3.106 \(\int \sin (e+f x) (a+b \tan ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=113 \[ \frac {3 b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 f}-\frac {\cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{f}+\frac {3 \sqrt {b} (a-b) \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{2 f} \]

[Out]

-cos(f*x+e)*(a-b+b*sec(f*x+e)^2)^(3/2)/f+3/2*(a-b)*arctanh(sec(f*x+e)*b^(1/2)/(a-b+b*sec(f*x+e)^2)^(1/2))*b^(1
/2)/f+3/2*b*sec(f*x+e)*(a-b+b*sec(f*x+e)^2)^(1/2)/f

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Rubi [A]  time = 0.08, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3664, 277, 195, 217, 206} \[ \frac {3 b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 f}-\frac {\cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{f}+\frac {3 \sqrt {b} (a-b) \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]*(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

(3*(a - b)*Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/(2*f) + (3*b*Sec[e + f*x]*S
qrt[a - b + b*Sec[e + f*x]^2])/(2*f) - (Cos[e + f*x]*(a - b + b*Sec[e + f*x]^2)^(3/2))/f

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 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 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 3664

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 \sin (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a-b+b x^2\right )^{3/2}}{x^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\cos (e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{f}+\frac {(3 b) \operatorname {Subst}\left (\int \sqrt {a-b+b x^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac {3 b \sec (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{2 f}-\frac {\cos (e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{f}+\frac {(3 (a-b) b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{2 f}\\ &=\frac {3 b \sec (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{2 f}-\frac {\cos (e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{f}+\frac {(3 (a-b) b) \operatorname {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 )}{2 f}\\ &=\frac {3 (a-b) \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{2 f}+\frac {3 b \sec (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{2 f}-\frac {\cos (e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{f}\\ \end {align*}

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Mathematica [A]  time = 1.33, size = 170, normalized size = 1.50 \[ \frac {\sec (e+f x) \sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (6 \sqrt {2} \sqrt {b} (a-b) \cos ^2(e+f x) \tanh ^{-1}\left (\frac {\sqrt {(a-b) \cos (2 (e+f x))+a+b}}{\sqrt {2} \sqrt {b}}\right )-2 ((a-b) \cos (2 (e+f x))+a-2 b) \sqrt {(a-b) \cos (2 (e+f x))+a+b}\right )}{4 \sqrt {2} f \sqrt {(a-b) \cos (2 (e+f x))+a+b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]*(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

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

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fricas [A]  time = 1.00, size = 268, normalized size = 2.37 \[ \left [-\frac {3 \, {\left (a - b\right )} \sqrt {b} \cos \left (f x + e\right ) \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \, {\left (2 \, {\left (a - b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, f \cos \left (f x + e\right )}, -\frac {3 \, {\left (a - b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) \cos \left (f x + e\right ) + {\left (2 \, {\left (a - b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, f \cos \left (f x + e\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/4*(3*(a - b)*sqrt(b)*cos(f*x + e)*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) + 2*(2*(a - b)*cos(f*x + e)^2 - b)*sqrt(((a - b)*cos(f*
x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e)), -1/2*(3*(a - b)*sqrt(-b)*arctan(sqrt(-b)*sqrt(((a - b)*cos(f*
x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b)*cos(f*x + e) + (2*(a - b)*cos(f*x + e)^2 - b)*sqrt(((a - b)*cos(
f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e))]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check si
gn: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)abs(f)*((-a*sqrt(a*f^2*(-cos(f*x+exp(1))/f)^2-b*f^2*(-cos(f*x+exp(1))/f)^2+b)*sign(cos(f*x+exp(1)))*sign(f
)+b*sqrt(a*f^2*(-cos(f*x+exp(1))/f)^2-b*f^2*(-cos(f*x+exp(1))/f)^2+b)*sign(cos(f*x+exp(1)))*sign(f))/f^2-1/2*(
b^2*sqrt(a*f^2*(-cos(f*x+exp(1))/f)^2-b*f^2*(-cos(f*x+exp(1))/f)^2+b)*sign(cos(f*x+exp(1)))*sign(f)-a*b*sqrt(a
*f^2*(-cos(f*x+exp(1))/f)^2-b*f^2*(-cos(f*x+exp(1))/f)^2+b)*sign(cos(f*x+exp(1)))*sign(f))/f^2/(a*f^2*(-cos(f*
x+exp(1))/f)^2-b*f^2*(-cos(f*x+exp(1))/f)^2)-1/2*(-3*b^2*sign(cos(f*x+exp(1)))*sign(f)+3*a*b*sign(cos(f*x+exp(
1)))*sign(f))*atan(sqrt(a*f^2*(-cos(f*x+exp(1))/f)^2-b*f^2*(-cos(f*x+exp(1))/f)^2+b)/sqrt(-b))/sqrt(-b)/f^2)

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maple [B]  time = 0.32, size = 357, normalized size = 3.16 \[ \frac {\left (\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\cos \left (f x +e \right )^{2}}\right )^{\frac {3}{2}} \cos \left (f x +e \right ) \left (-\left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{\frac {3}{2}} \left (\cos ^{2}\left (f x +e \right )\right ) a +\left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{\frac {3}{2}} \left (\cos ^{2}\left (f x +e \right )\right ) b +3 b^{\frac {3}{2}} \ln \left (\frac {2 \sqrt {b}\, \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}+2 b}{\cos \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right ) a -3 b^{\frac {5}{2}} \ln \left (\frac {2 \sqrt {b}\, \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}+2 b}{\cos \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+\left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{\frac {5}{2}}-3 \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}\, \left (\cos ^{2}\left (f x +e \right )\right ) a b +3 \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}\, \left (\cos ^{2}\left (f x +e \right )\right ) b^{2}\right )}{2 f \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{\frac {3}{2}} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)*(a+b*tan(f*x+e)^2)^(3/2),x)

[Out]

1/2/f*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(3/2)*cos(f*x+e)*(-(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^(3
/2)*cos(f*x+e)^2*a+(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^(3/2)*cos(f*x+e)^2*b+3*b^(3/2)*ln(2*(b^(1/2)*(a*cos(f*x+e
)^2-cos(f*x+e)^2*b+b)^(1/2)+b)/cos(f*x+e))*cos(f*x+e)^2*a-3*b^(5/2)*ln(2*(b^(1/2)*(a*cos(f*x+e)^2-cos(f*x+e)^2
*b+b)^(1/2)+b)/cos(f*x+e))*cos(f*x+e)^2+(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^(5/2)-3*(a*cos(f*x+e)^2-cos(f*x+e)^2
*b+b)^(1/2)*cos(f*x+e)^2*a*b+3*(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^(1/2)*cos(f*x+e)^2*b^2)/(a*cos(f*x+e)^2-cos(f
*x+e)^2*b+b)^(3/2)/b

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maxima [A]  time = 0.85, size = 176, normalized size = 1.56 \[ -\frac {4 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} {\left (a - b\right )} \cos \left (f x + e\right ) - \frac {2 \, {\left (a b - b^{2}\right )} \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{{\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} \cos \left (f x + e\right )^{2} - b} + \frac {3 \, {\left (a b - b^{2}\right )} \log \left (\frac {\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right )}{\sqrt {b}}}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

-1/4*(4*sqrt(a - b + b/cos(f*x + e)^2)*(a - b)*cos(f*x + e) - 2*(a*b - b^2)*sqrt(a - b + b/cos(f*x + e)^2)*cos
(f*x + e)/((a - b + b/cos(f*x + e)^2)*cos(f*x + e)^2 - b) + 3*(a*b - b^2)*log((sqrt(a - b + b/cos(f*x + e)^2)*
cos(f*x + e) - sqrt(b))/(sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e) + sqrt(b)))/sqrt(b))/f

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sin \left (e+f\,x\right )\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)*(a + b*tan(e + f*x)^2)^(3/2),x)

[Out]

int(sin(e + f*x)*(a + b*tan(e + f*x)^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \sin {\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)*(a+b*tan(f*x+e)**2)**(3/2),x)

[Out]

Integral((a + b*tan(e + f*x)**2)**(3/2)*sin(e + f*x), x)

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