3.428 \(\int \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=201 \[ \frac{\sqrt{-b+i a} (A+i B) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{(a B+2 A b) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{b} d}-\frac{\sqrt{b+i a} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d} \]

[Out]

(Sqrt[I*a - b]*(A + I*B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + ((2*A*b + a*
B)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[b]*d) - (Sqrt[I*a + b]*(A - I*B)*ArcT
anh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + (B*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c +
d*x]])/d

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Rubi [A]  time = 1.54709, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.257, Rules used = {3610, 3655, 6725, 63, 217, 206, 93, 205, 208} \[ \frac{\sqrt{-b+i a} (A+i B) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{(a B+2 A b) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{b} d}-\frac{\sqrt{b+i a} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]]*(A + B*Tan[c + d*x]),x]

[Out]

(Sqrt[I*a - b]*(A + I*B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + ((2*A*b + a*
B)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[b]*d) - (Sqrt[I*a + b]*(A - I*B)*ArcT
anh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + (B*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c +
d*x]])/d

Rule 3610

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(B*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(f*(m + n)), x] +
Dist[1/(m + n), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*A*c*(m + n) - B*(b*c*m +
a*d*n) + (A*b*c + a*B*c + a*A*d - b*B*d)*(m + n)*Tan[e + f*x] + (A*b*d*(m + n) + B*(a*d*m + b*c*n))*Tan[e + f
*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 +
d^2, 0] && LtQ[0, m, 1] && LtQ[0, n, 1]

Rule 3655

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x
]}, Dist[ff/f, Subst[Int[((a + b*ff*x)^m*(c + d*ff*x)^n*(A + B*ff*x + C*ff^2*x^2))/(1 + ff^2*x^2), x], x, Tan[
e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] &&
NeQ[c^2 + d^2, 0]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 63

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx &=\frac{B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}+\int \frac{-\frac{a B}{2}+(a A-b B) \tan (c+d x)+\frac{1}{2} (2 A b+a B) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx\\ &=\frac{B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{a B}{2}+(a A-b B) x+\frac{1}{2} (2 A b+a B) x^2}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}+\frac{\operatorname{Subst}\left (\int \left (\frac{2 A b+a B}{2 \sqrt{x} \sqrt{a+b x}}-\frac{A b+a B-(a A-b B) x}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}-\frac{\operatorname{Subst}\left (\int \frac{A b+a B-(a A-b B) x}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}+\frac{(2 A b+a B) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}-\frac{\operatorname{Subst}\left (\int \left (\frac{a A-b B+i (A b+a B)}{2 (i-x) \sqrt{x} \sqrt{a+b x}}+\frac{-a A+b B+i (A b+a B)}{2 \sqrt{x} (i+x) \sqrt{a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d}+\frac{(2 A b+a B) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}+\frac{((a-i b) (A-i B)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} (i+x) \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac{((a+i b) (A+i B)) \operatorname{Subst}\left (\int \frac{1}{(i-x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{(2 A b+a B) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}\\ &=\frac{(2 A b+a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{b} d}+\frac{B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}+\frac{((a-i b) (A-i B)) \operatorname{Subst}\left (\int \frac{1}{i-(-a+i b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{((a+i b) (A+i B)) \operatorname{Subst}\left (\int \frac{1}{i-(a+i b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}\\ &=\frac{\sqrt{i a-b} (A+i B) \tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{(2 A b+a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{b} d}-\frac{\sqrt{i a+b} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d}\\ \end{align*}

Mathematica [A]  time = 3.00635, size = 241, normalized size = 1.2 \[ \frac{-\sqrt [4]{-1} \sqrt{-a-i b} (A+i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )-\sqrt [4]{-1} \sqrt{a-i b} (A-i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )+\frac{(a B+2 A b) \sqrt{a+b \tan (c+d x)} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \sqrt{\frac{b \tan (c+d x)}{a}+1}}+B \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]]*(A + B*Tan[c + d*x]),x]

[Out]

(-((-1)^(1/4)*Sqrt[-a - I*b]*(A + I*B)*ArcTanh[((-1)^(1/4)*Sqrt[-a - I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c
+ d*x]]]) - (-1)^(1/4)*Sqrt[a - I*b]*(A - I*B)*ArcTanh[((-1)^(1/4)*Sqrt[a - I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a +
b*Tan[c + d*x]]] + B*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]] + ((2*A*b + a*B)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c
+ d*x]])/Sqrt[a]]*Sqrt[a + b*Tan[c + d*x]])/(Sqrt[a]*Sqrt[b]*Sqrt[1 + (b*Tan[c + d*x])/a]))/d

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Maple [B]  time = 0.712, size = 2178530, normalized size = 10838.5 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )} \sqrt{b \tan \left (d x + c\right ) + a} \sqrt{\tan \left (d x + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)), x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (A + B \tan{\left (c + d x \right )}\right ) \sqrt{a + b \tan{\left (c + d x \right )}} \sqrt{\tan{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**(1/2)*(a+b*tan(d*x+c))**(1/2)*(A+B*tan(d*x+c)),x)

[Out]

Integral((A + B*tan(c + d*x))*sqrt(a + b*tan(c + d*x))*sqrt(tan(c + d*x)), x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

Exception raised: RuntimeError