3.13.31 \(\int (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)} \, dx\) [1231]
Optimal. Leaf size=122 \[ -\frac {(i a+b) \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {(i a-b) \sqrt {c+i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 b \sqrt {c+d \tan (e+f x)}}{f} \]
[Out]
-(I*a+b)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))*(c-I*d)^(1/2)/f+(I*a-b)*arctanh((c+d*tan(f*x+e))^(1/2)/
(c+I*d)^(1/2))*(c+I*d)^(1/2)/f+2*b*(c+d*tan(f*x+e))^(1/2)/f
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Rubi [A]
time = 0.15, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3609, 3620,
3618, 65, 214} \begin {gather*} -\frac {(b+i a) \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {(-b+i a) \sqrt {c+i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 b \sqrt {c+d \tan (e+f x)}}{f} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(a + b*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]],x]
[Out]
-(((I*a + b)*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f) + ((I*a - b)*Sqrt[c + I*d]*ArcT
anh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/f + (2*b*Sqrt[c + d*Tan[e + f*x]])/f
Rule 65
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]
Rule 214
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]
Rule 3609
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]
Rule 3618
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[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]
Rule 3620
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(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]
Rubi steps
\begin {align*} \int (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)} \, dx &=\frac {2 b \sqrt {c+d \tan (e+f x)}}{f}+\int \frac {a c-b d+(b c+a d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx\\ &=\frac {2 b \sqrt {c+d \tan (e+f x)}}{f}+\frac {1}{2} ((a-i b) (c-i d)) \int \frac {1+i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx+\frac {1}{2} ((a+i b) (c+i d)) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx\\ &=\frac {2 b \sqrt {c+d \tan (e+f x)}}{f}+\frac {(i (a-i b) (c-i d)) \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 f}-\frac {((i a-b) (c+i d)) \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 f}\\ &=\frac {2 b \sqrt {c+d \tan (e+f x)}}{f}-\frac {((a-i b) (c-i d)) \text {Subst}\left (\int \frac {1}{-1-\frac {i c}{d}+\frac {i x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{d f}-\frac {((a+i b) (c+i d)) \text {Subst}\left (\int \frac {1}{-1+\frac {i c}{d}-\frac {i x^2}{d}} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{d f}\\ &=-\frac {(i a+b) \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {(i a-b) \sqrt {c+i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 b \sqrt {c+d \tan (e+f x)}}{f}\\ \end {align*}
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Mathematica [A]
time = 0.15, size = 120, normalized size = 0.98 \begin {gather*} \frac {-i (a-i b) \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+i (a+i b) \sqrt {c+i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+2 b \sqrt {c+d \tan (e+f x)}}{f} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]],x]
[Out]
((-I)*(a - I*b)*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]] + I*(a + I*b)*Sqrt[c + I*d]*ArcT
anh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]] + 2*b*Sqrt[c + d*Tan[e + f*x]])/f
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(631\) vs.
\(2(102)=204\).
time = 0.45, size = 632, normalized size = 5.18
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method |
result |
size |
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derivativedivides |
\(\frac {2 b \sqrt {c +d \tan \left (f x +e \right )}+\frac {\frac {\left (-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}\, a +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a c -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b d \right ) \ln \left (d \tan \left (f x +e \right )+c +\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (-2 \sqrt {c^{2}+d^{2}}\, b d -\frac {\left (-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}\, a +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a c -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b d \right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{2 d}+\frac {-\frac {\left (-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}\, a +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a c -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b d \right ) \ln \left (\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-d \tan \left (f x +e \right )-c -\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (2 \sqrt {c^{2}+d^{2}}\, b d +\frac {\left (-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}\, a +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a c -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b d \right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-2 \sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{2 d}}{f}\) |
\(632\) |
default |
\(\frac {2 b \sqrt {c +d \tan \left (f x +e \right )}+\frac {\frac {\left (-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}\, a +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a c -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b d \right ) \ln \left (d \tan \left (f x +e \right )+c +\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (-2 \sqrt {c^{2}+d^{2}}\, b d -\frac {\left (-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}\, a +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a c -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b d \right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{2 d}+\frac {-\frac {\left (-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}\, a +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a c -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b d \right ) \ln \left (\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-d \tan \left (f x +e \right )-c -\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (2 \sqrt {c^{2}+d^{2}}\, b d +\frac {\left (-\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}\, a +\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, a c -\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, b d \right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-2 \sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{2 d}}{f}\) |
\(632\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e)),x,method=_RETURNVERBOSE)
[Out]
1/f*(2*b*(c+d*tan(f*x+e))^(1/2)+1/2/d*(1/2*(-(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a+(2*(c^2+d^2)^(1/2
)+2*c)^(1/2)*a*c-(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b*d)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2
)+2*c)^(1/2)+(c^2+d^2)^(1/2))+2*(-2*(c^2+d^2)^(1/2)*b*d-1/2*(-(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a+
(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*c-(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b*d)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^
2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1
/2)))+1/2/d*(-1/2*(-(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a+(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*c-(2*(c^2+
d^2)^(1/2)+2*c)^(1/2)*b*d)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1
/2))+2*(2*(c^2+d^2)^(1/2)*b*d+1/2*(-(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a+(2*(c^2+d^2)^(1/2)+2*c)^(1
/2)*a*c-(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b*d)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan
(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))))
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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e)),x, algorithm="maxima")
[Out]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(d-c>0)', see `assume?` for mor
e details)Is
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 8653 vs.
\(2 (100) = 200\).
time = 11.89, size = 8653, normalized size = 70.93 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e)),x, algorithm="fricas")
[Out]
1/4*(4*sqrt(2)*f^5*sqrt(-((2*a*b*d - (a^2 - b^2)*c)*f^2*sqrt(((a^4 + 2*a^2*b^2 + b^4)*c^2 + (a^4 + 2*a^2*b^2 +
b^4)*d^2)/f^4) - (a^4 + 2*a^2*b^2 + b^4)*c^2 - (a^4 + 2*a^2*b^2 + b^4)*d^2)/(4*a^2*b^2*c^2 + 4*(a^3*b - a*b^3
)*c*d + (a^4 - 2*a^2*b^2 + b^4)*d^2))*sqrt((4*a^2*b^2*c^2 + 4*(a^3*b - a*b^3)*c*d + (a^4 - 2*a^2*b^2 + b^4)*d^
2)/f^4)*(((a^4 + 2*a^2*b^2 + b^4)*c^2 + (a^4 + 2*a^2*b^2 + b^4)*d^2)/f^4)^(3/4)*arctan(((2*(a^7*b + 3*a^5*b^3
+ 3*a^3*b^5 + a*b^7)*c^3 + (a^8 + 2*a^6*b^2 - 2*a^2*b^6 - b^8)*c^2*d + 2*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^
7)*c*d^2 + (a^8 + 2*a^6*b^2 - 2*a^2*b^6 - b^8)*d^3)*f^4*sqrt((4*a^2*b^2*c^2 + 4*(a^3*b - a*b^3)*c*d + (a^4 - 2
*a^2*b^2 + b^4)*d^2)/f^4)*sqrt(((a^4 + 2*a^2*b^2 + b^4)*c^2 + (a^4 + 2*a^2*b^2 + b^4)*d^2)/f^4) + (2*(a^9*b +
4*a^7*b^3 + 6*a^5*b^5 + 4*a^3*b^7 + a*b^9)*c^4 + (a^10 + 3*a^8*b^2 + 2*a^6*b^4 - 2*a^4*b^6 - 3*a^2*b^8 - b^10)
*c^3*d + 2*(a^9*b + 4*a^7*b^3 + 6*a^5*b^5 + 4*a^3*b^7 + a*b^9)*c^2*d^2 + (a^10 + 3*a^8*b^2 + 2*a^6*b^4 - 2*a^4
*b^6 - 3*a^2*b^8 - b^10)*c*d^3)*f^2*sqrt((4*a^2*b^2*c^2 + 4*(a^3*b - a*b^3)*c*d + (a^4 - 2*a^2*b^2 + b^4)*d^2)
/f^4) + sqrt(2)*((2*(a^3*b^2 + a*b^4)*c + (a^4*b - b^5)*d)*f^7*sqrt((4*a^2*b^2*c^2 + 4*(a^3*b - a*b^3)*c*d + (
a^4 - 2*a^2*b^2 + b^4)*d^2)/f^4)*sqrt(((a^4 + 2*a^2*b^2 + b^4)*c^2 + (a^4 + 2*a^2*b^2 + b^4)*d^2)/f^4) + (2*(a
^5*b^2 + 2*a^3*b^4 + a*b^6)*c^2 + (3*a^6*b + 5*a^4*b^3 + a^2*b^5 - b^7)*c*d + (a^7 + a^5*b^2 - a^3*b^4 - a*b^6
)*d^2)*f^5*sqrt((4*a^2*b^2*c^2 + 4*(a^3*b - a*b^3)*c*d + (a^4 - 2*a^2*b^2 + b^4)*d^2)/f^4))*sqrt(-((2*a*b*d -
(a^2 - b^2)*c)*f^2*sqrt(((a^4 + 2*a^2*b^2 + b^4)*c^2 + (a^4 + 2*a^2*b^2 + b^4)*d^2)/f^4) - (a^4 + 2*a^2*b^2 +
b^4)*c^2 - (a^4 + 2*a^2*b^2 + b^4)*d^2)/(4*a^2*b^2*c^2 + 4*(a^3*b - a*b^3)*c*d + (a^4 - 2*a^2*b^2 + b^4)*d^2))
*sqrt((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))*(((a^4 + 2*a^2*b^2 + b^4)*c^2 + (a^4 + 2*a^2*b^2 + b^4)*
d^2)/f^4)^(3/4) + sqrt(2)*(b*f^7*sqrt((4*a^2*b^2*c^2 + 4*(a^3*b - a*b^3)*c*d + (a^4 - 2*a^2*b^2 + b^4)*d^2)/f^
4)*sqrt(((a^4 + 2*a^2*b^2 + b^4)*c^2 + (a^4 + 2*a^2*b^2 + b^4)*d^2)/f^4) + ((a^2*b + b^3)*c + (a^3 + a*b^2)*d)
*f^5*sqrt((4*a^2*b^2*c^2 + 4*(a^3*b - a*b^3)*c*d + (a^4 - 2*a^2*b^2 + b^4)*d^2)/f^4))*sqrt(-((2*a*b*d - (a^2 -
b^2)*c)*f^2*sqrt(((a^4 + 2*a^2*b^2 + b^4)*c^2 + (a^4 + 2*a^2*b^2 + b^4)*d^2)/f^4) - (a^4 + 2*a^2*b^2 + b^4)*c
^2 - (a^4 + 2*a^2*b^2 + b^4)*d^2)/(4*a^2*b^2*c^2 + 4*(a^3*b - a*b^3)*c*d + (a^4 - 2*a^2*b^2 + b^4)*d^2))*sqrt(
((4*(a^4*b^2 + a^2*b^4)*c^4 + 4*(a^5*b - a*b^5)*c^3*d + (a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*c^2*d^2 + 4*(a^5*b
- a*b^5)*c*d^3 + (a^6 - a^4*b^2 - a^2*b^4 + b^6)*d^4)*f^2*sqrt(((a^4 + 2*a^2*b^2 + b^4)*c^2 + (a^4 + 2*a^2*b^
2 + b^4)*d^2)/f^4)*cos(f*x + e) + sqrt(2)*((4*a^2*b^3*c^3 + 4*(2*a^3*b^2 - a*b^4)*c^2*d + (5*a^4*b - 6*a^2*b^3
+ b^5)*c*d^2 + (a^5 - 2*a^3*b^2 + a*b^4)*d^3)*f^3*sqrt(((a^4 + 2*a^2*b^2 + b^4)*c^2 + (a^4 + 2*a^2*b^2 + b^4)
*d^2)/f^4)*cos(f*x + e) + (4*(a^4*b^3 + a^2*b^5)*c^4 + 4*(a^5*b^2 - a*b^6)*c^3*d + (a^6*b + 3*a^4*b^3 + 3*a^2*
b^5 + b^7)*c^2*d^2 + 4*(a^5*b^2 - a*b^6)*c*d^3 + (a^6*b - a^4*b^3 - a^2*b^5 + b^7)*d^4)*f*cos(f*x + e))*sqrt(-
((2*a*b*d - (a^2 - b^2)*c)*f^2*sqrt(((a^4 + 2*a^2*b^2 + b^4)*c^2 + (a^4 + 2*a^2*b^2 + b^4)*d^2)/f^4) - (a^4 +
2*a^2*b^2 + b^4)*c^2 - (a^4 + 2*a^2*b^2 + b^4)*d^2)/(4*a^2*b^2*c^2 + 4*(a^3*b - a*b^3)*c*d + (a^4 - 2*a^2*b^2
+ b^4)*d^2))*sqrt((c*cos(f*x + e) + d*sin(f*x + e))/cos(f*x + e))*(((a^4 + 2*a^2*b^2 + b^4)*c^2 + (a^4 + 2*a^2
*b^2 + b^4)*d^2)/f^4)^(1/4) + (4*(a^6*b^2 + 2*a^4*b^4 + a^2*b^6)*c^5 + 4*(a^7*b + a^5*b^3 - a^3*b^5 - a*b^7)*c
^4*d + (a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*c^3*d^2 + 4*(a^7*b + a^5*b^3 - a^3*b^5 - a*b^7)*c^2*d^3
+ (a^8 - 2*a^4*b^4 + b^8)*c*d^4)*cos(f*x + e) + (4*(a^6*b^2 + 2*a^4*b^4 + a^2*b^6)*c^4*d + 4*(a^7*b + a^5*b^3
- a^3*b^5 - a*b^7)*c^3*d^2 + (a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*c^2*d^3 + 4*(a^7*b + a^5*b^3 - a
^3*b^5 - a*b^7)*c*d^4 + (a^8 - 2*a^4*b^4 + b^8)*d^5)*sin(f*x + e))/((c^2 + d^2)*cos(f*x + e)))*(((a^4 + 2*a^2*
b^2 + b^4)*c^2 + (a^4 + 2*a^2*b^2 + b^4)*d^2)/f^4)^(3/4))/(4*(a^10*b^2 + 4*a^8*b^4 + 6*a^6*b^6 + 4*a^4*b^8 + a
^2*b^10)*c^4*d + 4*(a^11*b + 3*a^9*b^3 + 2*a^7*b^5 - 2*a^5*b^7 - 3*a^3*b^9 - a*b^11)*c^3*d^2 + (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)*c^2*d^3 + 4*(a^11*b + 3*a^9*b^3 + 2*a^7*b^5 -
2*a^5*b^7 - 3*a^3*b^9 - a*b^11)*c*d^4 + (a^12 + 2*a^10*b^2 - a^8*b^4 - 4*a^6*b^6 - a^4*b^8 + 2*a^2*b^10 + b^12
)*d^5)) + 4*sqrt(2)*f^5*sqrt(-((2*a*b*d - (a^2 - b^2)*c)*f^2*sqrt(((a^4 + 2*a^2*b^2 + b^4)*c^2 + (a^4 + 2*a^2*
b^2 + b^4)*d^2)/f^4) - (a^4 + 2*a^2*b^2 + b^4)*c^2 - (a^4 + 2*a^2*b^2 + b^4)*d^2)/(4*a^2*b^2*c^2 + 4*(a^3*b -
a*b^3)*c*d + (a^4 - 2*a^2*b^2 + b^4)*d^2))*sqrt((4*a^2*b^2*c^2 + 4*(a^3*b - a*b^3)*c*d + (a^4 - 2*a^2*b^2 + b^
4)*d^2)/f^4)*(((a^4 + 2*a^2*b^2 + b^4)*c^2 + (a^4 + 2*a^2*b^2 + b^4)*d^2)/f^4)^(3/4)*arctan(-((2*(a^7*b + 3*a^
5*b^3 + 3*a^3*b^5 + a*b^7)*c^3 + (a^8 + 2*a^6*b^2 - 2*a^2*b^6 - b^8)*c^2*d + 2*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5
+ a*b^7)*c*d^2 + (a^8 + 2*a^6*b^2 - 2*a^2*b^6 -...
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \tan {\left (e + f x \right )}\right ) \sqrt {c + d \tan {\left (e + f x \right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*tan(f*x+e))**(1/2)*(a+b*tan(f*x+e)),x)
[Out]
Integral((a + b*tan(e + f*x))*sqrt(c + d*tan(e + f*x)), x)
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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e)),x, algorithm="giac")
[Out]
Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choi
ce was done
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Mupad [B]
time = 7.65, size = 845, normalized size = 6.93 \begin {gather*} 2\,\mathrm {atanh}\left (\frac {32\,b^2\,d^4\,\sqrt {\frac {b^2\,c}{4\,f^2}-\frac {\sqrt {-b^4\,d^2\,f^4}}{4\,f^4}}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{\frac {16\,b\,d^4\,\sqrt {-b^4\,d^2\,f^4}}{f^3}+\frac {16\,b\,c^2\,d^2\,\sqrt {-b^4\,d^2\,f^4}}{f^3}}-\frac {32\,c\,d^2\,\sqrt {\frac {b^2\,c}{4\,f^2}-\frac {\sqrt {-b^4\,d^2\,f^4}}{4\,f^4}}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-b^4\,d^2\,f^4}}{\frac {16\,b\,d^4\,\sqrt {-b^4\,d^2\,f^4}}{f}+\frac {16\,b\,c^2\,d^2\,\sqrt {-b^4\,d^2\,f^4}}{f}}\right )\,\sqrt {-\frac {\sqrt {-b^4\,d^2\,f^4}-b^2\,c\,f^2}{4\,f^4}}-2\,\mathrm {atanh}\left (\frac {32\,b^2\,d^4\,\sqrt {\frac {\sqrt {-b^4\,d^2\,f^4}}{4\,f^4}+\frac {b^2\,c}{4\,f^2}}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{\frac {16\,b\,d^4\,\sqrt {-b^4\,d^2\,f^4}}{f^3}+\frac {16\,b\,c^2\,d^2\,\sqrt {-b^4\,d^2\,f^4}}{f^3}}+\frac {32\,c\,d^2\,\sqrt {\frac {\sqrt {-b^4\,d^2\,f^4}}{4\,f^4}+\frac {b^2\,c}{4\,f^2}}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-b^4\,d^2\,f^4}}{\frac {16\,b\,d^4\,\sqrt {-b^4\,d^2\,f^4}}{f}+\frac {16\,b\,c^2\,d^2\,\sqrt {-b^4\,d^2\,f^4}}{f}}\right )\,\sqrt {\frac {\sqrt {-b^4\,d^2\,f^4}+b^2\,c\,f^2}{4\,f^4}}-\mathrm {atanh}\left (\frac {f^3\,\left (\frac {16\,\left (a^2\,d^4-a^2\,c^2\,d^2\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{f^2}+\frac {16\,c\,d^2\,\left (\sqrt {-a^4\,d^2\,f^4}+a^2\,c\,f^2\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{f^4}\right )\,\sqrt {-\frac {\sqrt {-a^4\,d^2\,f^4}+a^2\,c\,f^2}{f^4}}}{16\,\left (a^3\,c^2\,d^3+a^3\,d^5\right )}\right )\,\sqrt {-\frac {\sqrt {-a^4\,d^2\,f^4}+a^2\,c\,f^2}{f^4}}-\mathrm {atanh}\left (\frac {f^3\,\left (\frac {16\,\left (a^2\,d^4-a^2\,c^2\,d^2\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{f^2}-\frac {16\,c\,d^2\,\left (\sqrt {-a^4\,d^2\,f^4}-a^2\,c\,f^2\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{f^4}\right )\,\sqrt {\frac {\sqrt {-a^4\,d^2\,f^4}-a^2\,c\,f^2}{f^4}}}{16\,\left (a^3\,c^2\,d^3+a^3\,d^5\right )}\right )\,\sqrt {\frac {\sqrt {-a^4\,d^2\,f^4}-a^2\,c\,f^2}{f^4}}+\frac {2\,b\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b*tan(e + f*x))*(c + d*tan(e + f*x))^(1/2),x)
[Out]
2*atanh((32*b^2*d^4*((b^2*c)/(4*f^2) - (-b^4*d^2*f^4)^(1/2)/(4*f^4))^(1/2)*(c + d*tan(e + f*x))^(1/2))/((16*b*
d^4*(-b^4*d^2*f^4)^(1/2))/f^3 + (16*b*c^2*d^2*(-b^4*d^2*f^4)^(1/2))/f^3) - (32*c*d^2*((b^2*c)/(4*f^2) - (-b^4*
d^2*f^4)^(1/2)/(4*f^4))^(1/2)*(c + d*tan(e + f*x))^(1/2)*(-b^4*d^2*f^4)^(1/2))/((16*b*d^4*(-b^4*d^2*f^4)^(1/2)
)/f + (16*b*c^2*d^2*(-b^4*d^2*f^4)^(1/2))/f))*(-((-b^4*d^2*f^4)^(1/2) - b^2*c*f^2)/(4*f^4))^(1/2) - 2*atanh((3
2*b^2*d^4*((-b^4*d^2*f^4)^(1/2)/(4*f^4) + (b^2*c)/(4*f^2))^(1/2)*(c + d*tan(e + f*x))^(1/2))/((16*b*d^4*(-b^4*
d^2*f^4)^(1/2))/f^3 + (16*b*c^2*d^2*(-b^4*d^2*f^4)^(1/2))/f^3) + (32*c*d^2*((-b^4*d^2*f^4)^(1/2)/(4*f^4) + (b^
2*c)/(4*f^2))^(1/2)*(c + d*tan(e + f*x))^(1/2)*(-b^4*d^2*f^4)^(1/2))/((16*b*d^4*(-b^4*d^2*f^4)^(1/2))/f + (16*
b*c^2*d^2*(-b^4*d^2*f^4)^(1/2))/f))*(((-b^4*d^2*f^4)^(1/2) + b^2*c*f^2)/(4*f^4))^(1/2) - atanh((f^3*((16*(a^2*
d^4 - a^2*c^2*d^2)*(c + d*tan(e + f*x))^(1/2))/f^2 + (16*c*d^2*((-a^4*d^2*f^4)^(1/2) + a^2*c*f^2)*(c + d*tan(e
+ f*x))^(1/2))/f^4)*(-((-a^4*d^2*f^4)^(1/2) + a^2*c*f^2)/f^4)^(1/2))/(16*(a^3*d^5 + a^3*c^2*d^3)))*(-((-a^4*d
^2*f^4)^(1/2) + a^2*c*f^2)/f^4)^(1/2) - atanh((f^3*((16*(a^2*d^4 - a^2*c^2*d^2)*(c + d*tan(e + f*x))^(1/2))/f^
2 - (16*c*d^2*((-a^4*d^2*f^4)^(1/2) - a^2*c*f^2)*(c + d*tan(e + f*x))^(1/2))/f^4)*(((-a^4*d^2*f^4)^(1/2) - a^2
*c*f^2)/f^4)^(1/2))/(16*(a^3*d^5 + a^3*c^2*d^3)))*(((-a^4*d^2*f^4)^(1/2) - a^2*c*f^2)/f^4)^(1/2) + (2*b*(c + d
*tan(e + f*x))^(1/2))/f
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