3.1.95 \(\int \csc (e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\) [95]

Optimal. Leaf size=84 \[ -\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{f}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{f} \]

[Out]

-arctanh(sec(f*x+e)*a^(1/2)/(a-b+b*sec(f*x+e)^2)^(1/2))*a^(1/2)/f+arctanh(sec(f*x+e)*b^(1/2)/(a-b+b*sec(f*x+e)
^2)^(1/2))*b^(1/2)/f

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Rubi [A]
time = 0.06, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3745, 399, 223, 212, 385, 213} \begin {gather*} \frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{f}-\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

-((Sqrt[a]*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/f) + (Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[
e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/f

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 213

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

Rule 223

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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 399

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Dist[b/d, Int[(a + b*x^n)^(p - 1), x]
, x] - Dist[(b*c - a*d)/d, Int[(a + b*x^n)^(p - 1)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c
- a*d, 0] && EqQ[n*(p - 1) + 1, 0] && IntegerQ[n]

Rule 3745

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a - b + b*ff^2*x^2)^p/x^(m
+ 1)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \csc (e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx &=\frac {\text {Subst}\left (\int \frac {\sqrt {a-b+b x^2}}{-1+x^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac {a \text {Subst}\left (\int \frac {1}{\left (-1+x^2\right ) \sqrt {a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{f}+\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac {a \text {Subst}\left (\int \frac {1}{-1+a x^2} \, dx,x,\frac {\sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{f}+\frac {b \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{f}\\ &=-\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{f}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{f}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(300\) vs. \(2(84)=168\).
time = 4.68, size = 300, normalized size = 3.57 \begin {gather*} \frac {\cos (e+f x) \left (4 \sqrt {b} \tanh ^{-1}\left (\frac {-\sqrt {a} \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{2 \sqrt {b}}\right )+\sqrt {a} \left (2 \tanh ^{-1}\left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\frac {\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{\sqrt {a}}\right )+\log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )\right )\right ) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{2 f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

(Cos[e + f*x]*(4*Sqrt[b]*ArcTanh[(-(Sqrt[a]*(-1 + Tan[(e + f*x)/2]^2)) + Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 +
Tan[(e + f*x)/2]^2)^2])/(2*Sqrt[b])] + Sqrt[a]*(2*ArcTanh[Tan[(e + f*x)/2]^2 - Sqrt[4*b*Tan[(e + f*x)/2]^2 +
a*(-1 + Tan[(e + f*x)/2]^2)^2]/Sqrt[a]] + Log[a - 2*b - a*Tan[(e + f*x)/2]^2 + Sqrt[a]*Sqrt[4*b*Tan[(e + f*x)/
2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]]))*Sec[(e + f*x)/2]^2*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*
x]^2])/(2*f*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[(e + f*x)/2]^4])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(720\) vs. \(2(72)=144\).
time = 0.38, size = 721, normalized size = 8.58

method	result	size
default	\(-\frac {\sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\cos \left (f x +e \right )^{2}}}\, \sqrt {4}\, \cos \left (f x +e \right ) \left (\cos \left (f x +e \right )-1\right ) \left (2 \arctanh \left (\frac {\left (\cos \left (f x +e \right )-1\right ) \left (\sqrt {4}\, \cos \left (f x +e \right )-\sqrt {4}-2 \cos \left (f x +e \right )-2\right ) \sqrt {b}\, \sqrt {4}}{8 \sin \left (f x +e \right )^{2} \sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right ) \sqrt {a}\, b +2 \ln \left (-\frac {2 \left (\cos \left (f x +e \right )-1\right ) \left (\cos \left (f x +e \right ) \sqrt {a}\, \sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right ) a +b \cos \left (f x +e \right )+\sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {a}+b \right )}{\sin \left (f x +e \right )^{2} \sqrt {a}}\right ) b^{\frac {3}{2}}-2 b^{\frac {3}{2}} \ln \left (-\frac {4 \left (\cos \left (f x +e \right )-1\right ) \left (\cos \left (f x +e \right ) \sqrt {a}\, \sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right ) a +b \cos \left (f x +e \right )+\sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {a}+b \right )}{\sin \left (f x +e \right )^{2} \sqrt {a}}\right )-\ln \left (-\frac {2 \left (\cos \left (f x +e \right )-1\right ) \left (\cos \left (f x +e \right ) \sqrt {a}\, \sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right ) a +b \cos \left (f x +e \right )+\sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {a}+b \right )}{\sin \left (f x +e \right )^{2} \sqrt {a}}\right ) a \sqrt {b}-a \ln \left (-\frac {4 \left (\cos \left (f x +e \right ) \sqrt {a}\, \sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+\sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {a}+\cos \left (f x +e \right ) a -b \cos \left (f x +e \right )+b \right )}{\cos \left (f x +e \right )-1}\right ) \sqrt {b}\right )}{4 f \sin \left (f x +e \right )^{2} \sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {a}\, \sqrt {b}}\)	\(721\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/4/f*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(1/2)*4^(1/2)*cos(f*x+e)*(cos(f*x+e)-1)*(2*arctanh(1/8
*(cos(f*x+e)-1)*(4^(1/2)*cos(f*x+e)-4^(1/2)-2*cos(f*x+e)-2)/sin(f*x+e)^2/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(c
os(f*x+e)+1)^2)^(1/2)*b^(1/2)*4^(1/2))*a^(1/2)*b+2*ln(-2*(cos(f*x+e)-1)*(cos(f*x+e)*a^(1/2)*((a*cos(f*x+e)^2-c
os(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x
+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2/a^(1/2))*b^(3/2)-2*b^(3/2)*ln(-4*(cos(f*x+e)-1)*(cos(f*x+e)*a^(1/2)*((
a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)
^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2/a^(1/2))-ln(-2*(cos(f*x+e)-1)*(cos(f*x+e)*a^(1/2)*((a*
cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2
*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2/a^(1/2))*a*b^(1/2)-a*ln(-4*(cos(f*x+e)*a^(1/2)*((a*cos(f
*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a
^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)-1))*b^(1/2))/sin(f*x+e)^2/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(
cos(f*x+e)+1)^2)^(1/2)/a^(1/2)/b^(1/2)

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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(csc(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*tan(f*x + e)^2 + a)*csc(f*x + e), x)

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Fricas [A]
time = 2.57, size = 546, normalized size = 6.50 \begin {gather*} \left [\frac {\sqrt {a} \log \left (-\frac {2 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) + \sqrt {b} \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right )}{2 \, f}, -\frac {2 \, \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) - \sqrt {a} \log \left (-\frac {2 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right )}{2 \, f}, \frac {2 \, \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a}\right ) + \sqrt {b} \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right )}{2 \, f}, \frac {\sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a}\right ) - \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right )}{f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/2*(sqrt(a)*log(-2*((a - b)*cos(f*x + e)^2 - 2*sqrt(a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos
(f*x + e) + a + b)/(cos(f*x + e)^2 - 1)) + sqrt(b)*log(-((a - b)*cos(f*x + e)^2 + 2*sqrt(b)*sqrt(((a - b)*cos(
f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2))/f, -1/2*(2*sqrt(-b)*arctan(sqrt(-b)*sqrt(
((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b) - sqrt(a)*log(-2*((a - b)*cos(f*x + e)^2 - 2*sqrt
(a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + a + b)/(cos(f*x + e)^2 - 1)))/f, 1/2*(2*s
qrt(-a)*arctan(sqrt(-a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/a) + sqrt(b)*log(-((a -
b)*cos(f*x + e)^2 + 2*sqrt(b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x +
e)^2))/f, (sqrt(-a)*arctan(sqrt(-a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/a) - sqrt(
-b)*arctan(sqrt(-b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b))/f]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a + b \tan ^{2}{\left (e + f x \right )}} \csc {\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)*(a+b*tan(f*x+e)**2)**(1/2),x)

[Out]

Integral(sqrt(a + b*tan(e + f*x)**2)*csc(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(csc(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(t_

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{\sin \left (e+f\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(e + f*x)^2)^(1/2)/sin(e + f*x),x)

[Out]

int((a + b*tan(e + f*x)^2)^(1/2)/sin(e + f*x), x)

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