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

Optimal. Leaf size=138 \[ -\frac{B \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{B \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{B \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{B \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d} \]

[Out]

-((B*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d)) + (B*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2
]*d) + (B*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) - (B*Log[1 + Sqrt[2]*Sqrt[Tan[c +
d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d)

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Rubi [A]  time = 0.0991125, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {21, 3476, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac{B \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{B \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{B \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{B \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((B*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d)) + (B*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2
]*d) + (B*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) - (B*Log[1 + Sqrt[2]*Sqrt[Tan[c +
d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{\sqrt{\tan (c+d x)} (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx &=B \int \sqrt{\tan (c+d x)} \, dx\\ &=\frac{B \operatorname{Subst}\left (\int \frac{\sqrt{x}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{(2 B) \operatorname{Subst}\left (\int \frac{x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{B \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}+\frac{B \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{B \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 d}+\frac{B \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 d}+\frac{B \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} d}+\frac{B \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} d}\\ &=\frac{B \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}-\frac{B \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}+\frac{B \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{B \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}\\ &=-\frac{B \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{B \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{B \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}-\frac{B \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}\\ \end{align*}

Mathematica [C]  time = 0.019991, size = 36, normalized size = 0.26 \[ \frac{2 B \tan ^{\frac{3}{2}}(c+d x) \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},-\tan ^2(c+d x)\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*B*Hypergeometric2F1[3/4, 1, 7/4, -Tan[c + d*x]^2]*Tan[c + d*x]^(3/2))/(3*d)

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Maple [A]  time = 0.035, size = 104, normalized size = 0.8 \begin{align*}{\frac{B\sqrt{2}}{4\,d}\ln \left ({ \left ( 1-\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }+{\frac{B\sqrt{2}}{2\,d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }+{\frac{B\sqrt{2}}{2\,d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/d*B*2^(1/2)*ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+1/2*B*arct
an(1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)+1/2*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)

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Maxima [A]  time = 1.55446, size = 147, normalized size = 1.07 \begin{align*} \frac{{\left (2 \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) - \sqrt{2} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt{2} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} B}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) -
2*sqrt(tan(d*x + c)))) - sqrt(2)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + sqrt(2)*log(-sqrt(2)*sq
rt(tan(d*x + c)) + tan(d*x + c) + 1))*B/d

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Fricas [B]  time = 2.29129, size = 1283, normalized size = 9.3 \begin{align*} -\sqrt{2} \left (\frac{B^{4}}{d^{4}}\right )^{\frac{1}{4}} \arctan \left (-\frac{\sqrt{2} B^{3} d \left (\frac{B^{4}}{d^{4}}\right )^{\frac{1}{4}} \sqrt{\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right )}} + B^{4} - \sqrt{2} d \left (\frac{B^{4}}{d^{4}}\right )^{\frac{1}{4}} \sqrt{\frac{\sqrt{2} B^{3} d^{3} \left (\frac{B^{4}}{d^{4}}\right )^{\frac{3}{4}} \sqrt{\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) + B^{4} d^{2} \sqrt{\frac{B^{4}}{d^{4}}} \cos \left (d x + c\right ) + B^{6} \sin \left (d x + c\right )}{\cos \left (d x + c\right )}}}{B^{4}}\right ) - \sqrt{2} \left (\frac{B^{4}}{d^{4}}\right )^{\frac{1}{4}} \arctan \left (-\frac{\sqrt{2} B^{3} d \left (\frac{B^{4}}{d^{4}}\right )^{\frac{1}{4}} \sqrt{\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right )}} - B^{4} - \sqrt{2} d \left (\frac{B^{4}}{d^{4}}\right )^{\frac{1}{4}} \sqrt{-\frac{\sqrt{2} B^{3} d^{3} \left (\frac{B^{4}}{d^{4}}\right )^{\frac{3}{4}} \sqrt{\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - B^{4} d^{2} \sqrt{\frac{B^{4}}{d^{4}}} \cos \left (d x + c\right ) - B^{6} \sin \left (d x + c\right )}{\cos \left (d x + c\right )}}}{B^{4}}\right ) - \frac{1}{4} \, \sqrt{2} \left (\frac{B^{4}}{d^{4}}\right )^{\frac{1}{4}} \log \left (\frac{\sqrt{2} B^{3} d^{3} \left (\frac{B^{4}}{d^{4}}\right )^{\frac{3}{4}} \sqrt{\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) + B^{4} d^{2} \sqrt{\frac{B^{4}}{d^{4}}} \cos \left (d x + c\right ) + B^{6} \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right ) + \frac{1}{4} \, \sqrt{2} \left (\frac{B^{4}}{d^{4}}\right )^{\frac{1}{4}} \log \left (-\frac{\sqrt{2} B^{3} d^{3} \left (\frac{B^{4}}{d^{4}}\right )^{\frac{3}{4}} \sqrt{\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - B^{4} d^{2} \sqrt{\frac{B^{4}}{d^{4}}} \cos \left (d x + c\right ) - B^{6} \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(2)*(B^4/d^4)^(1/4)*arctan(-(sqrt(2)*B^3*d*(B^4/d^4)^(1/4)*sqrt(sin(d*x + c)/cos(d*x + c)) + B^4 - sqrt(2
)*d*(B^4/d^4)^(1/4)*sqrt((sqrt(2)*B^3*d^3*(B^4/d^4)^(3/4)*sqrt(sin(d*x + c)/cos(d*x + c))*cos(d*x + c) + B^4*d
^2*sqrt(B^4/d^4)*cos(d*x + c) + B^6*sin(d*x + c))/cos(d*x + c)))/B^4) - sqrt(2)*(B^4/d^4)^(1/4)*arctan(-(sqrt(
2)*B^3*d*(B^4/d^4)^(1/4)*sqrt(sin(d*x + c)/cos(d*x + c)) - B^4 - sqrt(2)*d*(B^4/d^4)^(1/4)*sqrt(-(sqrt(2)*B^3*
d^3*(B^4/d^4)^(3/4)*sqrt(sin(d*x + c)/cos(d*x + c))*cos(d*x + c) - B^4*d^2*sqrt(B^4/d^4)*cos(d*x + c) - B^6*si
n(d*x + c))/cos(d*x + c)))/B^4) - 1/4*sqrt(2)*(B^4/d^4)^(1/4)*log((sqrt(2)*B^3*d^3*(B^4/d^4)^(3/4)*sqrt(sin(d*
x + c)/cos(d*x + c))*cos(d*x + c) + B^4*d^2*sqrt(B^4/d^4)*cos(d*x + c) + B^6*sin(d*x + c))/cos(d*x + c)) + 1/4
*sqrt(2)*(B^4/d^4)^(1/4)*log(-(sqrt(2)*B^3*d^3*(B^4/d^4)^(3/4)*sqrt(sin(d*x + c)/cos(d*x + c))*cos(d*x + c) -
B^4*d^2*sqrt(B^4/d^4)*cos(d*x + c) - B^6*sin(d*x + c))/cos(d*x + c))

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

[Out]

B*Integral(sqrt(tan(c + d*x)), x)

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Giac [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+b*B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

Timed out