\(\int f^{a+c x^2} \sin ^2(d+f x^2) \, dx\) [89]

Optimal result

Integrand size = 20, antiderivative size = 140 \[ \int f^{a+c x^2} \sin ^2\left (d+f x^2\right ) \, dx=\frac {f^a \sqrt {\pi } \text {erfi}\left (\sqrt {c} x \sqrt {\log (f)}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {e^{-2 i d} f^a \sqrt {\pi } \text {erf}\left (x \sqrt {2 i f-c \log (f)}\right )}{8 \sqrt {2 i f-c \log (f)}}-\frac {e^{2 i d} f^a \sqrt {\pi } \text {erfi}\left (x \sqrt {2 i f+c \log (f)}\right )}{8 \sqrt {2 i f+c \log (f)}} \]

[Out]

1/4*f^a*erfi(x*c^(1/2)*ln(f)^(1/2))*Pi^(1/2)/c^(1/2)/ln(f)^(1/2)-1/8*f^a*erf(x*(2*I*f-c*ln(f))^(1/2))*Pi^(1/2)
/exp(2*I*d)/(2*I*f-c*ln(f))^(1/2)-1/8*exp(2*I*d)*f^a*erfi(x*(2*I*f+c*ln(f))^(1/2))*Pi^(1/2)/(2*I*f+c*ln(f))^(1
/2)

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4560, 2235, 2325, 2236} \[ \int f^{a+c x^2} \sin ^2\left (d+f x^2\right ) \, dx=-\frac {\sqrt {\pi } e^{-2 i d} f^a \text {erf}\left (x \sqrt {-c \log (f)+2 i f}\right )}{8 \sqrt {-c \log (f)+2 i f}}-\frac {\sqrt {\pi } e^{2 i d} f^a \text {erfi}\left (x \sqrt {c \log (f)+2 i f}\right )}{8 \sqrt {c \log (f)+2 i f}}+\frac {\sqrt {\pi } f^a \text {erfi}\left (\sqrt {c} x \sqrt {\log (f)}\right )}{4 \sqrt {c} \sqrt {\log (f)}} \]

[In]

Int[f^(a + c*x^2)*Sin[d + f*x^2]^2,x]

[Out]

(f^a*Sqrt[Pi]*Erfi[Sqrt[c]*x*Sqrt[Log[f]]])/(4*Sqrt[c]*Sqrt[Log[f]]) - (f^a*Sqrt[Pi]*Erf[x*Sqrt[(2*I)*f - c*Lo
g[f]]])/(8*E^((2*I)*d)*Sqrt[(2*I)*f - c*Log[f]]) - (E^((2*I)*d)*f^a*Sqrt[Pi]*Erfi[x*Sqrt[(2*I)*f + c*Log[f]]])
/(8*Sqrt[(2*I)*f + c*Log[f]])

Rule 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2236

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erf[(c + d*x)*Rt[(-b)*Log[F],
2]]/(2*d*Rt[(-b)*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 2325

Int[(u_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, Int[u*NormalizeIntegrand[E^z, x],
x] /; BinomialQ[z, x] || (PolynomialQ[z, x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, x]

Rule 4560

Int[(F_)^(u_)*Sin[v_]^(n_.), x_Symbol] :> Int[ExpandTrigToExp[F^u, Sin[v]^n, x], x] /; FreeQ[F, x] && (LinearQ
[u, x] || PolyQ[u, x, 2]) && (LinearQ[v, x] || PolyQ[v, x, 2]) && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{2} f^{a+c x^2}-\frac {1}{4} e^{-2 i d-2 i f x^2} f^{a+c x^2}-\frac {1}{4} e^{2 i d+2 i f x^2} f^{a+c x^2}\right ) \, dx \\ & = -\left (\frac {1}{4} \int e^{-2 i d-2 i f x^2} f^{a+c x^2} \, dx\right )-\frac {1}{4} \int e^{2 i d+2 i f x^2} f^{a+c x^2} \, dx+\frac {1}{2} \int f^{a+c x^2} \, dx \\ & = \frac {f^a \sqrt {\pi } \text {erfi}\left (\sqrt {c} x \sqrt {\log (f)}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {1}{4} \int \exp \left (-2 i d+a \log (f)-x^2 (2 i f-c \log (f))\right ) \, dx-\frac {1}{4} \int \exp \left (2 i d+a \log (f)+x^2 (2 i f+c \log (f))\right ) \, dx \\ & = \frac {f^a \sqrt {\pi } \text {erfi}\left (\sqrt {c} x \sqrt {\log (f)}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {e^{-2 i d} f^a \sqrt {\pi } \text {erf}\left (x \sqrt {2 i f-c \log (f)}\right )}{8 \sqrt {2 i f-c \log (f)}}-\frac {e^{2 i d} f^a \sqrt {\pi } \text {erfi}\left (x \sqrt {2 i f+c \log (f)}\right )}{8 \sqrt {2 i f+c \log (f)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.34 \[ \int f^{a+c x^2} \sin ^2\left (d+f x^2\right ) \, dx=\frac {1}{8} f^a \sqrt {\pi } \left (\frac {2 \text {erfi}\left (\sqrt {c} x \sqrt {\log (f)}\right )}{\sqrt {c} \sqrt {\log (f)}}+\frac {\sqrt [4]{-1} \left (\text {erf}\left (\sqrt [4]{-1} x \sqrt {2 f+i c \log (f)}\right ) \sqrt {2 f+i c \log (f)} (2 i f+c \log (f)) (\cos (2 d)-i \sin (2 d))+\text {erf}\left ((-1)^{3/4} x \sqrt {2 f-i c \log (f)}\right ) \sqrt {2 f-i c \log (f)} (2 f+i c \log (f)) (\cos (2 d)+i \sin (2 d))\right )}{4 f^2+c^2 \log ^2(f)}\right ) \]

[In]

Integrate[f^(a + c*x^2)*Sin[d + f*x^2]^2,x]

[Out]

(f^a*Sqrt[Pi]*((2*Erfi[Sqrt[c]*x*Sqrt[Log[f]]])/(Sqrt[c]*Sqrt[Log[f]]) + ((-1)^(1/4)*(Erf[(-1)^(1/4)*x*Sqrt[2*
f + I*c*Log[f]]]*Sqrt[2*f + I*c*Log[f]]*((2*I)*f + c*Log[f])*(Cos[2*d] - I*Sin[2*d]) + Erf[(-1)^(3/4)*x*Sqrt[2
*f - I*c*Log[f]]]*Sqrt[2*f - I*c*Log[f]]*(2*f + I*c*Log[f])*(Cos[2*d] + I*Sin[2*d])))/(4*f^2 + c^2*Log[f]^2)))
/8

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.76

method	result	size
risch	\(-\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-2 i d} \operatorname {erf}\left (x \sqrt {2 i f -c \ln \left (f \right )}\right )}{8 \sqrt {2 i f -c \ln \left (f \right )}}-\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{2 i d} \operatorname {erf}\left (\sqrt {-c \ln \left (f \right )-2 i f}\, x \right )}{8 \sqrt {-c \ln \left (f \right )-2 i f}}+\frac {f^{a} \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-c \ln \left (f \right )}\, x \right )}{4 \sqrt {-c \ln \left (f \right )}}\)	\(107\)

[In]

int(f^(c*x^2+a)*sin(f*x^2+d)^2,x,method=_RETURNVERBOSE)

[Out]

-1/8*Pi^(1/2)*f^a*exp(-2*I*d)/(2*I*f-c*ln(f))^(1/2)*erf(x*(2*I*f-c*ln(f))^(1/2))-1/8*Pi^(1/2)*f^a*exp(2*I*d)/(
-c*ln(f)-2*I*f)^(1/2)*erf((-c*ln(f)-2*I*f)^(1/2)*x)+1/4*f^a*Pi^(1/2)/(-c*ln(f))^(1/2)*erf((-c*ln(f))^(1/2)*x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.21 \[ \int f^{a+c x^2} \sin ^2\left (d+f x^2\right ) \, dx=-\frac {2 \, \sqrt {\pi } {\left (c^{2} \log \left (f\right )^{2} + 4 \, f^{2}\right )} \sqrt {-c \log \left (f\right )} f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right )} x\right ) - \sqrt {\pi } {\left (c^{2} \log \left (f\right )^{2} - 2 i \, c f \log \left (f\right )\right )} \sqrt {-c \log \left (f\right ) - 2 i \, f} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) - 2 i \, f} x\right ) e^{\left (a \log \left (f\right ) + 2 i \, d\right )} - \sqrt {\pi } {\left (c^{2} \log \left (f\right )^{2} + 2 i \, c f \log \left (f\right )\right )} \sqrt {-c \log \left (f\right ) + 2 i \, f} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) + 2 i \, f} x\right ) e^{\left (a \log \left (f\right ) - 2 i \, d\right )}}{8 \, {\left (c^{3} \log \left (f\right )^{3} + 4 \, c f^{2} \log \left (f\right )\right )}} \]

[In]

integrate(f^(c*x^2+a)*sin(f*x^2+d)^2,x, algorithm="fricas")

[Out]

-1/8*(2*sqrt(pi)*(c^2*log(f)^2 + 4*f^2)*sqrt(-c*log(f))*f^a*erf(sqrt(-c*log(f))*x) - sqrt(pi)*(c^2*log(f)^2 -
2*I*c*f*log(f))*sqrt(-c*log(f) - 2*I*f)*erf(sqrt(-c*log(f) - 2*I*f)*x)*e^(a*log(f) + 2*I*d) - sqrt(pi)*(c^2*lo
g(f)^2 + 2*I*c*f*log(f))*sqrt(-c*log(f) + 2*I*f)*erf(sqrt(-c*log(f) + 2*I*f)*x)*e^(a*log(f) - 2*I*d))/(c^3*log
(f)^3 + 4*c*f^2*log(f))

Sympy [F]

\[ \int f^{a+c x^2} \sin ^2\left (d+f x^2\right ) \, dx=\int f^{a + c x^{2}} \sin ^{2}{\left (d + f x^{2} \right )}\, dx \]

[In]

integrate(f**(c*x**2+a)*sin(f*x**2+d)**2,x)

[Out]

Integral(f**(a + c*x**2)*sin(d + f*x**2)**2, x)

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.22 (sec) , antiderivative size = 315, normalized size of antiderivative = 2.25 \[ \int f^{a+c x^2} \sin ^2\left (d+f x^2\right ) \, dx=\frac {\sqrt {\pi } \sqrt {2 \, c^{2} \log \left (f\right )^{2} + 8 \, f^{2}} {\left (f^{a} {\left (i \, \cos \left (2 \, d\right ) + \sin \left (2 \, d\right )\right )} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) + 2 i \, f} x\right ) + f^{a} {\left (-i \, \cos \left (2 \, d\right ) + \sin \left (2 \, d\right )\right )} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) - 2 i \, f} x\right )\right )} \sqrt {c \log \left (f\right ) + \sqrt {c^{2} \log \left (f\right )^{2} + 4 \, f^{2}}} \sqrt {-c \log \left (f\right )} - \sqrt {\pi } \sqrt {2 \, c^{2} \log \left (f\right )^{2} + 8 \, f^{2}} {\left (f^{a} {\left (\cos \left (2 \, d\right ) - i \, \sin \left (2 \, d\right )\right )} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) + 2 i \, f} x\right ) + f^{a} {\left (\cos \left (2 \, d\right ) + i \, \sin \left (2 \, d\right )\right )} \operatorname {erf}\left (\sqrt {-c \log \left (f\right ) - 2 i \, f} x\right )\right )} \sqrt {-c \log \left (f\right ) + \sqrt {c^{2} \log \left (f\right )^{2} + 4 \, f^{2}}} \sqrt {-c \log \left (f\right )} + 2 \, \sqrt {\pi } {\left ({\left (c^{2} f^{a} \log \left (f\right )^{2} + 4 \, f^{a + 2}\right )} \operatorname {erf}\left (x \overline {\sqrt {-c \log \left (f\right )}}\right ) + {\left (c^{2} f^{a} \log \left (f\right )^{2} + 4 \, f^{a + 2}\right )} \operatorname {erf}\left (\sqrt {-c \log \left (f\right )} x\right )\right )}}{16 \, {\left (c^{2} \log \left (f\right )^{2} + 4 \, f^{2}\right )} \sqrt {-c \log \left (f\right )}} \]

[In]

integrate(f^(c*x^2+a)*sin(f*x^2+d)^2,x, algorithm="maxima")

[Out]

1/16*(sqrt(pi)*sqrt(2*c^2*log(f)^2 + 8*f^2)*(f^a*(I*cos(2*d) + sin(2*d))*erf(sqrt(-c*log(f) + 2*I*f)*x) + f^a*
(-I*cos(2*d) + sin(2*d))*erf(sqrt(-c*log(f) - 2*I*f)*x))*sqrt(c*log(f) + sqrt(c^2*log(f)^2 + 4*f^2))*sqrt(-c*l
og(f)) - sqrt(pi)*sqrt(2*c^2*log(f)^2 + 8*f^2)*(f^a*(cos(2*d) - I*sin(2*d))*erf(sqrt(-c*log(f) + 2*I*f)*x) + f
^a*(cos(2*d) + I*sin(2*d))*erf(sqrt(-c*log(f) - 2*I*f)*x))*sqrt(-c*log(f) + sqrt(c^2*log(f)^2 + 4*f^2))*sqrt(-
c*log(f)) + 2*sqrt(pi)*((c^2*f^a*log(f)^2 + 4*f^(a + 2))*erf(x*conjugate(sqrt(-c*log(f)))) + (c^2*f^a*log(f)^2
+ 4*f^(a + 2))*erf(sqrt(-c*log(f))*x)))/((c^2*log(f)^2 + 4*f^2)*sqrt(-c*log(f)))

Giac [F]

\[ \int f^{a+c x^2} \sin ^2\left (d+f x^2\right ) \, dx=\int { f^{c x^{2} + a} \sin \left (f x^{2} + d\right )^{2} \,d x } \]

[In]

integrate(f^(c*x^2+a)*sin(f*x^2+d)^2,x, algorithm="giac")

[Out]

integrate(f^(c*x^2 + a)*sin(f*x^2 + d)^2, x)

Mupad [F(-1)]

Timed out. \[ \int f^{a+c x^2} \sin ^2\left (d+f x^2\right ) \, dx=\int f^{c\,x^2+a}\,{\sin \left (f\,x^2+d\right )}^2 \,d x \]

[In]

int(f^(a + c*x^2)*sin(d + f*x^2)^2,x)

[Out]

int(f^(a + c*x^2)*sin(d + f*x^2)^2, x)