^4*B^3 + 94499843726*A^3*B^4 + 30038132996*A^2*B^5 + 5236606595*A*B^6 + 385221725*B^7)*x^3 - 26*(
6285799130*A^7 + 24465611897*A^6*B + 40097508044*A^5*B^2 + 35982344267*A^4*B^3 + 19115816506*A^3*B^4 + 6009785
311*A^2*B^5 + 1033263520*A*B^6 + 74645725*B^7)*x^2 + 83317*(728858*A^7 + 2817665*A^6*B + 4579976*A^5*B^2 + 406
8863*A^4*B^3 + 2135026*A^3*B^4 + 660847*A^2*B^5 + 111340*A*B^6 + 7825*B^7)*x)*sqrt(4*A^4 + 12*A^3*B + 13*A^2*B
^2 + 6*A*B^3 + B^4)) + (289*A^4 + 612*A^3*B + 494*A^2*B^2 + 180*A*B^3 + 25*B^4)^(1/4)*(sqrt(35)*(484*(15723276
2*A^7 + 709896659*A^6*B + 1357218467*A^5*B^2 + 1427411336*A^4*B^3 + 893876296*A^3*B^4 + 333960955*A^2*B^5 + 69
043675*A*B^6 + 6102250*B^7)*x^7 - 333962059340*A^7 - 1517703207488*A^6*B - 2918486677691*A^5*B^2 - 30850530587
63*A^4*B^3 - 1940556409714*A^3*B^4 - 727869487774*A^2*B^5 - 151009290055*A*B^6 - 13388682775*B^7 - 132*(501410
0292*A^7 + 22653024620*A^6*B + 43334268219*A^5*B^2 + 45598611647*A^4*B^3 + 28567646154*A^3*B^4 + 10677391858*A
^2*B^5 + 2208246135*A*B^6 + 195232675*B^7)*x^6 + 4*(613357332190*A^7 + 2773103498689*A^6*B + 5308303203313*A^5
*B^2 + 5588896485784*A^4*B^3 + 3503234148632*A^3*B^4 + 1309950681977*A^2*B^5 + 271026274265*A*B^6 + 2397029195
0*B^7)*x^5 - 120*(41983825906*A^7 + 189975041633*A^6*B + 363921818398*A^5*B^2 + 383408095829*A^4*B^3 + 2404656
76886*A^3*B^4 + 89961959563*A^2*B^5 + 18621370010*A*B^6 + 1647594175*B^7)*x^4 + 65*(95416948628*A^7 + 43216170
4988*A^6*B + 828549443447*A^5*B^2 + 873550452491*A^4*B^3 + 548223512530*A^3*B^4 + 205214947546*A^2*B^5 + 42499
175395*A*B^6 + 3761974975*B^7)*x^3 - 2197*(2086784612*A^7 + 9461242424*A^6*B + 18155944997*A^5*B^2 + 191574415
01*A^4*B^3 + 12031320766*A^3*B^4 + 4506452650*A^2*B^5 + 933784825*A*B^6 + 82698625*B^7)*x^2 + 1083121*(1742092
*A^7 + 7907380*A^6*B + 15189289*A^5*B^2 + 16041157*A^4*B^3 + 10081934*A^3*B^4 + 3778838*A^2*B^5 + 783485*A*B^6
+ 69425*B^7)*x)*sqrt(289*A^4 + 612*A^3*B + 494*A^2*B^2 + 180*A*B^3 + 25*B^4)*sqrt(4*A^4 + 12*A^3*B + 13*A^2*B
^2 + 6*A*B^3 + B^4) - sqrt(35)*(4017086485204*A^9 + 23217940414552*A^8*B + 58930857497369*A^7*B^2 + 8638888668
5855*A^6*B^3 + 80763852173121*A^5*B^4 + 50022837369063*A^4*B^5 + 20557133078131*A^3*B^6 + 5412069730405*A^2*B^
7 + 829187950175*A*B^8 + 56382984125*B^9 - 484*(1574842498*A^9 + 9214075735*A^8*B + 23625881189*A^7*B^2 + 3493
0883699*A^6*B^3 + 32893594929*A^5*B^4 + 20500170921*A^4*B^5 + 8470101559*A^3*B^6 + 2240463145*A^2*B^7 + 344699
825*A*B^8 + 23526500*B^9)*x^7 + 132*(51217887436*A^9 + 299242286004*A^8*B + 766396055657*A^7*B^2 + 11320191853
59*A^6*B^3 + 1065126177441*A^5*B^4 + 6633534962...
________________________________________________________________________________________
Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A x + B}{\left (5 x^{2} - 18 x + 17\right ) \sqrt {10 x^{2} - 22 x + 13}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A*x+B)/(5*x**2-18*x+17)/(10*x**2-22*x+13)**(1/2),x)
[Out]
Integral((A*x + B)/((5*x**2 - 18*x + 17)*sqrt(10*x**2 - 22*x + 13)), x)
________________________________________________________________________________________
Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 629 vs.
\(2 (61) = 122\).
time = 0.57, size = 629, normalized size = 7.86 \begin {gather*} \frac {2 \, \sqrt {35} {\left (2 \, A^{2} + 3 \, A B + B^{2}\right )} \sqrt {A^{2} + 2 \, A B + B^{2}} {\left (\arctan \left (3\right ) + \arctan \left (-\frac {5 \, {\left (\sqrt {10} x - \sqrt {10 \, x^{2} - 22 \, x + 13}\right )} {\left (300 \, \sqrt {14} - 1129\right )} - 7658 \, \sqrt {35} + 14361 \, \sqrt {10}}{2329 \, \sqrt {35} - 4358 \, \sqrt {10}}\right )\right )}}{35 \, {\left (15 \, A^{2} + 14 \, A B + 3 \, B^{2} - \sqrt {289 \, A^{4} + 612 \, A^{3} B + 494 \, A^{2} B^{2} + 180 \, A B^{3} + 25 \, B^{4}}\right )}} - \frac {2 \, \sqrt {35} {\left (2 \, A^{2} + 3 \, A B + B^{2}\right )} \sqrt {A^{2} + 2 \, A B + B^{2}} {\left (\arctan \left (\frac {1}{7}\right ) + \arctan \left (-\frac {5 \, {\left (\sqrt {10} x - \sqrt {10 \, x^{2} - 22 \, x + 13}\right )} {\left (62556 \, \sqrt {14} + 245977\right )} - 1617962 \, \sqrt {35} - 3089577 \, \sqrt {10}}{496201 \, \sqrt {35} + 929846 \, \sqrt {10}}\right )\right )}}{35 \, {\left (15 \, A^{2} + 14 \, A B + 3 \, B^{2} - \sqrt {289 \, A^{4} + 612 \, A^{3} B + 494 \, A^{2} B^{2} + 180 \, A B^{3} + 25 \, B^{4}}\right )}} + \frac {1}{140} \, \sqrt {35} \sqrt {A^{2} + 2 \, A B + B^{2}} \log \left (25 \, {\left (546 \, \sqrt {14} {\left (\sqrt {10} x - \sqrt {10 \, x^{2} - 22 \, x + 13}\right )} + 2807 \, \sqrt {10} x - 234 \, \sqrt {35} \sqrt {14} - 1014 \, \sqrt {14} \sqrt {10} - 1203 \, \sqrt {35} - 5213 \, \sqrt {10} - 2807 \, \sqrt {10 \, x^{2} - 22 \, x + 13}\right )}^{2} + 25 \, {\left (78 \, \sqrt {14} {\left (\sqrt {10} x - \sqrt {10 \, x^{2} - 22 \, x + 13}\right )} + 401 \, \sqrt {10} x + 48 \, \sqrt {35} \sqrt {14} + 208 \, \sqrt {14} \sqrt {10} + 141 \, \sqrt {35} + 611 \, \sqrt {10} - 401 \, \sqrt {10 \, x^{2} - 22 \, x + 13}\right )}^{2}\right ) - \frac {1}{140} \, \sqrt {35} \sqrt {A^{2} + 2 \, A B + B^{2}} \log \left (625 \, {\left (18 \, \sqrt {14} {\left (\sqrt {10} x - \sqrt {10 \, x^{2} - 22 \, x + 13}\right )} - 75 \, \sqrt {10} x + 8 \, \sqrt {35} \sqrt {14} - 24 \, \sqrt {14} \sqrt {10} - 37 \, \sqrt {35} + 111 \, \sqrt {10} + 75 \, \sqrt {10 \, x^{2} - 22 \, x + 13}\right )}^{2} + 625 \, {\left (6 \, \sqrt {14} {\left (\sqrt {10} x - \sqrt {10 \, x^{2} - 22 \, x + 13}\right )} - 25 \, \sqrt {10} x + 6 \, \sqrt {35} \sqrt {14} - 18 \, \sqrt {14} \sqrt {10} - 25 \, \sqrt {35} + 75 \, \sqrt {10} + 25 \, \sqrt {10 \, x^{2} - 22 \, x + 13}\right )}^{2}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A*x+B)/(5*x^2-18*x+17)/(10*x^2-22*x+13)^(1/2),x, algorithm="giac")
[Out]
2/35*sqrt(35)*(2*A^2 + 3*A*B + B^2)*sqrt(A^2 + 2*A*B + B^2)*(arctan(3) + arctan(-(5*(sqrt(10)*x - sqrt(10*x^2
- 22*x + 13))*(300*sqrt(14) - 1129) - 7658*sqrt(35) + 14361*sqrt(10))/(2329*sqrt(35) - 4358*sqrt(10))))/(15*A^
2 + 14*A*B + 3*B^2 - sqrt(289*A^4 + 612*A^3*B + 494*A^2*B^2 + 180*A*B^3 + 25*B^4)) - 2/35*sqrt(35)*(2*A^2 + 3*
A*B + B^2)*sqrt(A^2 + 2*A*B + B^2)*(arctan(1/7) + arctan(-(5*(sqrt(10)*x - sqrt(10*x^2 - 22*x + 13))*(62556*sq
rt(14) + 245977) - 1617962*sqrt(35) - 3089577*sqrt(10))/(496201*sqrt(35) + 929846*sqrt(10))))/(15*A^2 + 14*A*B
+ 3*B^2 - sqrt(289*A^4 + 612*A^3*B + 494*A^2*B^2 + 180*A*B^3 + 25*B^4)) + 1/140*sqrt(35)*sqrt(A^2 + 2*A*B + B
^2)*log(25*(546*sqrt(14)*(sqrt(10)*x - sqrt(10*x^2 - 22*x + 13)) + 2807*sqrt(10)*x - 234*sqrt(35)*sqrt(14) - 1
014*sqrt(14)*sqrt(10) - 1203*sqrt(35) - 5213*sqrt(10) - 2807*sqrt(10*x^2 - 22*x + 13))^2 + 25*(78*sqrt(14)*(sq
rt(10)*x - sqrt(10*x^2 - 22*x + 13)) + 401*sqrt(10)*x + 48*sqrt(35)*sqrt(14) + 208*sqrt(14)*sqrt(10) + 141*sqr
t(35) + 611*sqrt(10) - 401*sqrt(10*x^2 - 22*x + 13))^2) - 1/140*sqrt(35)*sqrt(A^2 + 2*A*B + B^2)*log(625*(18*s
qrt(14)*(sqrt(10)*x - sqrt(10*x^2 - 22*x + 13)) - 75*sqrt(10)*x + 8*sqrt(35)*sqrt(14) - 24*sqrt(14)*sqrt(10) -
37*sqrt(35) + 111*sqrt(10) + 75*sqrt(10*x^2 - 22*x + 13))^2 + 625*(6*sqrt(14)*(sqrt(10)*x - sqrt(10*x^2 - 22*
x + 13)) - 25*sqrt(10)*x + 6*sqrt(35)*sqrt(14) - 18*sqrt(14)*sqrt(10) - 25*sqrt(35) + 75*sqrt(10) + 25*sqrt(10
*x^2 - 22*x + 13))^2)
________________________________________________________________________________________
Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {B+A\,x}{\left (5\,x^2-18\,x+17\right )\,\sqrt {10\,x^2-22\,x+13}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((B + A*x)/((5*x^2 - 18*x + 17)*(10*x^2 - 22*x + 13)^(1/2)),x)
[Out]
int((B + A*x)/((5*x^2 - 18*x + 17)*(10*x^2 - 22*x + 13)^(1/2)), x)
________________________________________________________________________________________