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

Optimal. Leaf size=186 \[ \frac{b (3 a-5 b) \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{2 f (a-b)}+\frac{\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{5/2}}{3 f (a-b)}-\frac{(3 a-5 b) \cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{3 f (a-b)}+\frac{\sqrt{b} (3 a-5 b) \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{2 f} \]

[Out]

((3*a - 5*b)*Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/(2*f) + ((3*a - 5*b)*b*Se
c[e + f*x]*Sqrt[a - b + b*Sec[e + f*x]^2])/(2*(a - b)*f) - ((3*a - 5*b)*Cos[e + f*x]*(a - b + b*Sec[e + f*x]^2
)^(3/2))/(3*(a - b)*f) + (Cos[e + f*x]^3*(a - b + b*Sec[e + f*x]^2)^(5/2))/(3*(a - b)*f)

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Rubi [A]  time = 0.164599, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3664, 453, 277, 195, 217, 206} \[ \frac{b (3 a-5 b) \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{2 f (a-b)}+\frac{\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{5/2}}{3 f (a-b)}-\frac{(3 a-5 b) \cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{3 f (a-b)}+\frac{\sqrt{b} (3 a-5 b) \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a - 5*b)*Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/(2*f) + ((3*a - 5*b)*b*Se
c[e + f*x]*Sqrt[a - b + b*Sec[e + f*x]^2])/(2*(a - b)*f) - ((3*a - 5*b)*Cos[e + f*x]*(a - b + b*Sec[e + f*x]^2
)^(3/2))/(3*(a - b)*f) + (Cos[e + f*x]^3*(a - b + b*Sec[e + f*x]^2)^(5/2))/(3*(a - b)*f)

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]

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

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 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 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])

Rubi steps

\begin{align*} \int \sin ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right ) \left (a-b+b x^2\right )^{3/2}}{x^4} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac{\cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{5/2}}{3 (a-b) f}+\frac{(3 a-5 b) \operatorname{Subst}\left (\int \frac{\left (a-b+b x^2\right )^{3/2}}{x^2} \, dx,x,\sec (e+f x)\right )}{3 (a-b) f}\\ &=-\frac{(3 a-5 b) \cos (e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{3 (a-b) f}+\frac{\cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{5/2}}{3 (a-b) f}+\frac{((3 a-5 b) b) \operatorname{Subst}\left (\int \sqrt{a-b+b x^2} \, dx,x,\sec (e+f x)\right )}{(a-b) f}\\ &=\frac{(3 a-5 b) b \sec (e+f x) \sqrt{a-b+b \sec ^2(e+f x)}}{2 (a-b) f}-\frac{(3 a-5 b) \cos (e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{3 (a-b) f}+\frac{\cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{5/2}}{3 (a-b) f}+\frac{((3 a-5 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 a-5 b) b \sec (e+f x) \sqrt{a-b+b \sec ^2(e+f x)}}{2 (a-b) f}-\frac{(3 a-5 b) \cos (e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{3 (a-b) f}+\frac{\cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{5/2}}{3 (a-b) f}+\frac{((3 a-5 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-5 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 a-5 b) b \sec (e+f x) \sqrt{a-b+b \sec ^2(e+f x)}}{2 (a-b) f}-\frac{(3 a-5 b) \cos (e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{3 (a-b) f}+\frac{\cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{5/2}}{3 (a-b) f}\\ \end{align*}

Mathematica [A]  time = 1.78038, size = 188, normalized size = 1.01 \[ \frac{\sec (e+f x) \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (\sqrt{(a-b) \cos (2 (e+f x))+a+b} (-8 (a-3 b) \cos (2 (e+f x))+(a-b) \cos (4 (e+f x))-9 a+37 b)+12 \sqrt{2} \sqrt{b} (3 a-5 b) \cos ^2(e+f x) \tanh ^{-1}\left (\frac{\sqrt{(a-b) \cos (2 (e+f x))+a+b}}{\sqrt{2} \sqrt{b}}\right )\right )}{24 \sqrt{2} f \sqrt{(a-b) \cos (2 (e+f x))+a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((12*Sqrt[2]*(3*a - 5*b)*Sqrt[b]*ArcTanh[Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]]/(Sqrt[2]*Sqrt[b])]*Cos[e + f*x
]^2 + Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]]*(-9*a + 37*b - 8*(a - 3*b)*Cos[2*(e + f*x)] + (a - b)*Cos[4*(e +
f*x)]))*Sec[e + f*x]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(24*Sqrt[2]*f*Sqrt[a + b + (a -
b)*Cos[2*(e + f*x)]])

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Maple [B]  time = 0.183, size = 1104, normalized size = 5.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12/f/a^(5/2)/b^(1/2)*(cos(f*x+e)-1)^3*(2*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(7/2)
*b^(1/2)*cos(f*x+e)^5-2*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(5/2)*b^(3/2)*cos(f*x+e)^
5+2*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(7/2)*b^(1/2)*cos(f*x+e)^4-2*((a*cos(f*x+e)^2
-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(5/2)*b^(3/2)*cos(f*x+e)^4-6*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(
cos(f*x+e)+1)^2)^(1/2)*a^(7/2)*b^(1/2)*cos(f*x+e)^3+14*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1
/2)*a^(5/2)*b^(3/2)*cos(f*x+e)^3-6*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(7/2)*b^(1/2)*
cos(f*x+e)^2+14*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(5/2)*b^(3/2)*cos(f*x+e)^2-6*ln(-
2/a^(1/2)*(cos(f*x+e)-1)*(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)-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)*b^(7/2
)*cos(f*x+e)^2*a+6*ln(-4/a^(1/2)*(cos(f*x+e)-1)*(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)-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)*b^(7/2)*cos(f*x+e)^2*a+9*arctanh(1/8*b^(1/2)*4^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*4^(1/2)-2*cos
(f*x+e)-4^(1/2)-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^(7/2)*cos(f*x+e)
^2*b+3*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(5/2)*b^(3/2)*cos(f*x+e)-15*arctanh(1/8*b^
(1/2)*4^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-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^(5/2)*cos(f*x+e)^2*b^2+3*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+
1)^2)^(1/2)*a^(5/2)*b^(3/2))*cos(f*x+e)*4^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(3/2)/((a*cos
(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(3/2)/sin(f*x+e)^6

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.51812, size = 768, normalized size = 4.13 \begin{align*} \left [-\frac{3 \,{\left (3 \, a - 5 \, 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 )^{4} - 2 \,{\left (3 \, a - 7 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \, f \cos \left (f x + e\right )}, -\frac{3 \,{\left (3 \, a - 5 \, 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 )^{4} - 2 \,{\left (3 \, a - 7 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{6 \, f \cos \left (f x + e\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*(3*(3*a - 5*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)^4 - 2*(3*a - 7*b)*cos(
f*x + e)^2 + 3*b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e)), -1/6*(3*(3*a - 5*b)*sqr
t(-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)^4 - 2*(3*a - 7*b)*cos(f*x + e)^2 + 3*b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f
*cos(f*x + e))]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError